美国数学月刊问题18-10-31
Problem 12067 - 08 - P. Bracken (USA).
对于正整数$n$.令$$\beta_n=6n+12n^2(\gamma-\gamma_n),$$
其中$\gamma_n=H_n-\ln n$, $\displaystyle H_n=\sum_{j=1}^n\frac{1}{j}$为第$n$个调和数(Harmonic number),且$\gamma$为Euler常数.证明:对所有$n$,均有$\beta_{n+1}>\beta_n$.
Problem 12066 - 08 - Xiang-Qian Chang (USA).
设$n$和$k$是大于$1$的正整数, $A$是$n\times n$正定Hermite矩阵.证明
$$\left( \det A \right) ^{1/n}\le \left( \frac{\mathrm{tr}^k\left( A \right) -\mathrm{tr}\left( A^k \right)}{n^k-n} \right) ^{1/k}.$$
Problem 12064 - 08 - C. A. Hernandez Melo (Brazil).
设$f$为凸的,从$[1,\infty)$到$\mathbb{R}$的连续可微函数,使得对所有$x\geq 1$,均有$f'(x)>0$.证明反常积分$$\int_1^\infty\frac{dx}{f'(x)}$$收敛当且仅当级数$$\sum_{n=1}^\infty\left(f^{-1}(f(n)+\varepsilon)-n\right)$$对所有$\varepsilon>0$均收敛.
Problem 12063 - 08 - H. Ohtsuka (Japan).
设$p$和$q$为实数,且$p>0,q>-p^2/4$.令$U_0=0,U_1=1$,而且对于$n\geq 0$,有$U_{n+2}=pU_{n+1}+qU_n$.计算
$$\lim_{n\rightarrow \infty}\sqrt{U_{1}^{2}+\sqrt{U_{2}^{2}+\sqrt{U_{4}^{2}+\sqrt{\cdots +\sqrt{U_{2^{n-1}}^{2}}}}}}.$$
Problem 12060 - 07 - O. Furdui and A. Sintamarian (Romania).
证明$$\sum_{n=2}^{\infty}{\frac{H_nH_{n+1}}{n^3-n}}=\frac{5}{2}-\frac{\pi ^2}{24}-\zeta \left( 3 \right),$$
其中$\displaystyle H_n=\sum_{j=1}^n\frac{1}{j}$为第$n$个调和数(Harmonic number).
Problem 12057 - 07 - P. Korus (Hungary).
(a)设$a_1=1,a_2=2$且对任意正整数$k$,有
$$a_{2k+1}=\frac{a_{2k-1}+a_{2k}}{2},\quad a_{2k+2}=\sqrt{a_{2k}a_{2k+1}}.$$
求数列$\{a_n\}$的极限.
(b)设$b_1=1,b_2=2$且对任意正整数$k$,有
$$b_{2k+1}=\frac{b_{2k-1}+b_{2k}}{2},\quad b_{2k+2}=\frac{2b_{2k}b_{2k+1}}{b_{2k}+b_{2k+1}}.$$
求数列$\{b_n\}$的极限.
Problem 12054 - 06 - C. I. Valean (Romania).
证明$$\int_0^1{\frac{\arctan x}{x}\ln \frac{1+x^2}{\left( 1-x \right) ^2}dx}=\frac{\pi ^3}{16}.$$
Problem 12051 - 06 - P. Ribeiro (Portugal).
证明$$
\sum_{n=0}^{\infty}{\left( \begin{array}{c}
2n\\
n\\
\end{array} \right) \frac{1}{4^n\left( 2n+1 \right) ^3}}=\frac{\pi ^3}{48}+\frac{\pi \ln ^22}{4}.
$$
Problem 12049 - 06 - Z. K. Silagadze (Russia).
对所有满足$m\leq n$的非负整数$m$和$n$.证明
$$
\sum_{k=m}^n{\frac{\left( -1 \right) ^{k+m}}{2k+1}\left( \begin{array}{c}
n+k\\
n-k\\
\end{array} \right) \left( \begin{array}{c}
2k\\
k-m\\
\end{array} \right)}=\frac{1}{2n+1}.
$$
https://math.stackexchange.com/q/876106/165013
prove that this integral
$$\int_{0}^{\infty}\dfrac{dx}{(1+x^2)(1+r^2x^2)(1+r^4x^2)(1+r^6x^2)\cdots}=
\dfrac{\pi}{2(1+r+r^3+r^6+r^{10}+\cdots}$$
for this integral,I can't find it.and I don't know how deal this such strange integral.
and this problem is from china QQ (someone ask it)
before I ask this question:
https://math.stackexchange.com/questions/671964/how-find-this-integral-fy-int-infty-infty-fracdx1x21xy2?rq=1
If we set
$$ f(x)=\prod_{n=0}^{+\infty}(1+r^{2n}x^2) $$
we have:
$$\int_{0}^{+\infty}\frac{dx}{f(x)}=\pi i\sum_{m=0}^{+\infty}\operatorname{Res}\left(f(z),z=\frac{i}{r^m}\right)=\frac{\pi}{2}\sum_{m=0}^{+\infty}\frac{1}{r^m}\prod_{n\neq m}(1-r^{2n-2m})^{-1}\tag{1}$$
but since
$$ \prod_{n=0}^{+\infty}(1-x^n z)^{-1}=\sum_{n=0}^{+\infty}\frac{z^n}{(1-x)\cdot\ldots\cdot(1-x^n)}$$
is one of the Euler's partitions identities, and:
$$\frac{\pi}{2}\sum_{m=0}^{+\infty}\frac{1}{r^m}\prod_{n\neq m}(1-r^{2n-2m})^{-1}=\frac{\pi}{2}\prod_{n=1}^{+\infty}(1-r^{2n})^{-1}\sum_{m=0}^{+\infty}\frac{(1/r)^m}{(1-(1/r^2))\cdot\ldots\cdot(1-(1/r^2)^m)}$$
we have:
$$\int_{0}^{+\infty}\frac{dz}{f(z)}=\frac{\pi}{2}\prod_{n=1}^{+\infty}(1-r^{2n})^{-1}\prod_{m=0}^{+\infty}\left(1-\frac{1}{r^{2m+1}}\right)^{-1}\tag{2}$$
and the claim follows from the Jacobi triple product identity:
$$\sum_{k=-\infty}^{+\infty}s^k q^{\binom{k+1}{2}}=\prod_{m\geq 1}(1-q^m)(1+s q^m)(1+s^{-1}q^{m-1}).$$
Find this integral
$$F(y)=\int_{-\infty}^{\infty}\dfrac{dx}{(1+x^2)(1+(x+y)^2)}$$
How about using the residue theorem? We have
$$F(y)= 2\pi i \sum_{x^*} \mathop{\rm Res}_{x=x^*} f(x),$$
where the sum ranges over all the poles of the integrand $$f(x)= \frac1{(1+x^2)[1+(x+y)^2]}$$ in the upper half-plane. These poles are situated at $x^*= i, i-y$ (assuming $y$ to be real).
As the poles are simple poles, we obtain the residues by
$$ \mathop{\rm Res}_{x=x^*} f(x) =\lim_{x\to x^*} (x-x^*)f(x) .$$
Thus, we have
$$ \mathop{\rm Res}_{x=i} f(x) = \frac{1}{2 i [1+(i + y)^2]}$$
and
$$ \mathop{\rm Res}_{x=i-y} f(x) = \frac{1}{2 i [1+(i-y)^2]}.$$
So, we obtain
$$ F(y) = \pi \left[\frac{1}{1+(i + y)^2} +
\frac{1}{1+(i-y)^2} \right] = \frac{2 \pi}{4+y^2}$$
as the final result.
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=114058
Fejer三角多项式不等式
https://math.stackexchange.com/q/2122045/165013
In a letter to Hardy, Ramanujan described a simple identity valid for $0<a<b+\frac 12$:
> $$\small\int\limits_{0}^{\infty}\frac {1+\dfrac {x^2}{(b+1)^2}}{1+\dfrac {x^2}{a^2}}\dfrac {1+\dfrac {x^2}{(b+2)^2}}{1+\dfrac {x^2}{(a+1)^2}}\dfrac {1+\dfrac {x^2}{(b+3)^2}}{1+\dfrac {x^2}{(a+2)^2}}\cdots \, dx=\dfrac {\sqrt{\pi}}2\dfrac {\Gamma\left(a+\frac 12\right)\Gamma\left(b+1\right)\Gamma(b-a+1)}{\Gamma(a)\Gamma\left(b+\frac 12\right)\Gamma\left(b-a+\frac 12\right)}\tag1$$
Which I find remarkable.
>**Questions:**
1. Has anyone discovered a way to prove $(1)$? If so, how do you prove it?
2. Where did Ramanujan learn all of his integrational-calculus material (It doesn't appear in the *Synopsis book*)?
3. Does anyone know a pdf or book where I can start learning advanced integration?
----------
I'm wondering how you would prove $(1)$ and if there are similar identities that can be made. Wikipedia doesn't have any information.
>**Proposition 1 :**$$\color{blue}{\displaystyle \sum\limits_{k=1}^\infty \dfrac{(-1)^{k-1}}{k}\;\zeta_H(k,a)x^k \; =\; \ln\left(\dfrac{\Gamma(a)}{\Gamma(a+x)}\right)\tag1}$$
**Proof :**
$$\begin{align*} & \displaystyle \sum\limits_{k=0}^{\infty}(-x)^{k}\zeta_H(k+1,a)\\ & \displaystyle = \sum\limits_{k,n=0}^{\infty} \dfrac{(-x)^k}{(n+a)^{k+1}}\\ & \displaystyle = \sum\limits_{n=0}^\infty \dfrac{1}{n+a}\sum\limits_{k=0}^\infty \left(\dfrac{-x}{n+a}\right)^k \\ & \displaystyle = \sum\limits_{n=0}^\infty \dfrac{1}{n+a+x} \\ & \displaystyle = -\psi(a+x)\end{align*}$$
Now integrating,
$$\displaystyle \sum\limits_{k=1}^{\infty} \dfrac{(-1)^{k-1}}{k}\zeta_H(k,a)x^k =\ln\left(\dfrac{\Gamma(a)}{\Gamma(a+x)}\right)\\ $$
>**Proposition 2 :**$$\color{blue}{\displaystyle \prod\limits_{n=0}^{\infty} \left(1+\dfrac{x^2}{(n+a)^2}\right)\;=\; \dfrac{\Gamma^2 (a)}{\Gamma(a+ix)\Gamma(a-ix)}\tag 2} $$
**Proof :**
It is sufficient to evaluate the series,
$$\begin{align*} & \displaystyle\sum\limits_{n=0}^{\infty}\ln\left(1+\dfrac{x}{n+a}\right) \\ & = \displaystyle \sum\limits_{k=1}^\infty\sum\limits_{n=0}^\infty \dfrac{(-1)^{k-1}}{k}x^k \dfrac{1}{(n+a)^k} \\ & =\displaystyle \sum\limits_{k=1}^{\infty} \dfrac{(-1)^{k-1}}{k}\zeta_H(k,a)x^k \\ & =\displaystyle \ln\left(\dfrac{\Gamma(a)}{\Gamma(a+x)}\right)\end{align*}$$
$$\displaystyle \prod\limits_{n=0}^{\infty} \left(1+\dfrac{x}{n+a}\right) = \dfrac{\Gamma(a)}{\Gamma(a+x)}$$
Therefore,
$$\displaystyle \prod\limits_{n=0}^{\infty} \left(1+\dfrac{x^2}{(n+a)^2}\right)\;=\; \dfrac{\Gamma^2 (a)}{\Gamma(a+ix)\Gamma(a-ix)}$$
>Proposition 3 : If $\; \displaystyle F(s)=\int\limits_0^\infty x^{s-1}f(x)\; dx\; $ then $$\color{blue}{\displaystyle \int\limits_{-\infty}^{\infty} |F(ix)|^2 \; dx \;= \; 2\pi\int\limits_0^\infty \dfrac{|f(x)|^2}{x}\; dx\tag 3} $$
**Proof :**
$$\displaystyle F(it)=\int\limits_0^\infty x^{it}\dfrac{f(x)}{x}\; dx$$
Set $x=e^y$ ,
$$\displaystyle F(it)=\int\limits_0^\infty e^{ixt}f(e^x)\; dx$$
Now by properties of Fourier Transform,
$$\begin{align*} & \displaystyle f(e^t)=\int\limits_{-\infty}^\infty g(x)e^{-ixt}\; dx \\ & \displaystyle g(t)=\dfrac{1}{2\pi}\int\limits_{-\infty}^\infty f(e^x)e^{ixt}\; dx\\\end{align*}$$
$$\begin{align*}\displaystyle F(it) & =2\pi g(t) \\ \displaystyle \int\limits_{-\infty}^\infty |F(it)|^2\; dt & = 4\pi^2 \int\limits_{-\infty}^\infty |g(t)|^2\; dt \\ \displaystyle \int\limits_{-\infty}^\infty |F(it)|^2\; dt & = 2\pi \int\limits_{-\infty}^\infty g(t) \int\limits_{-\infty}^\infty e^{ixt}f(e^x)\; dx\; dt \\ \displaystyle \int\limits_{-\infty}^\infty |F(it)|^2\; dt & = 2\pi \int\limits_{-\infty}^\infty f(e^x) \int\limits_{-\infty}^\infty e^{ixt}g(t)\; dt\; dx \\ \displaystyle \int\limits_{-\infty}^\infty |F(it)|^2\; dt & = 2\pi \int\limits_{-\infty}^\infty f(e^x)\overline{f(e^x)}\; dx =2\pi\int\limits_{-\infty}^\infty |f(e^x)|^2\; dx\end{align*}$$
Now by setting $e^x=t$ we get our result,
$$\displaystyle \int\limits_{-\infty}^\infty |F(it)|^2\; dt = 2\pi\int\limits_{-\infty}^\infty \dfrac{|f(t)|^2}{t}\; dt$$
>**Main Problem:** $$\color{blue}{\displaystyle \int\limits_{0}^{\infty}\frac {1+\dfrac {x^2}{(b+1)^2}}{1+\dfrac {x^2}{a^2}}\dfrac {1+\dfrac {x^2}{(b+2)^2}}{1+\dfrac {x^2}{(a+1)^2}}\dfrac {1+\dfrac {x^2}{(b+3)^2}}{1+\dfrac {x^2}{(a+2)^2}}\cdots \, dx=\dfrac {\sqrt{\pi}}2\dfrac {\Gamma\left(a+\frac 12\right)\Gamma\left(b+1\right)\Gamma(b-a+1)}{\Gamma(a)\Gamma\left(b+\frac 12\right)\Gamma\left(b-a+\frac 12\right)}} $$
**Proof :** If we denote the integral by $I$ then using $(2)$ it can be rewritten as,
$$\displaystyle I = \dfrac{\Gamma^2 (b+1)}{\Gamma^2 (a)}\dfrac{1}{2} \int\limits_{-\infty}^\infty \dfrac{|\Gamma(a+ix)|^2}{|\Gamma(b+1+ix)|^2}\; dx$$
Now by defining $\displaystyle h(x)=\dfrac{x^a (1-x)^{b-a}}{\Gamma(b-a+1)}$ for $x\in[0,1]$ and $0$ for $\forall x\notin[0,1]$ (Just like done in the link in the comment) we can conclude that $\displaystyle F(s)=M[h(x)]=\dfrac{\Gamma(s+a)}{\Gamma(s+b+1)}$ and from $(3)$ it follows that,
$$\displaystyle I = \dfrac{\Gamma^2 (b+1)}{\Gamma^2 (a)}\dfrac{1}{2} \int\limits_{0}^1 \dfrac{|h(x)|^2}{x}\; dx = \dfrac {\sqrt{\pi}}2\dfrac {\Gamma\left(a+\frac 12\right)\Gamma\left(b+1\right)\Gamma(b-a+1)}{\Gamma(a)\Gamma\left(b+\frac 12\right)\Gamma\left(b-a+\frac 12\right)}$$
where last line follows from the Duplication formula , and we are done !
$$\large \color{red}{\color{blue}{\boxed{\mathfrak{PROVED}}}} $$
https://mathoverflow.net/questions/66812/ramanujans-eccentric-integral-formula
The wikipedia page on [Srinivasa Ramanujan][1] gives a very strange formula:
> **Ramanujan:** If $0 < a < b + \frac{1}{2}$ then, $$\int\limits_{0}^{\infty} \frac{ 1 + x^{2}/(b+1)^{2}}{ 1 + x^{2}/a^{2}} \times \frac{ 1 + x^{2}/(b+2)^{2}}{ 1 + x^{2}/(a+1)^{2}} \times \cdots \ \textrm{dx} = \frac{\sqrt{\pi}}{2} \small{\frac{ \Gamma(a+\frac{1}{2}) \cdot \Gamma(b+1)\: \Gamma(b-a+\frac{1}{2})}{\Gamma(a) \cdot \Gamma(b+\frac{1}{2}) \cdot \Gamma(b-a+1)}}$$
- Question I would like to pose to this community is: What could be the Intuition behind discovering this formula.
- Next, I see that Ramanujan has discovered a lot of formulas for expressing $\pi$ as series. May I know what is the advantage of having a same number expressed as a series in a different way. Is it useful at all?
- From what I know Ramanujan basically worked on *Infinite series, Continued fractions,* $\cdots$ etc. I have never seen applications of *continued fractions*, in the real world. I would also like to know if continued fractions has any applications.
Hope I haven't asked too many questions. As I was posting this question the last question on *application of continued fractions* popped up and I thought it would be a good idea to pose it here, instead of posing it as a new question.
[1]: http://en.wikipedia.org/wiki/Srinivasa_Ramanujan
This is one of those precious cases when Ramanujan himself provided (a sketch of) a proof. The identity was published in his paper ["Some definite integrals"][1] (*Mess. Math.* 44 (1915), pp. 10-18) together with several related formulae.
It might be instructive to look first at the simpler identity (i.e. the limiting case when $b\to\infty$; the identity mentioned in the original question can be obtained by a similar approach):
$$\int\limits_{0}^{\infty} \prod_{k=0}^{\infty}\frac{1}{ 1 + x^{2}/(a+k)^{2}}dx = \frac{\sqrt{\pi}}{2} \frac{ \Gamma(a+\frac{1}{2})}{\Gamma(a)},\quad a>0.\qquad\qquad\qquad(1)$$
Ramanujan derives (1) by using a partial fraction decomposition of the product $\prod_{k=0}^{n}\frac{1}{ 1 + x^{2}/(a+k)^{2}}$, integrating term-wise, and passing to the limit $n\to\infty$. He also indicates that alternatively (1) is implied by the factorization
$$\prod_{k=0}^{\infty}\left[1+\frac{x^2}{(a+k)^2}\right] = \frac{ [\Gamma(a)]^2}{\Gamma(a+ix)\Gamma(a-ix)},$$
which follows readily from Euler's product formula for the gamma function. Thus (1) is equivalent to the formula
$$\int\limits_{0}^{\infty}\Gamma(a+ix)\Gamma(a-ix)dx=\frac{\sqrt{\pi}}{2} \Gamma(a)\Gamma\left(a+\frac{1}{2}\right).$$
------------------------------------------------------------------
There is a nice paper ["Wallis-Ramanujan-Schur-Feynman"][2] by Amdeberhan et al (*American Mathematical Monthly* 117 (2010), pp. 618-632) that discusses interesting combinatorial aspects of formula (1) and its generalizations.
[1]: http://www.imsc.res.in/~rao/ramanujan/CamUnivCpapers/Cpaper11/page1.htm
[2]: http://arminstraub.com/files/publications/ws.pdf
https://math.stackexchange.com/q/434933/165013
Wikipedia [informs][1] me that
$$S = \vartheta(0;i)=\sum_{n=-\infty}^\infty e^{-\pi n^2} = \frac{\sqrt[4] \pi}{\Gamma\left(\frac 3 4\right)}$$
I tried considering $f(x,n) = e^{-x n^2}$ so that its Mellin transform becomes $\mathcal{M}_x(f)=n^{-2z} \Gamma(z)$ so inverting and summing
$$\frac{1}{2}(S-1)=\sum_{n=1}^\infty f(\pi,n)=\sum_{n=1}^\infty \frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}n^{-2z} \Gamma(z)\pi^{-z}\,dz = \frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\zeta(2z) \Gamma(z) \pi^{-z}\,dz$$
However, this last integral (whose integrand has poles at $z=0,\frac{1}{2}$ with respective residues of $-\frac 1 2$ and $\frac 1 2$) is hard to evaluate due to the behavior of the function as $\Re(z)\to \pm\infty$ which makes a classic infinite contour over the entire left/right plane impossible.
How does one go about evaluating this sum?
[1]: http://en.wikipedia.org/wiki/Theta_function#Explicit_values
This one is a direct evaluation of elliptic integrals. Jacobi's theta function $\vartheta_{3}(q)$ is defined via the equation $$\vartheta_{3}(q) = \sum_{n = -\infty}^{\infty}q^{n^{2}}\tag{1}$$ Let $0 < k < 1$ and $k' = \sqrt{1 - k^{2}}$ and we define elliptic integrals $K, K'$ via $$K(k) = \int_{0}^{\pi/2}\frac{dx}{\sqrt{1 - k^{2}\sin^{2}x}}, K = K(k), K' = K(k')\tag{2}$$ Then it is [almost a miracle][1] that we can get $k$ in terms of $K, K'$ via the variable $q = e^{-\pi K'/K}$ using equations $$k = \frac{\vartheta_{2}^{2}(q)}{\vartheta_{3}^{2}(q)}\tag{3}$$ where $\vartheta_{2}(q)$ is another theta function of Jacobi defined by $$\vartheta_{2}(q) = \sum_{n = -\infty}^{\infty}q^{(n + (1/2))^{2}}\tag{4}$$ Also the function $\vartheta_{3}(q)$ is directly related to $K$ via $$\vartheta_{3}(q) = \sqrt{\frac{2K}{\pi}}\tag{5}$$ The proofs of $(3)$ and $(5)$ are given in the linked post on my blog.
--------
The sum in the question is $\vartheta_{3}(e^{-\pi})$ so that we have $q = e^{-\pi}$. This implies that $K'/K = 1$ so that $k = k'$ and from $k^{2} + k'^{2} = 1$ we get $k^{2} = 1/2$. And then $$\vartheta_{3}(q) = \sqrt{\frac{2K}{\pi}} = \sqrt{\frac{2}{\pi}\cdot\frac{\Gamma^{2}(1/4)}{4\sqrt{\pi}}} = \frac{\Gamma(1/4)}{\pi^{3/4}\sqrt{2}}$$ Now using $\Gamma(1/4)\Gamma(3/4) = \pi/\sin(\pi/4) = \pi\sqrt{2}$ we get $$\sum_{n = -\infty}^{\infty}e^{-\pi n^{2}} = \vartheta_{3}(e^{-\pi}) = \frac{\sqrt[4]{\pi}}{\Gamma(3/4)}$$ The value of $K = K(1/\sqrt{2})$ in terms of $\Gamma(1/4)$ is evaluated in [this answer][2].
[1]: http://paramanands.blogspot.com/2010/10/the-magic-of-theta-functions.html
[2]: https://math.stackexchange.com/a/1793756/72031
I found an interesting series
$$\sum_{k=1}^\infty \log \left(\tanh \frac{\pi k}{2} \right)=\log(\vartheta_4(e^{-\pi}))=\log \left(\frac{\pi^{\frac{1}{4}}}{2^{\frac{1}{4}}\Gamma \left( \frac{3}{4}\right)} \right)$$
- Does anybody know how to approach this series using Jacobi Theta Function?
- Also, can any one suggest any good papers/books on Jacobi theta functions and Jacobi Elliptic functions?
Thank you very much!
We have $$\vartheta_{4}(q) = \prod_{n = 1}^{\infty}(1 - q^{2n})(1 - q^{2n - 1})^{2}\tag{1}$$ It is easily seen that the above product can be written as $$\prod_{n = 1}^{\infty}(1 - q^{2n})\cdot\frac{(1 - q^{n})^{2}}{(1 - q^{2n})^{2}} = \prod_{n = 1}^{\infty}\frac{(1 - q^{n})^{2}}{1 - q^{2n}} = \prod_{n = 1}^{\infty}\frac{1 - q^{n}}{1 + q^{n}}\tag{2}$$ Putting $q = e^{-\pi}$ and noting that $$\tanh(n\pi/2) = \frac{e^{n\pi/2} - e^{n\pi/2}}{e^{n\pi/2} + e^{n\pi/2}} = \frac{1 - e^{-n\pi}}{1 + e^{-n\pi}} = \frac{1 - q^{n}}{1 + q^{n}}\tag{3}$$ we have via virtue of equations $(1), (2), (3)$ $$\vartheta_{4}(e^{-\pi}) = \prod_{n = 1}^{\infty}\tanh\left(\frac{n\pi}{2}\right)$$ so that the first equality is proved.
Now to calculate the value of $\vartheta_{4}(q)$ for $q = e^{-\pi}$ note that $$\vartheta_{4}(q) = \frac{\vartheta_{4}(q)}{\vartheta_{3}(q)}\cdot \vartheta_{3}(q) = \sqrt{k'}\sqrt{\frac{2K}{\pi}}\tag{4}$$ For $q = e^{-\pi}$ we have $k = k' = 1/\sqrt{2}$ and $$K = \int_{0}^{\pi/2}\frac{dx}{\sqrt{1 - (1/2)\sin^{2}x}} = \frac{1}{4\sqrt{\pi}}\Gamma^{2}\left(\frac{1}{4}\right)\tag{5}$$ and hence $$\vartheta_{4}(e^{-\pi}) = 2^{-3/4}\pi^{-3/4}\Gamma(1/4)$$ and noting that $$\Gamma(1/4)\Gamma(3/4) = \frac{\pi}{\sin(\pi/4)} = \sqrt{2}\pi$$ we get $$\vartheta_{4}(e^{-\pi}) = \frac{\pi^{1/4}}{2^{1/4}\Gamma(3/4)}$$ so that the second equality of the question is also proved.
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The integral in $(5)$ is easily evaluated via the use of beta and gamma functions. Thus
\begin{align}
K &= K(1/\sqrt{2}) = \int_{0}^{\pi/2}\frac{dx}{\sqrt{1 - (1/2)\sin^{2}x}}\notag\\
&= \sqrt{2}\int_{0}^{\pi/2}\frac{dx}{\sqrt{2 - \sin^{2}x}}\notag\\
&= \sqrt{2}\int_{0}^{\pi/2}\frac{dx}{\sqrt{1 + \cos^{2}x}}\notag\\
&= \sqrt{2}\int_{0}^{1}\frac{dt}{\sqrt{1 - t^{4}}}\text{ (by putting }t = \cos x)\notag\\
&= \frac{\sqrt{2}}{4}\int_{0}^{1}x^{-3/4}(1 - x)^{-1/2}\,dx\text{ (by putting }t^{4} = x)\notag\\
&= \frac{\sqrt{2}}{4}B\left(\frac{1}{4}, \frac{1}{2}\right)\notag\\
&= \frac{\sqrt{2}}{4}\cdot\dfrac{\Gamma\left(\frac{1}{4}\right)\Gamma\left(\frac{1}{2}\right)}{\Gamma\left(\frac{3}{4}\right)}\notag\\
&= \frac{\sqrt{2\pi}}{4}\cdot\dfrac{\Gamma\left(\frac{1}{4}\right)}{\Gamma\left(\frac{3}{4}\right)}\notag\\
&= \frac{\sqrt{2\pi}}{4}\cdot\dfrac{\Gamma^{2}\left(\frac{1}{4}\right)}{\Gamma\left(\frac{1}{4}\right)\Gamma\left(\frac{3}{4}\right)}\notag\\
&= \frac{\sqrt{2\pi}}{4}\cdot\dfrac{\Gamma^{2}\left(\frac{1}{4}\right)}{\dfrac{\pi}{\sin(\pi/4)}}\notag\\
&= \frac{1}{4\sqrt{\pi}}\Gamma^{2}\left(\frac{1}{4}\right)\notag
\end{align}
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**Update**: Apparently I forgot to shed some light on the second question asked by OP namely any references/books on the theory of Jacobi Elliptic and Theta functions. This I deal with now.
The theory of Elliptic functions and theta functions is a very fascinating one. My own sources of study in this field are the following books (order of the books listed is not important here):
- *A Course of Modern Analysis* by Whittaker and Watson: A definitive resource for many many topics apart from theta functions. The focus here is on using the methods of complex analysis to develop a theory of elliptic and theta functions.
- *An Elementary Treatise on Elliptic Functions* by Cayley: This book develops the theory of elliptic functions in a very elementary fashion and I learnt most of the topic from this book. The presentation is easy to follow and does not require any deep skills apart from a good knowledge of calculus.
- *Elliptic Functions* by Armitage and Eberlein: This is another good book from Cambridge University Press which I read. It uses both elementary techniques as well as complex analysis to develop the theory of elliptic functions.
- *Pi and the AGM* by Borwein and Borwein: This is a truly singular book which connects elliptic integrals with Arithmetic Geometric Mean and is an interesting but somewhat difficult read.
- *Ramanujan Notebooks Vol 3* by Berndt: Ramanujan developed his own theory of theta functions independently of Jacobi and went far ahead, but his techniques are mostly unknown and Berndt has tried to discern his methods (to some extent) and establish most of his formulas in the theory of theta functions.
- *Collected Papers of Ramanujan* : There are some papers dealing with some applications of theory of theta functions. The value of this book is to get an insight into some of the techniques Ramanujan used. In particular one must read his monumental paper "Modular Equations and Approximations to $\pi$".
- *Fundamenta Nova* by Jacobi: This is a very good book but unfortunately written in Latin. I did manage to study some parts with the help of google translate and you can give it a try if you are willing to put extra effort of translation.
All the above books combined together cover most of the elementary and some advanced topics related to elliptic and theta functions *except its link with the algebraic number theory* (mainly the link with imaginary quadratic extensions of $\mathbb{Q}$). This is perhaps the most important and deep topic which is now famous by the name of *modular forms*. Unfortunately I don't have much knowledge on this topic.
The references listed above can be found online for free (if you search enough). I have tried to extract material from these references and present a coherent theory of elliptic integrals/functions and theta functions in [my blog posts][1] (in fact I started blogging only to document whatever I had learnt about these mysterious theta functions). The blog also contains the Ramanujan's theory of theta functions and the Borwein's approach via Arithmetic Geometric Mean. The advantage of the blog posts is that they are concise and are written in a particular order such that pre-requisites for proving a result are discussed before presenting the result.
[1]: https://paramanands.blogspot.in/p/archives.html
>**Prove That :**
>$$ S^2 = 1 + 4 \sum_{n=0}^{\infty} \frac{(-1)^n}{e^{(2n+1)\pi} - 1} $$
>**where** $\displaystyle S = \sum_{n=-\infty}^{\infty} e^{- \pi n^2}$
In [this answer](https://math.stackexchange.com/questions/434933/proving-sum-n-infty-infty-e-pi-n2-frac-sqrt4-pi-gamma-left) by the user [@Sangchul Lee](https://math.stackexchange.com/users/9340/sangchul-lee), the above identity is claimed. Can we prove it without calculating the two sums separately?
Any help will be appreciated.
Thanks in advance.
Let we set
$$ r_2(n) = \left|\{(a,b)\in\mathbb{Z}\times\mathbb{Z}: a^2+b^2=n \}\right|\tag{1}$$
as the arithmetic function that counts the number of representation as a sum of two integer squares. By Cauchy convolution we clearly have (by setting $x=e^{-\pi}$)
$$ S^2 = \sum_{n\geq 0} r_2(n)\,x^n =1+\sum_{n\geq 1}r_2(n)\,x^n\tag{2}$$
while the other series mentioned in the question is a [Lambert series][1]. By setting
$$ d_1(n) = |\{d\mid n : d\equiv 1\pmod{4}\}|,\\ d_3(n) = |\{d\mid n : d\equiv 3\pmod{4}\}|\phantom{\,}\tag{3} $$
the question boils down to proving that
$$ r_2(n) = 4\left(d_1(n)-d_3(n)\right) = 4(\chi * 1)(n) = 4\sum_{d\mid n}\chi(d) \tag{4}$$
where $\chi$ is the non-principal [Dirichlet character][2] $\!\!\pmod{4}$.<br>
By [Lagrange's identity][3], the set of numbers equal to a sum of two integer squares is a semigroup. $r_2(n)$ depends on the number of ways for writing $n$ as $z\cdot\bar{z}$, with $z=a+ib$. $\mathbb{Z}[i]$ is a Euclidean domain and so a UFD, and every integer prime $p$ splits in $\mathbb{Z}[i]$ as $z\cdot \bar{z}$ iff $p=2$ or $p\equiv 1\pmod{4}$ (that can be proved by Fermat's descent, for instance). It follows that $r_2(n)$ is a constant times a multiplicative function, given by the convolution between $\chi$ and $1$, and the previous constant is exactly the number of invertible elements in $\mathbb{Z}[i]$, namely $4$ ($1,-1,i,-i$).<br>
This is not the only way for proving the claim. Another chance is given by [Jacobi triple product][4] and differentiation. See, for instance, [Varouchas, *Démonstration Élémentaire d'une Identité de Lorenz*][5].
[1]: https://en.wikipedia.org/wiki/Lambert_series
[2]: https://en.wikipedia.org/wiki/Dirichlet_character
[3]: https://en.wikipedia.org/wiki/Lagrange%27s_identity
[4]: https://en.wikipedia.org/wiki/Jacobi_triple_product
[5]: http://link.springer.com/article/10.1023/A:1009732927027
This is a standard property of theta functions. Let $\vartheta_{3}(q)$ denote one of the Jacobi theta functions defined by $$\vartheta_{3}(q) = \sum_{n = -\infty}^{\infty}q^{n^{2}}\tag{1}$$ Then it is possible to show that $$
\vartheta_{3}^{2}(q) = 1 + 4\sum_{n = 1}^{\infty}\frac{q^{n}}{1 + q^{2n}}\tag{2}$$ and the sum on right can be written as a double sum $$\sum_{n = 1}^{\infty}\frac{q^{n}}{1 + q^{2n}} = \sum_{n = 1}^{\infty}q^{n}\sum_{m = 0}^{\infty}(-1)^{m}q^{2mn}$$ and interchanging order of summation we get $$\sum_{n = 1}^{\infty}\sum_{m = 0}^{\infty}(-1)^{m}q^{(2m + 1)n} = \sum_{m = 0}^{\infty}(-1)^{m}\sum_{n = 1}^{\infty}q^{(2m + 1)n} = \sum_{m = 0}^{\infty}(-1)^{m}\frac{q^{2m + 1}}{1 - q^{2m + 1}}\tag{3}$$ and hence from $(2)$ and $(3)$ we get $$\vartheta_{3}^{2}(q) = 1 + 4\sum_{n = 0}^{\infty}(-1)^{n}\frac{q^{2n + 1}}{1 - q^{2n + 1}}\tag{4}$$ Using $(1)$ and putting $q = e^{-\pi}$ we get $$\left(\sum_{n = -\infty}^{\infty}e^{-\pi n^{2}}\right)^{2} = 1 + 4\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{e^{(2n + 1)\pi} - 1}$$ So we are left with proving the fundamental identity $(2)$. This is easily proved by using the [Fourier series for elliptic function $\operatorname{dn}(u, k)$][1] given by $$\operatorname{dn}(u, k) = \frac{\pi}{2K} + \frac{2\pi}{K}\sum_{n = 1}^{\infty}\frac{q^{n}\cos (n\pi u/K)}{1 + q^{2n}}\tag{5}$$ and putting $u = 0$ and noting that $\operatorname{dn}(u, k) = 1$ and $2K/\pi = \vartheta_{3}^{2}(q)$. The identity $(4)$ i.e. $$\left(\sum_{n = -\infty}^{\infty}q^{n^{2}}\right)^{2} = 1 + 4\sum_{n = 0}^{\infty}(-1)^{n}\frac{q^{2n + 1}}{1 - q^{2n + 1}}$$ can also be [proved directly by using Jacobi's Triple Product][2].
------
The answer by Jack D'Aurizio proves the number theoretic interpretation of identity $(4)$ and thereby establishes $(4)$. Also note that Ramanujan was able to square the identity $(4)$ by direct algebraical manipulation to prove that $$\left(\sum_{n = -\infty}^{\infty}q^{n^{2}}\right)^{4} = 1 + 8\sum_{n = 1}^{\infty}\frac{nq^{n}}{(1 + (-q)^{n})}\tag{6}$$ (put $\theta = \pi/2$ in equation $(16)$ of [this post][3]). This identity also has number theoretic interpretation that $$r_{4}(n) = 8\sum_{d\mid n}d$$ if $n$ is odd and $$r_{4}(n) = 24\sum_{d \text{ odd }d\mid n}d$$ when $n$ is even where $r_{4}(n)$ is the number of ways in which a positive integer $n$ can be expressed as the sum of squares of $4$ integers (counting order as well as sign of integers). A simple corollary is that every positive integer $n$ can be expressed as a sum of four squares.
[1]: http://paramanands.blogspot.com/2011/02/elliptic-functions-fourier-series.html
[2]: https://math.stackexchange.com/a/737894/72031
[3]: http://paramanands.blogspot.com/2013/05/certain-lambert-series-identities-and-their-proof-via-trigonometry-part-1.html
How to prove the following identity
$$\sum_{k=0}^\infty \frac{1}{16^k} \left(\frac{120k^2 + 151k + 47}{512k^4 + 1024k^3 + 712k^2 + 194k + 15}\right) = \pi$$
I am totally clueless in this one. Would you help me, please? Any help would be appreciated. Thanks in advance.
This is the now famous Bailey–Borwein–Plouffe formula, see
http://en.wikipedia.org/wiki/Bailey%E2%80%93Borwein%E2%80%93Plouffe_formula,
http://crd-legacy.lbl.gov/~dhbailey/dhbpapers/digits.pdf.
Here is a summary of the proof given by Bailey, Borwein, Borwein, and Plouffe in [The Quest for Pi](http://crd-legacy.lbl.gov/~dhbailey/dhbpapers/pi-quest.pdf) in only a few lines of integration.
To begin they note the following definite integrals as summations, $n=1,\ldots,7$:
$$ \int_0^{\frac{1}{\sqrt{2}}} \frac{x^{n-1}}{1-x^8} dx =
\int_0^{\frac{1}{\sqrt{2}}} \sum_{k=0}^\infty x^{n-1+8k} dx =
\frac{1}{2^{n/2}} \sum_{k=0}^\infty \frac{1}{16^k(8k+n)} $$
If the fractional factor of the summation in the Question is expanded by partial fractions:
$$ \frac{120k^2 + 151k + 47}{512k^4 + 1024k^3 + 712k^2 + 194k + 15} =
\frac{4}{8k+1} - \frac{2}{8k+4} - \frac{1}{8k+5} - \frac{1}{8k+6} $$
then the integrals above can be applied to give:
$$ \sum_{k=0}^\infty \frac{1}{16^k} \left( \frac{4}{8k+1} - \frac{2}{8k+4}
- \frac{1}{8k+5} - \frac{1}{8k+6} \right) $$
$$ = \int_0^{\frac{1}{\sqrt{2}}}\frac{4\sqrt{2} -8x^3 -4\sqrt{2}x^4 -8x^5}{1-x^8} dx
$$
At this point the authors claim a substitution of $y= x \sqrt{2}$, making:
$$ \int_0^{\frac{1}{\sqrt{2}}}\frac{4\sqrt{2} -8x^3 -4\sqrt{2}x^4 -8x^5}{1-x^8} dx =
\int_0^1 \frac{16y-16}{y^4-2y^3+4y-4} dy $$
Finally the last integral may be expanded by partial fractions to give:
$$ \int_0^1 \frac{4y}{y^2-2} dy - \int_0^1 \frac{4y-8}{y^2-2y+2} dy = \pi $$
By way of explanation the authors point out that this rigorous proof was sought only after the discovery of the apparent [integer relations](https://en.wikipedia.org/wiki/Integer_relation_algorithm) among summations and $\pi$ via the [PSLQ algorithm](http://www.ams.org/journals/mcom/1999-68-225/S0025-5718-99-00995-3/S0025-5718-99-00995-3.pdf).
https://math.stackexchange.com/a/1703019/165013
How to calculate this relation?
$$I=\int_{0}^{1}\frac{(\arctan x)^2}{1+x^{2}}\ln\left ( 1+x^{2} \right )\mathrm{d}x=\frac{\pi^3}{96}\ln{2}-\frac{3\pi\zeta{(3)}}{128}-\frac{\pi^2G}{16}+\frac{\beta{(4)}}{2}$$
Where G is the Catalan's constant, and $$\beta(x)=\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{(2k-1)^{x}}$$ is the Dirichlet's beta function.
Integrate by parts $$u=(\arctan{\,x})^2\ln{(1+x^2)}$$ $$v=\arctan{\,x}$$
We have $$3I=\frac{\pi^3}{64}\ln{2-2\int_0^{1}x(\arctan{\,x})^3}\frac{dx}{1+x^2}$$
But how to calculate the latter integral?Could anybody please help by offering useful hints or solutions?Ithing very difficult to prove.
Take $\arctan\left(x\right)=u,\,\frac{dx}{1+x^{2}}=du
$. Then $$I=\int_{0}^{1}\frac{\left(\arctan\left(x\right)\right)^{2}}{1+x^{2}}\log\left(1+x^{2}\right)dx=\int_{0}^{\pi/4}u^{2}\log\left(1+\tan^{2}\left(u\right)\right)du
$$ $$=-2\int_{0}^{\pi/4}u^{2}\log\left(\cos\left(u\right)\right)du
$$ and now using the Fourier series $$\log\left(\cos\left(u\right)\right)=-\log\left(2\right)-\sum_{k\geq1}\frac{\left(-1\right)^{k}\cos\left(2ku\right)}{k},\,0\leq x<\frac{\pi}{2}
$$ we have $$I=\frac{\log\left(2\right)\pi^{3}}{96}+2\sum_{k\geq1}\frac{\left(-1\right)^{k}}{k}\int_{0}^{\pi/4}u^{2}\cos\left(2ku\right)du
$$ and the last integral is trivial to estimate $$\int_{0}^{\pi/4}u^{2}\cos\left(2ku\right)du=\frac{\pi^{2}\sin\left(\frac{\pi k}{2}\right)}{32k}-\frac{\sin\left(\frac{\pi k}{2}\right)}{4k^{3}}+\frac{\pi\cos\left(\frac{\pi k}{2}\right)}{8k^{2}}
$$ so we have $$I=\frac{\log\left(2\right)\pi^{3}}{96}+\pi^{2}\sum_{k\geq1}\frac{\left(-1\right)^{k}\sin\left(\frac{\pi k}{2}\right)}{16k^{2}}-\sum_{k\geq1}\frac{\left(-1\right)^{k}\sin\left(\frac{\pi k}{2}\right)}{2k^{4}}+\pi\sum_{k\geq1}\frac{\left(-1\right)^{k}\cos\left(\frac{\pi k}{2}\right)}{4k^{3}}
$$ and now observing that $$\cos\left(\frac{\pi k}{2}\right)=\begin{cases}
-1, & k\equiv2\,\mod\,4\\
1, & k\equiv0\,\mod\,4\\
0, & \textrm{otherwise}
\end{cases}
$$ and $$ \sin\left(\frac{\pi k}{2}\right)=\begin{cases}
-1, & k\equiv3\,\mod\,4\\
1, & k\equiv1\,\mod\,4\\
0, & \textrm{otherwise}
\end{cases}
$$ we have $$I=\frac{\log\left(2\right)\pi^{3}}{96}-\frac{\pi^{2}}{16}K+\frac{\beta\left(4\right)}{2}-\frac{3\pi\zeta\left(3\right)}{128}\approx 0.064824$$ where the last sum is obtained using the relation between [Dirichlet eta function][1] and Riemann zeta function.
[1]:http://mathworld.wolfram.com/DirichletEtaFunction.html
This is a nice problem. I am trying to use nice methods to solve this integral, But I failed.
$$\int_{-1}^{1}\dfrac{\arctan{x}}{1+x}\ln{\left(\dfrac{1+x^2}{2}\right)}dx, $$
where $\arctan{x}=\tan^{-1}{x}$
mark: this integral is my favorite one. Thanks to whoever has nice methods.
I have proved the following:
$$\int_{-1}^{1}\dfrac{\arctan{x}}{1+x}\ln{\left(\dfrac{1+x^2}{2}\right)}dx=\sum_{n=1}^{\infty}\dfrac{2^{n-1}H^2_{n-1}}{nC_{2n}^{n}}=\dfrac{\pi^3}{96}$$
where $$C_{m}^{n}=\dfrac{m}{(m-n)!n!},H_{n}=1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots+\dfrac{1}{n}$$
I also have got a few by-products
$$\int_{-1}^{1}\dfrac{\arctan{x}}{1+x}\ln{\left(\dfrac{1+x^2}{2}\right)}dx=-I_{1}-2I_{2}$$
where $$I_{1}=\int_{0}^{1}\dfrac{\ln{(1-x^2)}}{1+x^2}\ln{\left(\dfrac{1+x^2}{2}\right)}dx=\dfrac{\pi}{4}\ln^2{2}+\dfrac{\pi^3}{32}-2K\times\ln{2}$$
and
$$I_{2}=\int_{0}^{1}\dfrac{x\arctan{x}}{1+x^2}\ln{(1-x^2)}dx=-\dfrac{\pi^3}{48}-\dfrac{\pi}{8}\ln^2{2}+K\times\ln{2}$$
and same methods,I have follow integral
$$\int_{0}^{1}\dfrac{\ln{(1-x^4)}\ln{x}}{1+x^2}dx=\dfrac{\pi^3}{16}-3K\times\ln{2}$$
where $ K $ denotes Catalan's Constant.
Here is a solution that only uses complex analysis:
Let $\epsilon$ > 0 and consider the truncated integral
$$ I_{\epsilon} = \int_{-1+\epsilon}^{1} \frac{\arctan x}{x+1} \log\left( \frac{1+x^2}{2} \right) \, dx. $$
By using the formula
$$ \arctan x = \frac{1}{2i} \log \left( \frac{1 + ix}{1 - ix} \right) = \frac{1}{2i} \left\{ \log \left( \frac{1+ix}{\sqrt{2}} \right) - \log \left( \frac{1-ix}{\sqrt{2}} \right) \right\}, $$
it follows that
$$ I_{\epsilon} = \Im \int_{-1+\epsilon}^{1} \frac{1}{x+1} \log^{2} \left( \frac{1+ix}{\sqrt{2}} \right) \, dx. $$
Now let $\omega = e^{i\pi/4}$ and make the change of variable $z = \frac{1+ix}{\sqrt{2}}$ to obtain
$$ I_{\epsilon} = \Im \int_{L_{\epsilon}} \frac{\log^2 z}{z - \bar{\omega}} \, dz, $$
where $L_{\epsilon}$ is the line segment joining from $\bar{\omega}_{\epsilon} := \bar{\omega} + \frac{i\epsilon}{\sqrt{2}}$ to $\omega$. Now we tweak this contour of integration according to the following picture:
![enter image description here][1]
That is, we first draw a clockwise circular arc $\gamma_{\epsilon}$ centered at $\bar{\omega}$ joining from $\bar{\omega}_{\epsilon}$ to some points on the unit circle, and draw a counter-clockwise circular arc $\Gamma_{\epsilon}$ joining from the endpoint of $\gamma_{\epsilon}$ to $\omega$. Then
$$ I_{\epsilon} = \Im \int_{\gamma_{\epsilon}} \frac{\log^2 z}{z - \bar{\omega}} \, dz + \Im \int_{\Gamma_{\epsilon}} \frac{\log^2 z}{z - \bar{\omega}} \, dz =: J_{\epsilon} + K_{\epsilon}. $$
It is easy to check that as $\epsilon \to 0^{+}$, the central angle of $\gamma_{\epsilon}$ converges to $\pi / 4$. Since $\gamma_{\epsilon}$ winds $\bar{\omega}$ clockwise, we have
$$ \lim_{\epsilon \to 0^{+}} J_{\epsilon} = \Im \left( -\frac{i \pi}{4} \mathrm{Res}_{z=\bar{\omega}} \frac{\log^2 z}{z - \bar{\omega}} \right) = \frac{3}{2} \frac{\pi^3}{96}. $$
Also, by applying the change of variable $z = e^{i\theta}$,
$$ K_{\epsilon} = -\Re \int_{-\frac{\pi}{4}+o(1)}^{\frac{\pi}{4}} \frac{\theta^2}{1 - \bar{\omega}e^{-i\theta}} \, d\theta = \int_{-\frac{\pi}{4}+o(1)}^{\frac{\pi}{4}} \frac{\theta^2}{2} \, d\theta. $$
Thus taking $\epsilon \to 0^{+}$, we have
$$ \lim_{\epsilon \to 0^{+}} K_{\epsilon} = - \int_{0}^{\frac{\pi}{4}} \theta^2 \, d\theta = - \frac{1}{2} \frac{\pi^3}{96}. $$
Combining these results, we have
$$ \int_{-1}^{1} \frac{\arctan x}{x+1} \log \left( \frac{x^2 + 1}{2} \right) \, dx = \frac{\pi^3}{96}. $$
The same technique shows that
$$ \int_{-1}^{1} \frac{\arctan (t x)}{x+1} \log \left( \frac{1 + x^2 t^2}{1 + t^2} \right) \, dx = \frac{2}{3} \arctan^{3} t, \quad t \in \Bbb{R} .$$
[1]: http://i.stack.imgur.com/Omsqm.png
https://math.stackexchange.com/q/175843/165013
I'm very curious about the ways I may compute the following integral. I'd be very glad to know your approaching ways for this integral:
$$
I_{n} \equiv
\int_{-\infty}^\infty
{1-\cos\left(x\right)\cos\left(2x\right)\ldots\cos\left(nx\right) \over x^{2}}
\,{\rm d}x
$$
According to W|A, $I_1=\pi$, $I_2=2\pi$, $I_3=3\pi$, and one may be tempted to think that it's about an arithmetical progression here, but things change (unfortunately) from $I_4$ that is $\frac{9 \pi}{2}$. This problem
came to my mind when I was working on a different problem.
First note that
$$\int_{-\infty}^{\infty} \frac{1-\cos ax}{x^2} \; dx
= \left[ -\frac{1-\cos ax}{x}\right]_{-\infty}^{\infty} + a \int_{-\infty}^{\infty} \frac{\sin ax}{x} \; dx
= \pi \, |a|,$$
by the [Dirichlet integral](http://en.wikipedia.org/wiki/Dirichlet_integral). Also, by mathematical induction we can easily prove that
$$ \prod_{k=1}^{n} \cos \theta_k = \frac{1}{2^n} \sum_{\mathrm{e}\in S} \cos\left( e_1 \theta_1 + \cdots + e_n \theta_n \right),$$
where the summation runs over the set $S = \{ -1, 1\}^n$. Thus we have
$$ \begin{align*}
I_n = \int_{-\infty}^{\infty} \frac{1-\cos x \cdots \cos nx}{x^2} \; dx
&= \frac{1}{2^n} \sum_{\mathrm{e}\in S} \int_{-\infty}^{\infty} \frac{1-\cos(e_1 x + \cdots + e_n nx)}{x^2} \; dx \\
&= \frac{\pi}{2^n} \sum_{\mathrm{e}\in S} \left|e_1 + \cdots + e_n n\right|.
\end{align*}$$
For example, if $n = 3$, we have $\left|\pm 1 \pm 2 \pm 3\right| = 0, 0, 2, 2, 4, 4, 6, 6$ and hence
$$I_3 = \frac{\pi}{8}(0 + 0 + 2 + 2 + 4 + 4 + 6 + 6) = 3\pi.$$
Let the summation part as
$$ A_n = \sum_{\mathrm{e}\in S} |e_1 + \cdots + e_n n|.$$
The first 10 terms of $(A_n)$ are given by
$$ \left(A_n\right) = (2, 8, 24, 72, 196, 500, 1232, 2968, 7016, 16280, \cdots ), $$
and thus the corresponding $(I_n)$ are given by
$$ \left(I_n\right) = \left( \pi ,2 \pi ,3 \pi ,\frac{9 \pi }{2},\frac{49 \pi }{8},\frac{125 \pi }{16},\frac{77 \pi }{8},\frac{371 \pi }{32},\frac{877 \pi
}{64},\frac{2035 \pi }{128} \right).$$
So far, I was unable to find a simple formula for $(A_n)$, and I guess that it is not easy to find such one.
----------
p.s. The probability distribution of $S_n = e_1 + \cdots + e_n n$ is bell-shaped, and fits quite well with the corresponding normal distribution $X_n \sim N(0, \mathbb{V}(S_n))$. Thus it is not bad to conjecture that
$$ \frac{A_n}{2^n} = \mathbb{E}|S_n| \approx \mathbb{E}|X_n| = \sqrt{\frac{n(n+1)(2n+1)}{3\pi}},$$
and hence
$$ I_n \approx \sqrt{\frac{\pi \, n(n+1)(2n+1)}{3}}.$$
Indeed, numerical experiment shows that
![Numerical Experiment][1]
----------
I was able to prove a much weaker statement:
$$ \lim_{n\to\infty} \frac{I_n}{n^{3/2}} = \sqrt{\frac{2\pi}{3}}. $$
First, we observe that for $|x| \leq 1$ we have
$$ \log \cos x = -\frac{x^2}{2} + O\left(x^4\right).$$
Thus in particular,
$$
\sum_{k=1}^{n} \log\cos\left(\frac{kx}{n}\right)
= \sum_{k=1}^{n}\left[-\frac{k^2 x^2}{2n^2} + O\left(\frac{k^4x^4}{n^4}\right)\right]
= -\frac{nx^2}{6} + O\left(x^2 \vee nx^4\right).$$
Now let
$$ \begin{align*}\frac{1}{n^{3/2}} \int_{-\infty}^{\infty} \frac{1 - \prod_{k=1}^{n}\cos (kx)}{x^2} \; dx
&= \frac{1}{\sqrt{n}} \int_{-\infty}^{\infty} \frac{1 - \prod_{k=1}^{n}\cos \left(\frac{kx}{n}\right)}{x^2} \; dx \qquad (nx \mapsto x) \\
&= \frac{1}{\sqrt{n}} \int_{|x|\leq 1} + \frac{1}{\sqrt{n}} \int_{|x| > 1}
=: J_n + K_n.
\end{align*}$$
For $K_n$, we have
$$ \left|K_n\right| \leq \frac{1}{\sqrt{n}} \int_{1}^{\infty} \frac{2}{x^2}\;dx = O\left(\frac{1}{\sqrt{n}}\right).$$
For $J_n$, the substitution $\sqrt{n} x \mapsto y$ gives
$$ \begin{align*}
J_n
&= \frac{1}{\sqrt{n}} \int_{|x|\leq 1} \left( 1 - \exp\left( -\frac{nx^2}{6} + O\left(x^2 \vee nx^4\right) \right) \right) \; \frac{dx}{x^2} \\
&= \int_{|y|\leq\sqrt{n}} \left( 1 - \exp\left( -\frac{y^2}{6} + O\left(\frac{y^2}{n}\right) \right) \right) \; \frac{dy}{y^2} \\
&\xrightarrow[]{n\to\infty} \int_{-\infty}^{\infty} \frac{1 - e^{-y^2/6}}{y^2} \; dy \\
&= \left[-\frac{1-e^{-y^2/6}}{y}\right]_{-\infty}^{\infty} + \frac{1}{3} \int_{-\infty}^{\infty} e^{-y^2/6} \; dy
= \sqrt{\frac{2\pi}{3}}.
\end{align*}$$
This completes the proof.
[1]: http://i.stack.imgur.com/P28na.png
I thought it would be worth mentioning that $$\int_{-\infty}^{\infty} \frac{1-\prod_{k=1}^{n}\cos (a_{k}x)}{x^{2}} = \pi a_{n} \tag{1}$$ if the $a_{k}$'s are positive parameters and $a_{n} \ge \sum_{k=1}^{n-1} a_{k}. $
This integral is somewhat similar to the [Borwein integral][1].
We can show $(1)$ by integrating the function $$f(z) = \frac{1-e^{ia_{n}z}\prod_{k=1}^{n-1}\cos(a_{k}z)}{z^{2}}$$ around an indented contour that consists of the real axis and the semicircle above it.
If $a_{n} \ge \sum_{k=1}^{n-1} a_{k}$, then $$1-e^{ia_{n}z}\prod_{k=1}^{n-1}\cos(a_{k}z) = 1- e^{ia_{n}x}\prod_{k=1}^{n-1}\frac{e^{ia_{k}z}-e^{-ia_{k}z}}{2} $$ is bounded in the upper half-plane. (If you expand, all of the exponentials are of the form $e^{ibz}$, where $b\ge 0$.)
So integrating around the contour, we get $$\operatorname{PV} \int_{-\infty}^{\infty} f(x) \, dx - i \pi \operatorname{Res} [f(z), 0] = 0, $$ where $$ \begin{align}\operatorname{Res}[f(z),0] &= \lim_{z \to 0} \frac{1-e^{ia_{n}z}\prod_{k=1}^{n-1}\cos(a_{k}z)}{z} \\ &= \lim_{z \to 0} \left(-ia_{n}e^{ia_{n}z}\prod_{k=1}^{n-1}\cos(a_{k}z)+e^{ia_{n}z} \sum_{k=1}^{n-1} a_{k} \sin(a_{k}z) \prod_{\underset{j \ne k}{j =1}}^{n-1} \cos(a_{j}z)\right) \\ &= -i a_{n}. \end{align}$$
The result follows by equating the real parts on both sides of the equation.
[1]: https://en.wikipedia.org/wiki/Borwein_integral
$I_n(x)$ is defined as the following.
$$ I_n(x) := \int_0^{\infty } \left(\prod _{k=1}^n \frac{\sin \left(\displaystyle\frac{t}{ k^x}\right)}{\displaystyle\frac{t}{k^x}}\right) \, \mathbb{d}t$$
We know
$$ I_1(1) = I_2(1) = I_3(1) = \frac{\pi}{2},$$
$$ I_4(1) = \frac{1727 \pi}{3456}, I_5(1) = \frac{20652479 \pi}{41472000},$$
$$ I_6(1) = \frac{2059268143 \pi}{4147200000}, I_7(1) = \frac{24860948333867803 \pi}{50185433088000000}, \cdots .$$
Now, prove
$$ I_n(x) = \frac{\pi}{2}$$
for $x \ge 2$.
Let's define $$\textrm{sinc}\left(x\right)=\begin{cases}
\frac{\sin\left(x\right)}{x}, & x\neq0\\
1, & x=0
\end{cases}
$$ we have the following :
>**Theorem:** Suppose that $\left\{ a_{n}\right\}
$ is a sequence of positive numbers. Let $s_{n}=\sum_{k=1}^{n}a_{k}
$ and $$\tau_{n}=\int_{0}^{\infty}\prod_{k=0}^{n}\textrm{sinc}\left(a_{k}t\right)dt
$$ then $$0<\tau_{n}\leq\frac{\pi}{a_{0}n}
$$ and the equality holds if $n=0
$ or $a_{0}\geq s_{n}
$ when $n\geq1$.
(See [here][1] for the proof) So if we define $$a_{k}=\frac{1}{\left(k+1\right)^{x}}$$ with $x\geq2$ we note that $$a_{0}=1\geq\sum_{k=1}^{\infty}\frac{1}{\left(k+1\right)^{x}}=\zeta\left(x\right)-1$$ and so we have $$I_{n}(x)=\int_{0}^{\infty}\prod_{k=1}^{n}\frac{\sin\left(t/k{}^{x}\right)}{t/k^{x}}dt=\int_{0}^{\infty}\prod_{k=0}^{n}\frac{\sin\left(t/\left(k+1\right)^{x}\right)}{t/\left(k+1\right)^{x}}dt=\frac{\pi}{2}.
$$ **Note:** I think your product must start from $1$ since $$\frac{\sin\left(a/x\right)}{a/x}\overset{x\rightarrow0}{\rightarrow}0
$$ and so all product becomes $0$.
[1]:http://www.thebigquestions.com/borweinintegrals.pdf
https://math.stackexchange.com/questions/11246/how-to-prove-that-tan3-pi-11-4-sin2-pi-11-sqrt11?rq=1
How can we prove the following trigonometric identity?
$$\displaystyle \tan(3\pi/11) + 4\sin(2\pi/11) =\sqrt{11}$$
This is a famous problem!
A proof, which I got from just googling, appears as a solution Problem 218 in the College Mathematics Journal.
Snapshot:
![alt text][1]
[1]: http://i.stack.imgur.com/PYq30.png
You should be able to find a couple of different proofs more and references here: http://arxiv.org/PS_cache/arxiv/pdf/0709/0709.3755v1.pdf
Another way to solve it using the following theorem found [here][1] (author B.Sury):
> Let $p$ be an odd prime, $p\equiv -1 \pmod 4$ and let $Q$ be the set of squares in $\mathbb{Z}_p^*$. Then, $$\sum_{a\in Q}\sin\left(\frac{2a\pi}{p}\right)=\frac{\sqrt{p}}{2}$$
[1]: http://www.isibang.ac.in/~sury/luckyoct10.pdf
You may also need to use $2\sin(x)\cos(y)=\sin(x+y)+\sin(x-y)$.
Since $\tan\frac{3\pi}{11}+4\sin\frac{2\pi}{11}>0$, it's enough to prove that
$$\left(\sin\frac{3\pi}{11}+4\sin\frac{2\pi}{11}\cos\frac{3\pi}{11}\right)^2=11\cos^2\frac{3\pi}{11}$$ or
$$\left(\sin\frac{3\pi}{11}+2\sin\frac{5\pi}{11}-2\sin\frac{\pi}{11}\right)^2=11\cos^2\frac{3\pi}{11}$$ or
$$1-\cos\frac{6\pi}{11}+4-4\cos\frac{10\pi}{11}+4-4\cos\frac{2\pi}{11}+4\cos\frac{2\pi}{11}-4\cos\frac{8\pi}{11}-$$
$$-4\cos\frac{2\pi}{11}+4\cos\frac{4\pi}{11}-8\cos\frac{4\pi}{11}+8\cos\frac{6\pi}{11}=11+11\cos\frac{6\pi}{11}$$ or
$$\sum_{k=1}^5\cos\frac{2k\pi}{11}=-\frac{1}{2}$$ or
$$\sum_{k=1}^52\sin\frac{\pi}{11}\cos\frac{2k\pi}{11}=-\sin\frac{\pi}{11}$$ or
$$\sum_{k=1}^5\left(\sin\frac{(2k+1)\pi}{11}-\sin\frac{(2k-1)\pi}{11}\right)=-\sin\frac{\pi}{11}$$ or
$$\sin\frac{11\pi}{11}-\sin\frac{\pi}{11}=-\sin\frac{\pi}{11}.$$
Done!
Similar to the proof from the College Mathematics Journal, but structured slightly differently.
Let $\omega=e^{i\pi /11}$. Then we get $\sin\dfrac{k\pi}{11}=\dfrac{\omega^{2k}-1}{2i\omega^k}$ and $\tan\dfrac{k\pi}{11}=\dfrac{\omega^{2k}-1}{i(\omega^{2k}+1)}$
Substitution followed by some algebraic manipulations should lead to $\displaystyle\sum_{i=0}^{10}\omega^{2i}=0$, which is certainly true.
A slightly more general one is
$$ (\tan 3x+4\sin 2x)^{2}= 11-\frac{\cos 8x(\tan 8x+\tan 3x)}{\sin x\cos 3x}.$$ The proof is similar, see e.g. on Mathlinks [here][1] or the attached file [on the bottom of this post][2].
[1]: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=494824#p494824
[2]: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=90737#p90737
**Nice Question:**
show that: The follow nice trigonometry
>$$\tan{\dfrac{2\pi}{13}}+4\sin{\dfrac{6\pi}{13}}=\sqrt{13+2\sqrt{13}}$$
This problem I have ugly solution, maybe someone have nice methods? Thank you
My ugly solution:
>let $$A=\tan{\dfrac{2\pi}{13}}+4\sin{\dfrac{6\pi}{13}},B=\tan{\dfrac{4\pi}{13}}+4\sin{\dfrac{\pi}{13}}$$
since
$$\tan{w}=2[\sin{(2w)}-\sin{(4w)}+\sin{(6w)}-\sin{(8w)}+\cdots\pm \sin{(n-1)w}]$$
where $n$ is odd,and $w=\dfrac{2k\pi}{n}$
so
>$$\tan{\dfrac{2\pi}{13}}=2\left(\sin{\dfrac{4\pi}{13}}-\sin{\dfrac{5\pi}{13}}+\sin{\dfrac{\pi}{13}}+\sin{\dfrac{3\pi}{13}}-\sin{\dfrac{6\pi}{13}}+\sin{\dfrac{2\pi}{13}}\right)$$
$$\tan{\dfrac{4\pi}{13}}=2\left(\sin{\dfrac{5\pi}{13}}-\sin{\dfrac{3\pi}{13}}-\sin{\dfrac{2\pi}{13}}-\sin{\dfrac{6\pi}{13}}-\sin{\dfrac{\pi}{13}}+\sin{\dfrac{4\pi}{13}}\right)$$
then
$$A^2-B^2=(A+B)(A-B)=16\left(\sin{\dfrac{\pi}{13}}+\sin{\dfrac{3\pi}{13}}+\sin{\dfrac{4\pi}{13}}\right)\left(\sin{\dfrac{2\pi}{13}}-\sin{\dfrac{5\pi}{13}}+\sin{\dfrac{6\pi}{13}}\right)=\cdots=4\sqrt{13}$$
$$AB=\cdots=6\left(\cos{\dfrac{\pi}{13}}+\cos{\dfrac{2\pi}{13}}+\cos{\dfrac{3\pi}{13}}-\cos{\dfrac{4\pi}{13}}-\cos{\dfrac{5\pi}{13}}+\cos{\dfrac{6\pi}{13}}\right)=\cdots=3\sqrt{3}$$
so
$$A=\sqrt{13+2\sqrt{13}},B=\sqrt{13-2\sqrt{13}}$$
Have other nice metods?
and I know this is simlar 1982 AMM problem: https://math.stackexchange.com/questions/11246/how-to-prove-that-tan3-pi-11-4-sin2-pi-11-sqrt11
But My problem is hard then AMM problem。Thank you very much!
Here's an approach using some number theory. I'm no cleaner than yours, but it does apply standard techniques that might be good to know (and pretty much always work, even when there's no short solution).
Let $\zeta = \exp\bigl(\frac{2\pi i}{13}\bigr)$. Then $\zeta$ solves the 12th-order equation $p(\zeta) = \zeta^{12} + \zeta^{11} + \dots + \zeta + 1 = 0$, and no polynomial with rational coefficients of lower degree. Note that $\tan \frac{2\pi}{13} = -i\frac{\zeta - \zeta^{-1}}{\zeta + \zeta^{-1}}$, and $\sin \frac{6\pi}{13} = -i \frac{\zeta^3 - \zeta^{-3}}{2}$. Thus the problem is equivalent to showing that
$$ \xi = \frac{\zeta - \zeta^{-1}}{\zeta + \zeta^{-1}} + 2 (\zeta^3 - \zeta^{-3}) $$
solves the equation $ \xi^2 = -13 - 2\sqrt{13} $.
Note that this is almost equivalent to the equation
$$(\xi^2 + 13)^2 = 52$$
The difference is the choice of $\pm \sqrt{13}$ above, which can in principal be fixed with some estimations.
This latter equation must follow purely from the algebraic equation $p(\zeta) = 0$; in particular, it must hold for all other roots of $p$. This suggests that the ideas of Galois theory could help.
So, let's take some time to document the Galois theory of the algebraic integer $\zeta$, and of the field $\mathbb Q[\zeta]$. We begin by calculating its Galois group. The roots of $p$ are $\zeta,\zeta^2,\dots,\zeta^{12}$; and so the Galois group $G = \mathrm{Aut}(\mathbb Q[\zeta])$ has order $12$. Each element $f\in G$ is of the form $f_m : \zeta \mapsto \zeta^m$; noting that $f_m f_n (\zeta) = f_m( \zeta^n) = (\zeta^m)^n = \zeta^{mn}$, we see that the Galois group is abelian. Noting that $3^3 \equiv 1 \pmod {13}$, we see that $f_3$ is an automorphism of order $3$. Finally, $2^4 \equiv 3 \pmod{13}$, so $(f_2)^4 = f_3$, from which it follows that $f_2$ is an element in $G$ of order $12$. In particular, $G \cong \mathbb Z/(12)$ is cyclic, generated by (for example) $f_2$.
This means the following. An element of $\mathbb Q[\zeta]$ — i.e. a polynomial in $\zeta$ — is rational iff it it invariant under $f_2$. Every element of $\mathbb Q[\zeta]$ solves a 12th-order polynomial. Since $f_3$ and its inverse $(f_3)^{-1} = (f_3)^2 = f_9$ are the only elements of $G$ of order $3$, an element of $\mathbb Q[\zeta]$ solves a 4th-order polynomial iff it is invariant under $f_3$. Note that $f_4$ generates the subgroup of order $6$, which has index $2$ in $\mathbb Z / 6$; therefore an element solves a quadratic equation iff it is invariant under $f_4$. And so on: the subgroup of order $2$ is generated by $f_{12} : \zeta \mapsto \zeta^{12} = \zeta^{-1}$, and so elements invariant under $f_{12}$, like $\zeta + \zeta^{-1}$, solve 6th-degree polynomials.
Returning to the $\xi$ at hand, let's suppose we don't know what polynomial it's supposed to solve, and try to find it. Before continuing, let's factor a copy of $1 + z^2$ out of $p(z) - 1$, to clear denominators in $\xi$:
$$ p(z) - 1 = (z+z^{-1})(z^2 + z^3 + z^6 + z^7 + z^{10} + z^{11}) $$
$$ \frac1{\zeta + \zeta^{-1}} = -\zeta^2 - \zeta^3 - \zeta^6 - \zeta^{-6} - \zeta^{-3} - \zeta^{-2} $$
$$ \xi = \zeta^1 + \zeta^2 - \zeta^3 - \zeta^4 + \zeta^5 + \zeta^6 - \zeta^{-6} - \zeta^{-5} + \zeta^{-4} + \zeta^{-3} - \zeta^{-2} - \zeta^{-1} + 2(\zeta^3 - \zeta^{-3}) = \zeta^1 + \zeta^2 + \zeta^3 - \zeta^4 + \zeta^5 + \zeta^6 - \zeta^{-6} - \zeta^{-5} + \zeta^{-4} - \zeta^{-3} - \zeta^{-2} - \zeta^{-1}$$
Note that the orbits under $f_3$ are $\{\zeta,\zeta^3,\zeta^4\}$, $\{\zeta^2, \zeta^6,\zeta^{18} = \zeta^5\}$, and two more formed from these by $\zeta \mapsto \zeta^{-1}$. Inspection then shows that $\xi$ is in fact invariant under $f_3$, hence solves a 4th-order equation. The four roots are necessarily given by $\xi, f_2(\xi), f_4(\xi), f_8(\xi)$. Thus the equation is $(z-\xi)(z-f_2\xi)(z - f_4\xi)(z - f_8\xi)$.
Rather than multiplying this out, let's note that $f_4(\xi) = -\xi$, and $f_8(\xi) = -f_2(\xi)$. This is because $\xi$ transforms by a factor of $-1$ under the action of $f_{12} = (f_4)^3$. (Put another way, $\xi$ is pure imaginary.) Thus we're looking for the polynomial
$$ q(z) = \bigl(z^2 - \xi^2\bigr)\bigr(z^2 - f_2(\xi)^2\bigr) $$
since it will the minimal polynomial with rational coefficients solved by $\xi$.
Let us write $\alpha = \zeta + \zeta^3 + \zeta^9$, so that $\xi = \alpha + f_2(\alpha) - f_4(\alpha) - f_8(\alpha)$, and $f_2(\xi) = - \alpha + f_2(\alpha) + f_4(\alpha) - f_8(\alpha)$. Note also that the defining equation is $\alpha + f_2(\alpha) + f_4(\alpha) + f_8(\alpha) + 1 = 0$. We are reduced to calculating two numbers: $b = \xi^2 + f_2(\xi)^2$ and $c = \xi^2f_2(\xi)^2$; then $q(z) = z^4 - bz^2 + c$. We note that
$$ \alpha^2 = f_2(\alpha) + 2\zeta^4 + 2\zeta^{10} + 2\zeta^{12} = f_2(\alpha) + 2f_4(\alpha) $$
$$ \alpha \ f_4(\alpha) = (\zeta + \zeta^3 + \zeta^{-4})(\zeta^4 + \zeta^{-1} + \zeta^{-3}) = 3 + \zeta^5 + \zeta^{-2} + \zeta^{-6} + \zeta^2 + \zeta^{-5} + \zeta^6 = 3 + f_2(\alpha) + f_8(\alpha) $$
Writing $\beta = \alpha - f_4(\alpha)$, we have:
$$ \xi = \beta + f_2(\beta), \quad f_2(\xi) = -\beta + f_2(\beta) $$
$$ b = \xi^2 + f_2(\xi)^2 = 2 (\beta^2 + f_2(\beta)^2) $$
$$ c = \bigl(\xi f_2(\xi)\bigr)^2 = \bigl(\beta^2 - f_2(\beta)^2\bigr)^2 $$
$$ \beta^2 = \alpha^2 + f_4(\alpha)^2 - 2\alpha \ f_4(\alpha) = f_2(\alpha) + 2f_4(\alpha) + f_8(\alpha) + 2 \alpha - 2(3 + f_2(\alpha) + f_8(\alpha)) = -6 + 2\alpha - f_2(\alpha) + 2f_4(\alpha) - f_8(\alpha)$$
$$ \beta^2 + f_2(\beta)^2 = -12 + (\alpha + f_2\alpha + f_4\alpha + f_8\alpha) = -13 $$
$$ \beta^2 - f_2(\beta)^2 = 3\alpha - 3f_2(\alpha) + 3f_4(\alpha) - 3f_8(\alpha) = 3(\alpha + f_4(\alpha) - f_2(\alpha) - f_8(\alpha)) $$
Therefore $b = -26$. Let $\gamma = \alpha + f_4\alpha$, so that $\gamma + f_2(\gamma) = -1$, and
$$ \gamma \ f_2(\gamma) = (\zeta + \zeta^3 + \zeta^4 + \zeta^{-4} + \zeta^{-3} + \zeta^{-2})(\zeta^2 + \zeta^5 + \zeta^6 + \zeta^{-6} + \zeta^{-5} + \zeta^{-2}) =
\zeta + \zeta^2 + \zeta^3 + \zeta^5 + 2\zeta^6 + \zeta^{-6} + 2\zeta^{-5} + 3\zeta^{-4}
+ 2\zeta^{-3} + 2\zeta^{-2} + 2\zeta^{-1} + (\zeta \leftrightarrow \zeta^{-1}) = 3(\zeta + \dots + \zeta^{-1}) = -3$$
The last thing to calculate is:
$$ c = 9(\gamma - f_2(\gamma))^2 = 9\bigl((\gamma + f_2\gamma)^2 - 4\gamma \ f_2\gamma\bigr) = 9\bigl( 1 - 4(-3)\bigr) = 9\times 13$$
Thus $q(z) = z^4 + 26 z^2 + 9\times 13 = (z^2 + 13)^2 - 4\times 13$, completing the proof.
The following argument is more or less a duplicate in this [paper](https://math.stackexchange.com/questions/11246/how-to-prove-that-tan3-pi-11-4-sin2-pi-11-sqrt11?rq=1):
Let $x=e^{2\pi i/13}$. Then $$i\tan{2\pi/13}=\frac{x^2-1}{x^2+1}=\frac{x^2-x^{26}}{x^2+1}$$
(recall that $x^{13}=1$)
$$=x^2(1-x^2)(1+x^4+x^8+x^{12}+x^3+x^7)$$
$$=(x+x^2+x^5+x^6+x^9+x^{10}-x^3-x^4-x^7-x^8-x^{11}-x^{12})$$
$$4i\sin{6\pi/13}=2(x^3-x^{10})$$
So $i\tan{2\pi/13}+4i\sin{6\pi/13}=(x+x^2+x^3+x^5+x^6+x^9-x^4-x^7-x^8-x^{10}-x^{11}-x^{12})$
Recall that $1+x+x^2+\cdots+x^{12}=0$.
After some tedious computation, we arrive at
$$(x+x^2+x^3+x^5+x^6+x^9)(x^4+x^7+x^8+x^{10}+x^{11}+x^{12})$$
$$=4+x+x^3+x^4+x^9+x^{10}+x^{12}$$
The key step in the deduction is the [famous exponential sum of Gauss](http://en.wikipedia.org/wiki/Quadratic_Gauss_sum), which gives,
$$1+2(x+x^4+x^9+x^3+x^{12}+x^{10})=\sqrt{13}.$$
Hence $$(x+x^2+x^3+x^5+x^6+x^9)(x^4+x^7+x^8+x^{10}+x^{11}+x^{12})=(7+\sqrt{13})/2$$
Recall our formula $1+x+x^2+\cdots+x^{12}=0$ again, and
$$(x+x^2+x^3+x^5+x^6+x^9-x^4-x^7-x^8-x^{10}-x^{11}-x^{12})^2=(-1)^2-4\times(7+\sqrt{13})/2$$
$$=-13-2\sqrt{13}$$
Hence $i\tan{2\pi/13}+4i\sin{6\pi/13}=\pm i\sqrt{13+2\sqrt{13}}$
and it is obvious that $\tan{2\pi/13}+4\sin{6\pi/13}=\sqrt{13+2\sqrt{13}}$, Q.E.D.
**P.S.** I have a strong feeling that a generalization of such an identity to all primes is possible, but I cannot work them out right now.
let $\lambda $ is a any complex numbers,and $\tau\in[0,1]$
show that
$$\left|\dfrac{\sin{(\sqrt{\lambda}\cdot\tau)}}{\sqrt{\lambda}}\right|\le e^{|\mathrm{Im}\sqrt{\lambda}|\cdot\tau}$$
my idea:
$$\left|\dfrac{\sin{(\sqrt{\lambda}\cdot\tau)}}{\sqrt{\lambda}}\right|\le \int_{0}^{\tau}|\cos{(\sqrt{\lambda}\cdot s)}|ds$$
follow I can't solve it,Thank you
For any real numbers $x,y\neq 0$,show that
$$\Big|x\sin{\dfrac{1}{x}}-y\sin{\dfrac{1}{y}}\Big|<2\sqrt{|x-y|}$$
I found this problem when I dealt with [this problem][1]. But I can't prove it. Maybe the constant $2$ on the right hand side can be replaced by the better constant $\sqrt{2}$?
Thank you.
[1]: https://math.stackexchange.com/questions/522163/how-prove-this-analysis-function-a-le-frac12
We'll first assume that $0 \le x < y$.
I need three kinds of estimates here.
1. $\left| x \sin \frac{1}{x} - y \sin \frac{1}{y} \right| \le x + y$
2. $\left| x \sin \frac{1}{x} - y \sin \frac{1}{y} \right| \le 2$
3. $\left| x \sin \frac{1}{x} - y \sin \frac{1}{y} \right| \le \frac{y}{x} - 1$
Estimates 1 and 2 are trivial: $\left| x \sin \frac{1}{x} \right| \le \min(x, 1)$
Estimate 3 is proved as follows.
- First we show that $\left| (x \sin \frac{1}{x})^\prime \right| \le \frac{1}{x}$. Indeed, $(x \sin \frac{1}{x})^{\prime\prime} = - \frac{1}{x^3} \sin \frac{1}{x}$, so local maxima and minima of $(x \sin \frac{1}{x})^\prime$ are located at $\frac{1}{\pi n}, n \in \mathbb{N}$, and its values there are $\frac{(-1)^n}{\pi n}$, so that $\sup_{z \ge x} \left| (z \sin \frac{1}{z})^\prime \right| \le \frac{1}{\pi n} \le \frac{1}{x}$, where $\frac{1}{\pi n}$ is the smallest one in $[x,+\infty)$.
- Now $\left| x \sin \frac{1}{x} - y \sin \frac{1}{y} \right| \le \intop_x^y \frac{1}{z} dz = \log \frac{y}{x} \le \frac{y}{x} - 1$
Now let's find the three regions corresponding to:
1. $x + y \le C \sqrt{y - x}$
2. $2 \le C \sqrt{y - x}$
3. $\frac{y}{x} - 1 \le C \sqrt{y - x}$
and choose the constant $C$ in such a way that these regions cover $\{0 < x < y\}$. Clearly, 2 is bounded by a line, 1 and 3 are bounded by parabolas. Miraculously, $C = 2$ is the unique value when the three boundaries intersect at a single point, namely $(x,y) = (1/2, 3/2)$...
Anyway, let's rewrite our regions for $C = 2$:
1. $y-x \ge \frac{1}{4} (x+y)^2$
2. $y-x \ge 1$
3. $y-x \le 4 x^2$
So to cover the whole space we have to prove $y-x \ge \frac{1}{4} (x+y)^2$ on $\{4 x^2 \le y-x \le 1\}$, for which it is sufficient to consider just the two boundary cases, namely $y-x = 1$ and $y-x = 4 x^2$, since a quadratic polynomial with positive leading term attains maximum on the boundary of a segment. And these cases are easy to verify. Indeed, the inequality in terms of $x$ and $z := y-x$ looks like $x^2 + xz + \frac{1}{4} z^2 \le z$; for $z=1$ it's equivalent to $(x + \frac{1}{2})^2 \le 1$, which is true, since $4 x^2 \le 1$; for $z = 4 x^2$ it's equivalent to $3 x^2 \ge 4 x^3 + 4 x^4$, which follows once again from $4 x^2 \le 1$ (since $x^2 \ge 2 x^3$ and $x^2 \ge 4 x^4$).
Now the case $x < 0 < y$ is even simpler. We only need analogs of estimates 1 and 2:
1. $\left| x \sin \frac{1}{x} - y \sin \frac{1}{y} \right| \le |x| + |y| \le C \sqrt{|x| + |y|}$ whenever $|x| + |y| \le C^2 = 4$
2. $\left| x \sin \frac{1}{x} - y \sin \frac{1}{y} \right| \le 2 \le C \sqrt{|x| + |y|}$ whenever $|x| + |y| \ge (\frac{2}{C})^2 = 1$.
This is a partial alternative solution for the case $\left|\frac{1}{x}-\frac{1}{y}\right|\leq 1$.
We can clearly assume that both $x$ and $y$ are positive, then, through the substitutions $x\to 1/x,y\to 1/y$, prove that:
$$ \forall x,y>0,x\neq y,\quad (y\sin x-x\sin y)^2 < 4xy|y-x|. $$
The LHS can be written as:
$$ \left((y-x)\sin x + x\,(\sin x-\sin y)\right)^2, $$
that, by the Cauchy-Schwarz inequality, satisfies:
$$ \left((y-x)\sin x + x\,(\sin x-\sin y)\right)^2 \leq (x^2+\sin^2 x)\left((y-x)^2+(\sin y-\sin x)^2\right), $$
and the RHS is less than $4x^2(y-x)^2$, since $\sin x$ is a $1$-Lipschitz function.
By exchanging $x$ and $y$, we have:
$$ (y\sin x-x\sin y)^2 < \min\left(4x^2(y-x)^2,4y^2(y-x)^2\right)<4xy(y-x)^2,$$
so the inequality is clearly true if $|y-x|\leq 1$.
https://math.stackexchange.com/q/501984/165013
show that
>$$\mathop {\lim }\limits_{n \to \infty } \left( {\int\limits_0^{\frac{\pi }{2}} {\left\vert\frac{{\sin \left( {2n + 1} \right)x}}{{\sin x}}\right\vert\,dx - \frac{{2\ln n}}{\pi }} } \right) = \frac{{6\ln 2}}{\pi } + \frac{{2\gamma }}{\pi } + \frac{2}{\pi }\sum\limits_{k = 1}^\infty {\frac{1}{{2k + 1}}\ln \left( {1 + \frac{1}{k}} \right)}\cdots (1) $$
I can prove $(1)$ it exsit it.and also it is well kown that
$$I_{n}=\int_{0}^{\frac{\pi}{2}}\dfrac{\sin{(2n+1)x}}{\sin{x}}dx=\dfrac{\pi}{2}$$
>proof:$$I_{n}-I_{n-1}=\int_{0}^{\frac{\pi}{2}}\dfrac{\sin{(2n+1)x}-\sin{(2n-1)x}}{\sin{x}}dx=2\int_{0}^{\frac{\pi}{2}}\cos{(2nx)}dx=0$$
so
$$I_{n}=I_{n-1}=\cdots=I_{0}=\dfrac{\pi}{2}$$
But I can't prove $(1)$,Thank you
Notice for any continuous function $f(x)$ on $[0,\frac{\pi}{2}]$, we have:
$$\lim_{n\to\infty} \int_0^{\frac{\pi}{2}} \Big|\sin((2n+1)x)\Big| f(x) dx = \frac{2}{\pi}\int_0^{\frac{\pi}{2}} f(x) dx$$
Apply this to $\frac{1}{\sin x} - \frac{1}{x}$, we get
$$\lim_{n\to\infty} \int_0^{\frac{\pi}{2}} \Big|\sin((2n+1)x)\Big| \Big(\frac{1}{\sin x} - \frac{1}{x} \Big) dx
= \frac{2}{\pi} \int_0^{\frac{\pi}{2}} \Big(\frac{1}{\sin x} - \frac{1}{x} \Big) dx\\
= \frac{2}{\pi} \left[\log\left(\frac{\tan(\frac{x}{2})}{x}\right)\right]_0^{\frac{\pi}{2}}
= \frac{2}{\pi} \left[\log\frac{2}{\pi} - \log{\frac12}\right] = \frac{2}{\pi} \log\frac{4}{\pi}
\tag{*1}$$
So it suffices to figure out the asymptotic behavior of following integral:
$$\int_0^{\frac{\pi}{2}} \frac{|\sin((2n+1)x)|}{x} dx
= \int_0^{\pi(n+\frac12)} \frac{|\sin x|}{x} dx = \int_0^{\pi n} \frac{|\sin x|}{x} dx + O(\frac{1}{n})
$$
We can rewrite the rightmost integral as
$$\int_0^{\pi} \sin x \Big( \sum_{k=0}^{n-1} \frac{1}{x+k\pi} \Big) dx
= \int_0^1 \sin(\pi x) \Big( \sum_{k=0}^{n-1} \frac{1}{x+k} \Big) dx\\
= \int_0^1 \sin(\pi x) \Big( \psi(x+n) - \psi(x) \Big) dx
\tag{*2}
$$
where $\displaystyle \psi(x) = \frac{\Gamma'(x)}{\Gamma(x)}$ is the
[digamma function](http://en.wikipedia.org/wiki/Digamma_function).
Using following asymptotic expansion of $\psi(x)$ for large $x$:
$$\psi(x) = \log x - \frac{1}{2x} + \sum_{k=1}^{\infty}\frac{\zeta(1-2k)}{x^{2k}}$$
It is easy to verify
$$\int_0^1 \sin(\pi x)\psi(x+n) dx = \frac{2}{\pi} \log n + O(\frac{1}{n})\tag{*3}$$.
Substitute $(*3)$ into $(*2)$ and combine it with $(*1)$, we get
$$\lim_{n\to\infty} \left(\int_0^{\frac{\pi}{2}} \left|\frac{\sin((2n+1)x)}{\sin x}\right| dx - \frac{2}{\pi} \log n\right) = \frac{2}{\pi} \log\frac{4}{\pi} - \int_0^1 \sin(\pi x)\psi(x) dx \tag{*4}$$
To compute the rightmost integral of $(*4)$, we first integrate it by part:
$$\int_0^1 \sin(\pi x)\psi(x) dx = \int_0^1 \sin(\pi x)\,d\log\Gamma(x) =
-\pi\int_0^1 \cos(\pi x)\log\Gamma(x) dx
$$
We then apply following result$\color{blue}{^{[1]}}$
> **Kummer (1847)** Fourier series for $\log\Gamma(x)$ for $x \in (0,1)$
> $$\log\Gamma(x) = \frac12\log\frac{\pi}{\sin(\pi x)} + (\gamma + \log(2\pi))(\frac12 - x) + \frac{1}{\pi}\sum_{k=2}^{\infty}\frac{\log k}{k}\sin(2\pi k x)$$
Notice
1. $\displaystyle \int_0^1 \cos(\pi x)\log \frac{\pi}{\sin(\pi x)} dx = 0\quad$ because of symmtry.
2. $\displaystyle \int_0^1 \cos(\pi x)\Big(\frac12 - x\Big) dx = \frac{2}{\pi^2}$
3. $\displaystyle \int_0^1 \cos(\pi x)\sin(2\pi k x) dx = \frac{4k}{(4k^2-1)\pi} $
We can evaluate RHS of $(*4)$ as
$$\begin{align}
\text{RHS}_{(*4)} = & \frac{2}{\pi}\log\frac{4}{\pi} + \pi \left[
\Big(\gamma + \log(2\pi)\Big)\frac{2}{\pi^2}
+ \frac{4}{\pi^2}\sum_{k=2}^{\infty}\frac{\log k}{4k^2-1}
\right]\\
= & \frac{2}{\pi}\left[\log 8 + \gamma + \sum_{k=2}^{\infty}\log k \left(\frac{1}{2k-1}-\frac{1}{2k+1}\right) \right]\\
= & \frac{6\log 2}{\pi} + \frac{2\gamma}{\pi} + \frac{2}{\pi}\sum_{k=1}\frac{\log(1+\frac{1}{k})}{2k+1}
\end{align}$$
***Notes***
$\color{blue}{[1]}$ For more infos about Kummer's Fourier series, please see
following [paper](http://arxiv.org/abs/0903.4323) by Donal F. Connon.
AMM Problem 11777, Vol.121, May 2014
令$n\geq 3$, $x_1,\ldots,x_n$为实数使得$\prod_{k=1}^{n}x_k=1$.证明
\[\sum_{k=1}^{\infty}{\frac{x_{k}^{2}}{x_{k}^{2}-2x_k\cos \left( 2\pi /n \right) +1}}\ge 1.\]
注:不等式对于$n=1,2$不成立,例如取$x_1=x_2=1$.
若$z_1,\ldots,z_n$和$w_1,\ldots,w_n$为复数,则由Lagrange恒等式可知
\[\left( \sum_{k=1}^n{\left| z_k \right|^2} \right) \left( \sum_{k=1}^n{\left| w_k \right|^2} \right) -\left| \sum_{k=1}^n{z_kw_k} \right|^2=\sum_{1\le k<j\le n}{\left| z_k\overline{w_j}-z_j\overline{w_k} \right|^2}.\]
令$w_k=c_k\in\mathbb{R}^+$, 令$z_k=c_ky_k$且$y_k\in\mathbb{C}$,由上述不等式可知
\[\left( \sum_{k=1}^n{c_{k}^{2}\left| y_k \right|^2} \right) \left( \sum_{k=1}^n{c_{k}^{2}} \right) \ge \sum_{1\le k<j\le n}{c_{k}^{2}c_{j}^{2}\left| y_k-y_j \right|^2}\ge \sum_{k=1}^n{c_{k}^{2}c_{k+1}^{2}\left| y_k-y_{k+1} \right|^2},\]
第二个不等式对$n\geq 3$成立当且仅当$c_{n+1}=c_1,y_{n+1}=y_1$.
设$y_1,\ldots,y_n$互异,并令$c_k=1/|y_k-y_{k+1}|>0$,我们有
\[\sum_{k=1}^n{\frac{\left| y_k \right|^2}{\left| y_k-y_{k+1} \right|^2}}\ge 1.\]
最后,令$y_{k+1}/y_k=e^{2\pi i/n}/x_k\neq 1$,则
\[\prod_{k=1}^n{\frac{y_{k+1}}{y_k}}=\frac{\left( e^{2\pi i/n} \right) ^n}{\prod_{k=1}^n{x_k}}=1.\]
因此
\begin{align*}
\sum_{k=1}^n{\frac{x_{k}^{2}}{x_{k}^{2}-2x_k\cos \left( 2\pi /n \right) +1}}&=\sum_{k=1}^n{\frac{x_{k}^{2}}{\left| x_k-e^{2\pi i/n} \right|^2}}
\\
&=\sum_{k=1}^n{\frac{1}{\left| 1-y_{k+1}/y_k \right|^2}}=\sum_{k=1}^n{\frac{\left| y_k \right|^2}{\left| y_k-y_{k+1} \right|^2}}\ge 1.
\end{align*}
