Pi和e的积分

Evaluate integral

$$\int_{0}^{1}{x^{-x}(1-x)^{x-1}\sin{\pi x}dx}$$

Well,I think we have

$$\int_{0}^{1}{x^{-x}(1-x)^{x-1}\sin{\pi x}dx}=\frac{\pi}{e}$$

 

and

 

$$\int_{0}^{1}{x^{x}(1-x)^{1-x}\sin{\pi x}dx}=\frac{e\pi}{24}$$ 

 

With such nice result of these integral,why isn't worth to evaluate it?

 

I found a solution about the second one,but I wonder it will work for the first one

Note

$$ S=\int_{0}^{1}{\sin{\pi x}x^{x}(1-x)^{1-x}dx}-\int_{0}^{1}{(1-x)e^{(i\pi+\ln{x}-\ln{(1-x)})x}dx} $$

Let $t=\ln{x}-\ln{(1-x)}$,$x=\frac{e^{t}}{1+e^{t}}$

Thus

\begin{align}S&=\int_{-\infty}^{+\infty}{\frac{1}{e^{t}+1}e^{(i\pi+t)\frac{e^{t}}{1+e^t}}\frac{e^{t}}{(1+e^{t})^{2}}dt}\\ &=\int_{-\infty+i\pi}^{-\infty-i\pi}{e^{\frac{te^{t}}{e^{t}-1} } \frac{e^{t}}{(e^{t}-1)^{3}}dt}\end{align}

Due to

$$ f(z)=e^{\frac{te^{t}}{e^{t}-1} } \frac{e^{t}}{(e^{t}-1)^{3}},\qquad D=\{Z\in C|-\pi\leq Im(z) \leq \pi\}$$

Therefore

$res(f,0)=-\frac{e}{24}$when $z=0$

with $ \zeta_{R}=\gamma_{R}+o_{R}+\tau_{R}$

$$\oint_{\zeta_{R}}{f(z)dz}=-2\pi i\cdot res(f,0)=\frac{2i\pi e}{24}$$

because

$$ \{z_{n}\}\subset D,\qquad |z_{n}|\rightarrow\infty $$

Therefore

$$ 2S=2\lim_{R\rightarrow \infty}\int_{\gamma_{R}}{f(z)dz} $$

gives

$$ \int_{0}^{1}{\sin{\pi x}x^{x}(1-x)^{1-x}dx}=Im(S)=\frac{e\pi}{24} $$

 

My friend tian_275461 told me he use a simliar method to deal with the first one to obtain the result $\frac{\pi}{e}$,but I am not figure it out.

---

第一个积分的解答:

Exactly the same method works for the other case.

$$\int_0^1 x^{-x} (1-x)^{x-1}\sin{\pi x} dx = \mathrm{Im}\left[\int_0^1 \frac{e^{(i\pi+\ln(1-x)-\ln x)x}}{1-x}dx\right]$$

Write $t=\ln((1-x)/x)$ and $z=t+i\pi$ as you did above to get

$$S = \int_0^1 \frac{e^{(i\pi+\ln(1-x)-\ln x)x}}{1-x}dx=\int_{-\infty+i\pi}^{\infty+i\pi} \frac{e^{\frac{z}{1-e^z}}}{1-e^z}dz$$

 

Then with $$f(z)=\frac{e^{\frac{z}{1-e^z}}}{1-e^z}$$

the only pole is at $z=0$, $res(f,0)=-\frac{1}{e}$ and in the limit $2S = \oint f(z)dz=-2\pi i \cdot res(f,0) = 2\pi i/e$ and your answer follows.

---

第二个积分的另一种求法:

This one can be done with "residue at infinity" calculation. This method is shown in the Example VI of http://en.wikipedia.org/wiki/Methods_of_contour_integration . 

 

First, we use $z^z = \exp ( z \log z )$ where $\log z$ is defined for $-\pi\leq \arg z < \pi$. 

 

For $(1-z)^{1-z} = \exp ( (1-z)\log (1-z) )$, we use $\log (1-z)$ defined for $0\leq \arg(1-z) <2\pi$. 

 

Then, let  $f(z)= \exp( i\pi z + z \log z + (1-z) \log (1-z) )$. 

 

As shown in the Ex VI in the wikipedia link, we can prove that $f$ is continuous on $(-\infty, 0)$ and $(1,\infty)$, so that the cut of $f(z)$ is $[0,1]$. 

 

We use the contour: (consisted of upper segment: slightly above $[0,1]$, lower segment: slightly below $[0,1]$, circle of small radius enclosing $0$, and circle of small radius enclosing $1$, that looks like a dumbbell having knobs at $0$ and $1$, can someone edit this and include a picture of it please? In fact, this is also the same contour as in Ex VI, with different endpoints.)

 

On the upper segment, the function $f$ gives, for $0\leq r \leq 1$, 

$$\exp(i\pi r) r^r (1-r)^{1-r} \exp( (1-r) 2\pi i ).$$ 

 

On the lower segment, the function $f$ gives, for $0\leq r \leq 1$, 

$$\exp(i\pi r) r^r(1-r)^{1-r}. $$

 

 

Since the functions are bounded, the integrals over circles vanishes when the radius tend to zero. 

 

Thus, the integral of $f(z)$ over the contour, is the integral over the upper and lower segments, which contribute to

 

$$\int_0^1 \exp(i\pi r) r^r (1-r)^{1-r} dr - \int_0^1 \exp(-i\pi r) r^r(1-r)^{1-r} dr$$

 

which is 

$$2i \int_0^1 \sin(\pi r) r^r (1-r)^{1-r} dr.$$

 

By the Cauchy residue theorem, the integral over the contour is

$$-2\pi i \textrm{Res}_{z=\infty} f(z) = 2\pi i \textrm{Res}_{z=0} \frac{1}{z^2} f(\frac 1 z).$$

 

From a long and tedious calculation of residue, it turns out that the value on the right is 

$$2i \frac{\pi e}{24}.$$

Then we have the result:

$$ \int_0^1 \sin(\pi r) r^r (1-r)^{1-r} dr = \frac{\pi e}{24}.$$

 

我们也可得到\begin{align*} \int_{0}^{1} e^{i \pi x} \, x^{x} (1-x)^{1-x} \, dx = i \, \frac{\pi e}{4!} \end{align*}

来自:http://math.stackexchange.com/questions/324647/integrate-int-01x-x1-xx-1-sin-pi-xdx

http://math.stackexchange.com/questions/958624/prove-that-int-01-sin-pi-xxx1-x1-x-dx-frac-pi-e24

---

 

[Torsten Carleman][1] $[2]$ proved in 1922 that
> $$
\sum_{n=1}^\infty\left(a_1a_2\cdots a_n\right)^{1/n} < e\sum_{n=1}^\infty a_n,
$$

 

where $a_n \geq 0$, $n=1,2,\dots$, and $0 < \sum_{n=1}^\infty a_n < \infty$. Thenceforth, this result is known as [Carleman's inequality][2]. There exists a number of refined versions of Carleman's original work $[3, 6]$. It has turned out that the following generalization is – from our point of view – important, which is proved by Yang $[7]$:
$$
\sum_{n=1}^\infty\left(a_1a_2\cdots a_n\right)^{1/n} < e\sum_{n=1}^\infty \left(1-\sum_{k=1}^6 \frac{b_k}{(n+1)^k}\right)a_n,
$$

 

with $b_1 = 1/2, b_2 = 1/24, b_3 = 1/48, b_4 = 73/5670, b_5 = 11/1280, b_6 = 1945/580608$.
On the last page of his paper, Yang $[7]$ conjectured that if
$$
\left(1+\frac{1}{x}\right)^x = e\left(1-\sum_{n=1}^\infty \frac{b_n}{(x+1)^n}\right), \quad x>0,
$$

 

then $b_n > 0$, $n=1,2,\dots.$ In fact, the constants $b_4$ and $b_6$ are not corrent in Yang's work, the correct values are $b_4 = 73/5760$ and $b_6 = 3625/580608$. Later, this conjecture was proved and discussed by Yang $[8]$, Gylletberg and Ping $[4]$, and Yue $[9]$. They are using the recurrence
$$
b_1 = \frac12, \quad b_n = \frac{1}{n}\left(\frac{1}{n+1} - \sum_{k=0}^{n-2} \frac{b_{n-k-1}}{k+2} \right), \quad n = 2,3,\dots.
$$
The recurrence is given in a somewhat more compact form in Finch's manuscript $[3]$, as the following:
> $$
b_0 = -1, \quad b_n = -\frac{1}{n}\sum_{k=1}^{n} \frac{b_{n-k}}{k+1}, \quad n = 1,2,\dots.
$$

 

The first ten values of the sequence are listed in the next table.
\begin{array} {|r|r|}
\hline
b_0 & -1 \\ \hline b_1 & 1/2 \\ \hline b_2 & 1/24 \\ \hline b_3 & 1/48 \\ \hline b_4 & 73/5760 \\ \hline b_5 & 11/1280 \\ \hline b_6 & 3625/580608 \\ \hline b_7 & 5525/1161216 \\ \hline b_8 & 5233001/1393459200 \\ \hline b_9 & 1212281/398131200 \\ \hline b_{10} & 927777937/367873228800 \\
\hline
\end{array}
The numerators are recorded as [A249276][3], and the denominators as [A249277][4] in the [OEIS][5]. I've calculated the $b_n$ sequence in the range $n=0,\dots,32$, the elements are listed [here][6].

 

The following theorem is proved in general in the paper by Hu and Mortici $[5]$, and for the special cases $n=0$ and $n=1$ in the paper by Alzer and Berg $[1]$.

 

For all integer $n \geq 0$, we have
> $$
\int_0^1 x^n\sin\left(\pi x\right)x^x\left(1-x\right)^{1-x}\,dx = b_{n+2}\pi e.
$$

 

The special case $n=0$ answers my question.

 

----------
**References**

 

1. H. Alzer, C. Berg, [*Some classes of completely monotonic functions*][7], Annales Academiæ Scientiarum Fennicæ Mathematica, 27, 2002, 445–460. ([pdf][8])
2. T. Carleman, [*Sur les fonctions quasi-analytiques*][9], Comptes rendus du Ve Congres des Mathematiciens Scandinaves, Helsinki (Helsingfors), 1922, 181–196.
3. S. Finch, [*Carleman's Inequality*][10], manuscript, 2013.
4. M. Gyllenberg, Y. Ping, [*On a conjecture by Yang*][11], Journal of Mathematical Analysis and Applications, 264(2), 2001, 687–690.
5. Y. Hu, C. Mortici, [*On the coefficients of an expansion of $(1+1/x)^x$ related to Carleman's inequality*][12], manuscript, arXiv:1401.2236, 2014.
6. M. Johansson, L.-E. Persson, A. Wedestig, [*Carleman's inequality - History, proofs and some new generalizations*][13], Journal of Inequalities in Pure and Applied Mathematics, 4(3), 2003.
7. X. Yang, [*On Carleman’s inequality*][14], Journal of Mathematical Analysis and Applications, 253(2), 2001, 691–694.
8. X. Yang, [*Approximations for constant $e$ and Their Applications*][15], Journal of Mathematical Analysis and Applications, 262(2), 2001, 651–659.
9. H. Yue, [*A Strengthened Carleman’s Inequality*][16], Communications in Mathematical Analysis, 1(2), 2006, 115–119. ([pdf][17])

 

----------
**Related**

 

This answer is related to the following stackexchange questions:

 

- [On the search for an explicit form of a particular integral][18]
- [Two curious “identities” on $x^x, e$, and $\pi$][19]
- [Evaluating an integral using real methods][20]

 

[1]: https://en.wikipedia.org/wiki/Torsten_Carleman
[2]: http://mathworld.wolfram.com/CarlemansInequality.html
[3]: http://oeis.org/A249276
[4]: http://oeis.org/A249277
[5]: http://oeis.org/
[6]: http://mathb.in/145272?key=677ade0f6738b4bc973f3955937b952544a82225
[7]: http://www.acadsci.fi/mathematica/Vol27/alzer.html
[8]: http://www.acadsci.fi/mathematica/Vol27/alzer.pdf
[9]: https://www.researchgate.net/publication/247679096_Sur_les_functions_quasi-analytiques
[10]: https://oeis.org/A219245/a219245.pdf
[11]: http://www.sciencedirect.com/science/article/pii/S0022247X01977029
[12]: https://arxiv.org/abs/1401.2236
[13]: https://www.researchgate.net/publication/237246073_Carleman%27s_inequality-history_proofs_and_some_new_generalizations
[14]: http://www.sciencedirect.com/science/article/pii/S0022247X00971555
[15]: http://www.sciencedirect.com/science/article/pii/S0022247X01975924
[16]: http://math-res-pub.org/cma/1/2/strengthened-carleman%E2%80%99s-inequality
[17]: https://www.ripublication.com/cma_files/cmav1n2_6.pdf
[18]: https://mathoverflow.net/questions/215816/on-the-search-for-an-explicit-form-of-a-particular-integral
[19]: https://math.stackexchange.com/questions/242587/two-curious-identities-on-xx-e-and-pi
[20]: https://mathoverflow.net/questions/226870/evaluating-an-integral-using-real-methods

---

 

This is something I am absolutely cautious to share, but I feel the need unveil anyway. I have lost some will to believe this is a significant result due to doubts expressed by other mathematicians who I have corresponded with, so this led me to construe this might not be important after all. I have read about these integrals supposedly popping up in the work of Ramanujan, though I have found no reliable source, and Bruce Berndt still has yet to get back to me.:/

This project started when I was curious what parametrizations would be needed to encapsulate impressive information about the following integrals:

\begin{align}
&\int_0^1 \sin(\pi x) x^x(1-x)^{1-x} \, dx &= \frac{\pi e}{24} \\
&\int_0^1 \frac{\sin(\pi x)}{x^x(1-x)^{1-x}}\, dx &= \frac{\pi }{e} \\
&\int_0^1 \frac{\sin(\pi x)}{x(1-x)}\frac{1}{x^x(1-x)^{1-x}}\, dx &= 2\pi
\end{align}

However, as it turns out, I was able to show they are related via the following theorem.

$\textbf{Theorem}$ For $m, q \in \mathbb{Z}$, and $m+q+1 \geq 0$,
$$ \int_0^1 x^m \sin\left(\pi q x \right) \left(x^x (1-x)^{1-x}\right)^q\ dx = (-1)^{q+1} \frac{d_{m+q+1}(q)}{(m+q+2)!_\mathbb{P}} \pi e^{q}$$
where $d_n(q)$ is a primitive polynomial of $\mathbb{Z}[x]$ of degree $n$, and $ n!_\mathbb{P}$ is the Bhargava factorial over the set of primes.

In addition, these rational numbers satisfy a neat recurrence relation, of which Carleman's inequality is a [special case][1] of:

$$\frac{d_{n}(q)}{(n+1)!_\mathbb{P}} = -\frac{q}{n} \sum_{k=1}^n \frac{d_{n-k}(q)}{(n-k+1)!_\mathbb{P}} \frac{1}{k+1}; \; d_0(q) = -1,\; \text{if} \,(q=0).$$

Using these results, we can unlock a whole class of crazy stuff:

\begin{align*}\sum_{j=1}^n A_j(1-\alpha_j)^{q\left(1-\frac{1}{\alpha_j}\right)}&= (-1)^q\int_0^1 \frac{\sin\left(\pi q x \right)}{\pi x} \frac{\left[x^x\left(1-x\right)^{1-x}\right]^q}{x^q} \prod_{j=1}^n \frac{1}{1-\alpha_j x}\ dx,
\end{align*}
\begin{align*}
A_j = \prod_{k=1, k\neq j}^n \frac{\alpha_j}{\alpha_j-\alpha_k}, \quad \alpha_j \in (0,1).
\end{align*}

Here are some special values:
\begin{align}
&\int_0^1 \frac{\sin\left( \pi x \right)}{ (1-x)\left[x^x\left(1-x\right)^{1-x}\right]} \ dx = \pi \quad \quad &\int_0^1 \frac{\sin\left( \pi x \right)}{(1-x^2)\left[x^x\left(1-x\right)^{1-x}\right]} \ dx &= \frac{5\pi}{8}
\end{align}

I don't want to reveal too much anyway. Enjoy!
[1]: http://www.people.fas.harvard.edu/~sfinch/csolve/crl.pdf

---

 

来源：https://math.stackexchange.com/questions/516001/what-is-the-most-surprising-result-that-you-have-personally-discovered/1884617#1884617

---

The Ramanujan Cos/Cosh Identity is stated [here](http://mathworld.wolfram.com/RamanujanCosCoshIdentity.html) as
$$\left[1+2\sum_{n=1}^{\infty}\frac{\cos n\theta}{\cosh n\pi}\right]^{-2}+
\left[1+2\sum_{n=1}^{\infty}\frac{\cosh n\theta}{\cosh n\pi}\right]^{-2}=
\frac{2\Gamma^4\left(\frac34\right)}{\pi}$$

Then there is a line:

> Equating coefficients of $\theta^0$, $\theta^4$, and $\theta^8$ gives
> some amazing identities for the hyperbolic secant.

Those identities are given [here](http://mathworld.wolfram.com/HyperbolicSecant.html).

So I have two questions:

1. How do we get those formulas from the Cos/Cosh identity?

2. Are there similar identities? (similar to Cos/cosh identity)

---

It will be helpful to start from an explanation of the origin and the proof of the Ramanujan identity. These are hidden (not very deeply) in the theory of elliptic functions.

Indeed, Jacobi elliptic function $\operatorname{dn}(z,k)$ [has Fourier series](http://dlmf.nist.gov/22.11)
$$\operatorname{dn}(z,k)=\frac{\pi}{2K}\left[1+2\sum_{n=1}^{\infty}\frac{\cos n\pi\frac{z}{K}}{\cosh n \pi \frac{K'}{K}}\right],$$
where $K(k)$ denotes complete elliptic integral and $K'(k)=K(\sqrt{1-k^2})$ the complementary one. The Ramanujan Cos/Cosh identity is thus equivalent to showing that
$$\operatorname{dn}^{-2}\left(\frac{K_1}{\pi}\theta,k_1\right)+\operatorname{dn}^{-2}\left(\frac{iK_1}{\pi}\theta,k_1\right)=\frac{8\Gamma^4\left(\frac34\right)K_1^2}{\pi^3},\tag{1}$$
where $k_1=\frac{1}{\sqrt2}$ is the first [elliptic integral singular value](http://mathworld.wolfram.com/EllipticIntegralSingularValue.html) and $K_1:=K(k_1)=K'(k_1)$.

The right hand side of (1) is independent on $\theta$ and is readily shown to be equal to $2$ using e.g. formula (3) from the [same page](http://mathworld.wolfram.com/EllipticIntegralSingularValue.html). Therefore it remains to show that for any $\sigma\in\mathbb{C}$ one has
$$\operatorname{dn}^{-2}\left(\sigma,k_1\right)+\operatorname{dn}^{-2}\left(i\sigma,k_1\right)=2.$$
I leave this last point to you as an exercise (hint: use [Jacobi's imaginary transformation](http://mathworld.wolfram.com/JacobisImaginaryTransformation.html)).

----------
Hopefully it is now clear that one can construct many generalizations of Ramanujan identity. Such constructions would involve two basic ingredients:

- Fourier series of elliptic functions,

- elliptic integral singular values.

Indeed, pick your favorite identity satisfied by the elliptic functions. The first ingredient will transform them into trigonometric series. The second one will allow to replace the elliptic modulus by algebraic numbers and the corresponding half-periods by misteriously-looking combinations of gamma functions of rational arguments.

----------
P.S. The first question is just Taylor expansion in $\theta$ (for instance, set $\theta=0$ in the Ramanujan identity and see what happens).

来源：https://math.stackexchange.com/questions/517409/extensions-of-ramanujans-cos-cosh-identity/955420#955420

---

I am Brian Diaz, and I am new to the math.stackexchange community.

I have been struggling with attempting to find a closed form of the following series:

$$ \varphi(\theta) = 1 + 2\sum_{n=1}^{\infty} \frac{\cosh(n\theta)}{\cosh(n\pi)} $$

Admittedly, I attempted to convert it to a "workable integral", but to no avail. Heck, in the process of converting it to an integral, I am not even sure interchanging the sum and the integral was valid. Nevertheless, this was my result.
$$\frac{1}{\pi}\int_{-\infty}^{\infty} \frac{\sin(x)}{\cosh(\theta) - \cos(x)} \frac{1}{\cosh(x)}dx $$

This was derived from a problem Ramanujan was working. For those who are interested in the source, you can visit http://mathworld.wolfram.com/RamanujanCosCoshIdentity.html. Note: Even if it does not have a closed form, I am still interested in valuable insight to the problem. In addition, I have been reported by my professor to consider applying residue theory, though he his not so sure what the result would be.

Thank you so much for your support, and I hope you do have a blessed day!

---

The closed form involves Jacobi elliptic function $\operatorname{dn}(z,k)$, which has [Fourier series](http://dlmf.nist.gov/22.11)
$$\operatorname{dn}(z,k)=\frac{\pi}{2K}\left[1+2\sum_{n=1}^{\infty}\frac{\cos n\pi\frac{z}{K}}{\cosh n \pi \frac{K'}{K}}\right],$$
where $K(k)$ denotes complete elliptic integral and $K'(k)=K(\sqrt{1-k^2})$ the complementary one.

Now if we denote $k_1=\frac{1}{\sqrt2}$ the first [elliptic integral singular value](http://mathworld.wolfram.com/EllipticIntegralSingularValue.html) and
$$K_1:=K(k_1)=K'(k_1)=\frac{\Gamma^2\left(\frac14\right)}{4\sqrt{\pi}},$$
the sum can be expressed as
$$\boxed{\displaystyle \quad \varphi\left(\theta\right):=1+2\sum_{n=1}^{\infty}\frac{\cosh n\theta}{\cosh n \pi}=\frac{2K_1}{\pi}\,\operatorname{dn}\left(\frac{iK_1\theta}{\pi},k_1\right)\quad}$$

**P.S.** To check the answer with Mathematica, note that the latter uses $k^2$ instead of $k$ in the arguments of $\mathrm{EllipticK[}\cdot\mathrm{]}$ and $\mathrm{JacobiDN[}z,\cdot\mathrm{]}$. For example, $K_1$ is evaluated with $\mathrm{EllipticK[}\frac12\mathrm{]}$.

**P.P.S** This transforms the proof of Ramanujan cos/cosh identity into a one-line calculation involving [Jacobi imaginary transformation](http://mathworld.wolfram.com/JacobisImaginaryTransformation.html) for $\operatorname{dn}(z,k)$, as explained [here](https://math.stackexchange.com/a/955420/73025).

来源：https://math.stackexchange.com/questions/946071/a-problem-of-ramanujans-interest-closed-form-of-1-2-sum-n-1-infty-fra

---

 

Rather than relying on the consequences of Schanuel's conjecture, I set about using the same ideas Apery had used to construct integer arguments converging fast enough to show $\zeta(3)$ is irrational in a form Beukers had introduced. I'm sure someone out there can crack what I have so far.

 

I will be using the following facts:

 

>**Theorem 1**: Suppose the complex-valued function $$\begin{align}
f{(z)} = \begin{cases}
-\left(\frac{1}{e}(1-z)^{1-\frac{1}{z}} \right)^q, & z\neq 0 \\
-1, & z = 0
\end{cases}
\end{align}$$ has a power series with positive radius of convergence of the form
$$f(z) = \sum_{n=0}^\infty b_n(q) z^n$$
Then
$$b_n(q) = -\frac{q}{n} \sum_{k=1}^n \frac{b_{n-k}(q)}{k+1}, \quad b_0(q) = -1$$

 

Note that $b_n(q)$ is a polynomial of degree $n$.
>**Theorem 2**: Let $m,q \in \mathbb{Z}$ and $m+q+1 \geq 0$; then $$\int_0^1 x^m \sin(\pi q x) \left(x^x (1-x)^{1-x}\right)^q \ dx = (-1)^{q+1} \pi e^q b_{m+q+1}(q)$$

 

The above can be shown by applying contour integration and residue theorem to the above function.

 

> **Theorem 3**: For $n \in \mathbb{N} \cup \{0\}$, $\mathbb{P}$ be the set of primes, and let $$(n+1)!_\mathbb{P} = \prod_{p \in \mathbb{P}} p ^{\sum_{k\geq 0} \left\lfloor \frac{n}{(p-1)p^k} \right\rfloor}$$
> Then, for integer $q$, $(n+1)!_\mathbb{P} \cdot b_n(q)$ is an integer for $n \geq 0$.

 

This factorial like function is borrowed from Manjul Bhargava's work on the general factorial function.
>**Theorem 4** Let $n \in \mathbb{N} \cup \{0\}$; then $$(n+1)!_\mathbb{P}\sim e^{n(C-\gamma+o(1))}n^n$$ where $C = \sum_{p \in \mathbb{P}} \frac{\ln p }{(p-1)^2}$ and $\gamma$ is the Euler-Mascheroni constant.

 

If we let $P_n(x)$ be a polynomial of degree $n$ with integer coefficients and let $$I_n = \int_0^1 P_n(x) \left( \sin(\pi q x) \left( x^x (1-x)^{1-x}\right)^q -1\right) \ dx$$

 

We have the following inequality, in the form of Dirichlet's irrationality criterion,

 

$$0 < \left|C_n \pi e^q - D_n \right| = \left|(n+q+2)!_\mathbb{P} (n+1) I_n \right|$$

 

where $C_n, D_n \in \mathbb{Z}$. Of course, we can apply Theorem 4, and have something more familiar to work with.

 

>Question: Can we construct a polynomial $P_n(x)$ such that, for large $n$, $$\left|(n+q+2)!_\mathbb{P} (n+1) I_n \right| \to 0 \text{?}$$

 

If there does exist one, then, for $q \geq -2, q \neq 0$, the number $\pi e^q$ is irrational. Letting $q = 1, -1$, and the result follows.

 

I've been at this problem for some time, with no further progress. Frankly, I don't know what to do at all. If it helps, I've considered the shifted Legendre polynomials, as Beukers had done, though to no avail.

 

Most of what I've seen regarding the nature of constructing a polynomial is that it belongs to the family of *orthogonal polynomials*.

 

God bless.

---

This isn't really an answer as much as it is an "expanded" comment.

Consider, for integer $a$, $$P_n(x) = \frac{1}{n!} \frac{d^n}{d x^n} x^n (1-ax)^n = \sum_{m=0}^n \binom{n}{m} \binom{n+m}{m} (-ax)^m$$

 

Given $$I_n = \int_0^1 P_n(x) \left( \sin(\pi q x) (x^x(1-x)^{1-x})^q - 1\right) \ dx$$
We have
$$I_n \leq \int_0^1 P_n(x) \left( \sin(\pi q x) a_q - 1\right) \ dx$$ where $a_q = \max_{x\in (0,1)}\{(x^x(1-x)^{1-x})^q\}$. Furthermore, we have
$$\left|\int_0^1 P_n(x) \left( \sin(\pi q x) a_q - 1\right) \ dx \right|= \left|\int_0^1 \frac{1}{n!} \frac{d^n}{d x^n} x^n (1-ax)^n\left( \sin(\pi q x) a_q - 1\right) \ dx\right|$$

$$= \left|\frac{1}{a^{n+1}n!} \int_{(0,1)\cup(1,a)} \frac{d^n}{d x^n} x^n (1-x)^n\left( \sin\left(\frac{\pi q x}{a}\right) a_q - 1\right) \ dx\right| $$
$$\leq \left|\left(\frac{\pi q}{4a^2}\right)^n \frac{a_q}{n! a}+\frac{1}{n!a^{n+1}}\int_1^a \frac{d^n}{d x^n} x^n (1-x)^n\left( \sin\left(\frac{\pi q x}{a}\right) a_q - 1\right) \ dx\right| $$

Let $$S_n = \int_1^a \left( \sin\left(\frac{\pi q x}{a}\right) a_q - 1\right)\frac{d^n}{d x^n} x^n (1-x)^n \ dx$$

So that we have
$$ = \left|\left(\frac{\pi q}{4a^2}\right)^n \frac{a_q}{n! a}+\frac{S_n}{n!a^{n+1}}\right|$$

Now, observing the bound in question, applying Theorem 4, and letting $A = C - \gamma + o(1)$, we have
$$\left| (n+q+2)!_\mathbb{P} (n+1) I_n \right|<\left|e^{An} e^{q+1} n^n \left(1+\frac{q+1}{n}\right)^n \left(1+\frac{q+1}{n}\right)^{q+1}n^{q+1}(n+1)\left(\left(\frac{\pi q}{4a^2}\right)^n \frac{a_q}{n! a}+\frac{S_n}{n!a^{n+1}}\right) \right| $$

If we ignore the $S_n$ term, we have that

$$\left| e^{q+1} \frac{n^n}{e^n n!} \left(1+\frac{q+1}{n}\right)^n \left(1+\frac{q+1}{n}\right)^{q+1}\frac{n^{q+1}(n+1)}{b^n}\left(\frac{e^{A+1}\pi bq}{4a^2}\right)^n \frac{a_q}{a} \right|$$

where $b > 1 $. If we consider $a$ such that $ a^2 > \frac{e^{A+1}\pi bq}{4}$, and applying Stirling's approximation to the left-most term (-ish), for large $n$, then the whole expression above tends to $0$. Now, it is left to consider the $S_n$ term, though I have a bad feeling about it. :/

来源：https://mathoverflow.net/questions/226875/proving-the-irrationality-of-pi-e-and-pi-e