椭圆积分

For the elliptic integral of first kind, $K(m)=\int_0^{\pi/2}\frac{d\theta}{\sqrt{1-m^2sin^2\theta}}$, it is well-known that $K(m)$ can be expressed in what Chowla and Selberg call "finite terms" (i.e. algebraic numbers and a finite product of Gamma functions of rational values) whenever $i\frac{K(\sqrt{1-m^2})}{K(m)}$ belongs to an imaginary quadratic field $\mathbb Q(\sqrt{d})$
(see Theorem 7 in S. Chowla and A. Selberg, [On Epstein's zeta function, J. reine angew. Math. 227, 86-110, 196][1]).

Examples for these so called **elliptic integral singular values** are given on [this Wolfram page][2] (with some small typos) and in the note of J.M. Borwein and I.J. Zucker, "Elliptic integral evaluation of the Gamma function at
rational values of small denominator," [IMA Journal on Numerical Analysis, 12 (1992), 519-
526][3].

See also what Tito Piezas has to say about this in his pleasant-to-read [Collection of Algebraic Identities][4].

The following question arises:

- **For these singular values, is there** (always, or, if not always: when?) **a polynomial $P(t)$ of degree 3 with *integer* coefficients such that $K(m)=c\int\limits_{t_0}^\infty\dfrac{dt}{\sqrt{P(t)}}$ with $c\in\mathbb Q$?**
(EDIT: After Noam Elkies' remark, introduced $t_0$, the biggest real zero of $P$, instead of $0$ as the lower limit. Only "complete" integrals make sense here.)

In particular for $d=-7$, we have by the [Carlson symmetric form][5] $$\frac12\int\limits_0^\infty\dfrac{dt}{\sqrt{t(t+1)(t+\frac{8+3\sqrt 7}{16})}} =K(k_7)=\dfrac1{7^{1/4}4\pi}\Gamma\left(\dfrac17\right)\Gamma\left(\dfrac27\right)\Gamma\left(\dfrac47\right),$$ on the other hand I have seen somewhere (I can't remember the reference) $$\int\limits_0^\infty\dfrac{dt}{\sqrt{t(t^2+21t+112)}} =\dfrac1{4\pi\sqrt{7}}\Gamma\left(\dfrac17\right)\Gamma\left(\dfrac27\right)\Gamma\left(\dfrac47\right)=\frac{K(k_7)}{7^{1/4}}.$$

I would like to get it straight at least for this example:

- **Can the polynomial in the first integral be transformed into one with integer coefficients? And is there any sort of relationship between both above polynomials?**
Note that the ratio of the discriminants of the two above polynomials is $-2^{24}\cdot7^3$, and both of them do *not* yield affirmative answers, as the second one would have to be divided by $\sqrt7$ to obtain $K(k_7)$ directly!

- **EDIT: Follow-up question: If $P(t)$ is an integer cubic polynomial such that $\int\limits_{t_0}^\infty\dfrac{dt}{\sqrt{P(t)}}$ (with $t_0$ its biggest real zero) can be written in "finite terms", is this value always an algebraic multiple of an elliptic integral singular value $K(m)$?**

[1]: http://www.ncbi.nlm.nih.gov/pmc/articles/PMC1063041/pdf/pnas01544-0031.pdf
[2]: http://mathworld.wolfram.com/EllipticIntegralSingularValue.html
[3]: http://carma.newcastle.edu.au/jon/Preprints/Books/EMA/Exercises/For%20others/K-beta.pdf
[4]: https://www.sites.google.com/site/tpiezas/0026
[5]: http://en.wikipedia.org/wiki/Carlson_symmetric_form

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The flurry of comments did not yet produce an answer to the question
concerning the complete elliptic integrals
$$I_1 := \frac12 \int\limits_0^\infty\dfrac{dt}{\sqrt{t(t+1)(t+\frac{8+3\sqrt{7}}{16})}}$$
and
$$I_2 := \int\limits_0^\infty\dfrac{dt}{\sqrt{t(t^2+21t+112)}}.$$
It turns out that (i) Yes, $I_1$ can be transformed to a complete
elliptic integral associated to a cubic with integer coefficients, and
(ii) The identity $I_1 = 7^{1/4} I_2$ can then be recovered via
a form of
<a href="http://en.wikipedia.org/wiki/Landen's_transformation">Landen's transformation</a>.

This could be surmised by calculating the $j$-invariants of the
corresponding elliptic curves
$$y^2 = x(x+1)(x+\frac{8+3\sqrt{7}}{16}) \phantom{\cong}\text{and}\phantom{\cong} y^2 = x(x^2+21x+112).$$
The first $j$-invariant is $j_1 = 16581375 = 255^3$, which is rational,
so there's a linear change of variable that transforms
$x(x+1)(x+\frac{8+3\sqrt{7}}{16})$ to a polynomial with integer coefficients.
The second $j$-invariant is $j_2 = -3375 = -15^3 \neq j_1$, so we can't
get immediately from $j_1$ to $j_2$. But $j_1$ and $j_2$ are still
related by a $2$-isogeny [indeed $j_1 = j(\sqrt{-7})$ and
$j_2 = j((1+\sqrt{-7})/2)$], so $I_1$ and $I_2$ are related by
a Landen transformation.

For (i), first translate $t$ by $(8+3\sqrt{7})/16$ to get
$$I_1 := \frac12 \int\limits_{\frac{8+3\sqrt{7}}{16}}^\infty \dfrac{dt}{\sqrt{t \bigl(t-\frac{8+3\sqrt{7}}{16}\bigr) \bigl(t-\frac{-8+3\sqrt{7}}{16}\bigr) }} \phantom{0}.$$
Then observe that
$$\bigl(t-\frac{8+3\sqrt{7}}{16}\bigr) \bigl(t-\frac{-8+3\sqrt{7}}{16}\bigr) = t^2 - \frac{3\sqrt{7}}{8} + \frac{3^2 7 - 8^2}{16^2} = t^2 - \frac{6\sqrt{7}}{16} - \frac1{16^2}.$$
Thus the change of variable $x = 16 \sqrt{7} \cdot t$ yields
$$I_1 = 2 \cdot 7^{1/4} \int\limits_{21+8\sqrt{7}}^\infty \dfrac{dx}{\sqrt{x (x^2-42x-7)}}$$
with the integrand having integer coefficients.

For (ii), we apply Landen's change of variable for elliptic integrals
$\int dt/\sqrt{t(t^2+at+b)}$ with $a,b>0$: if $x = (t^2+at+b)/t$ then
$dt/\sqrt{t(t^2+at+b)} = dx/\sqrt{x(x^2+Ax+B)}$ where $(A,B) = (-2a, a^2-4b)$.
Moreover the map $t \mapsto (t^2+at+b)/t$ maps the interval $0 < t < \infty$
to $a + 2\sqrt{b} \leq x < \infty$, with each $x$ value arising twice
except for the left endpoint $x = a+2\sqrt{b}$ which has a single preimage
$t = \sqrt{b}$ of multiplicity $2$. Therefore
$$\int\limits_0^\infty\dfrac{dt}{\sqrt{t(t^2+at+b)}} = 2 \int\limits_{a+2\sqrt{b}}^\infty\dfrac{dx}{\sqrt{x(x^2-AX-B)}}.$$
The $I_2$ coefficients $(a,b) = (21,112)$ yield $(A,B) = (-42,-7)$ and
$a + 2 \sqrt{b} = 21 + 8 \sqrt{7}$. Therefore $I_1 = 7^{1/4} I_2$ as desired.

 来源：这里

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For the first one,
$$\begin{align} \int_0^{\infty} (\operatorname{sech}x)^{2s}dx \\ &= \int_0^{\infty} (\operatorname{sech}^2x)^{s-1}\operatorname{sech}^2x\, dx \\ &= \int_0^{\infty} (1-\tanh^2x)^{s-1}\,\mathrm{d}(\tanh x)\\ &= \int_0^1 (1-x^2)^{s-1} \mathrm{d}x\\ &= \frac12 \int_0^1 (1-x)^{s-1} x^{-\frac12} \mathrm{d}x\\ &= \frac12 B(s,\frac12)=\frac{\sqrt{\pi}}{2}\frac{\Gamma(s)}{\Gamma(\frac12+s)} \end{align}$$
and so your conjecture is correct.
For the second integral, let $x=2t$:
$$\begin{align} \int_0^{\infty} \frac{1}{(1+\cosh x)^s}\mathrm{d}x \\ &= 2\int_0^{\infty} \frac{1}{(1+\cosh 2t)^s}\mathrm{d}t \\ &= 2\int_0^{\infty} \frac{1}{(2\cosh^2(t))^s}\mathrm{d}t\\ &= 2^{1-s} \int_0^{\infty} (\operatorname{sech}t)^{2s}\mathrm{d}t\\ &= 2^{-s} B(s,\frac12)=\frac{\sqrt{\pi}}{2^s}\frac{\Gamma(s)}{\Gamma(\frac12+s)} \end{align}$$

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Extending **@nospoon**'s idea, we notice that

$$a + \cosh 2x = (a+1)\cosh^2 x (1 - b \tanh^2 x), \qquad b =\frac{a-1}{a+1}.$$

If $a > -1$, then $b < 1$ and using the substitution $u = \tanh^2 x$ we get

$$\int_{0}^{\infty} \frac{dx}{(a + \cosh x)^{s}} \, dx = \frac{1}{(a+1)^{s}} \int_{0}^{1} \frac{(1 - u)^{s-1}}{(1 - b u)^s\sqrt{u}} \, du.$$

Making further substitution $v = \frac{1-u}{1-bu}$, we have

$$\int_{0}^{\infty} \frac{dx}{(a + \cosh x)^{s}} \, dx = \frac{1}{(a+1)^{s}} \int_{0}^{1} \frac{v^{s-1}}{\sqrt{(1-v)(1-bv)}} \, dv.$$

This is easily integrated when $b = 0$ or $b = -1$, each correspondingto $a = 1$ or $a = 0$, but I doubt that this integral has a nice closed form in general. For example, when $s = 1/2$, this becomes an elliptic integral and *Mathematica 11* gives

$$\int_{0}^{1} \frac{dv}{\sqrt{v(1-v)(1-bv)}} = 2 \left[ \frac{1}{\sqrt{b}} K \left( \sqrt{\tfrac{1}{b}} \right) + i K \left( \sqrt{1-b} \right) \right], \quad 0 < b < 1,$$

where $K(k)$ is the [*complete elliptic integral of the 1st kind*](https://en.wikipedia.org/wiki/Elliptic_integral#Complete_elliptic_integral_of_the_first_kind).

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> $$I:=\int_{0}^{\infty}\frac{\ln{(x)}}{\sqrt{x}\,\sqrt{x+1}\,\sqrt{2x+1}}\mathrm{d}x.$$

After first multiplying and dividing the integrand by 2, substitute $x=\frac{t}{2}$:

$$I=\int_{0}^{\infty}\frac{2\ln{(x)}}{\sqrt{2x}\,\sqrt{2x+2}\,\sqrt{2x+1}}\mathrm{d}x=\int_{0}^{\infty}\frac{\ln{\left(\frac{t}{2}\right)}}{\sqrt{t}\,\sqrt{t+2}\,\sqrt{t+1}}\mathrm{d}t.$$

Next, substituting $t=\frac{1}{u}$ yields:

$$\begin{align} I &=-\int_{0}^{\infty}\frac{\ln{(2u)}}{\sqrt{u}\sqrt{u+1}\sqrt{2u+1}}\mathrm{d}u\\ &=-\int_{0}^{\infty}\frac{\ln{(2)}}{\sqrt{u}\sqrt{u+1}\sqrt{2u+1}}\mathrm{d}u-\int_{0}^{\infty}\frac{\ln{(u)}}{\sqrt{u}\sqrt{u+1}\sqrt{2u+1}}\mathrm{d}u\\ &=-\int_{0}^{\infty}\frac{\ln{(2)}}{\sqrt{u}\sqrt{u+1}\sqrt{2u+1}}\mathrm{d}u-I\\ \implies I&=-\frac{\ln{(2)}}{2}\int_{0}^{\infty}\frac{\mathrm{d}x}{\sqrt{x}\sqrt{x+1}\sqrt{2x+1}}. \end{align}$$

Making the sequence of substitutions $x=\frac{u-1}{2}$, then $u=\frac{1}{t}$, and finally $t=\sqrt{w}$, puts this integral into the form of a beta function:

$$\begin{align} \int_{0}^{\infty}\frac{\mathrm{d}x}{\sqrt{x}\sqrt{x+1}\sqrt{2x+1}} &=\int_{1}^{\infty}\frac{\mathrm{d}u}{\sqrt{u-1}\sqrt{u+1}\sqrt{u}}\\ &=\int_{1}^{\infty}\frac{\mathrm{d}u}{\sqrt{u^2-1}\sqrt{u}}\\ &=\int_{1}^{0}\frac{t^{3/2}}{\sqrt{1-t^2}}\frac{(-1)}{t^2}\mathrm{d}t\\ &=\int_{0}^{1}\frac{\mathrm{d}t}{\sqrt{t}\,\sqrt{1-t^2}}\\ &=\frac12\int_{0}^{1}\frac{\mathrm{d}w}{w^{3/4}\,\sqrt{1-w}}\\ &=\frac12\operatorname{B}{\left(\frac14,\frac12\right)}\\ &=\frac12\frac{\Gamma{\left(\frac12\right)}\Gamma{\left(\frac14\right)}}{\Gamma{\left(\frac34\right)}}\\ &=\frac{\pi^{3/2}}{2^{1/2}\Gamma^2{\left(\frac34\right)}} \end{align}$$

Hence,

$$I=-\frac{\ln{(2)}}{2}\frac{\pi^{3/2}}{2^{1/2}\Gamma^2{\left(\frac34\right)}}=-\frac{\pi^{3/2}\,\ln{(2)}}{2^{3/2}\,\Gamma^2{\left(\frac34\right)}}.~~~\blacksquare$$

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**Possible Alternative:** You could also derive the answer from the complete elliptic integral of the first kind instead of from the beta function by making the substitution $t=z^2$ instead of $t=\sqrt{w}$.

$$\begin{align} \int_{0}^{\infty}\frac{\mathrm{d}x}{\sqrt{x}\sqrt{x+1}\sqrt{2x+1}} &=\int_{1}^{\infty}\frac{\mathrm{d}u}{\sqrt{u-1}\sqrt{u+1}\sqrt{u}}\\ &=\int_{1}^{\infty}\frac{\mathrm{d}u}{\sqrt{u^2-1}\sqrt{u}}\\ &=\int_{1}^{0}\frac{t^{3/2}}{\sqrt{1-t^2}}\frac{(-1)}{t^2}\mathrm{d}t\\ &=\int_{0}^{1}\frac{\mathrm{d}t}{\sqrt{t}\,\sqrt{1-t^2}}\\ &=2\int_{0}^{1}\frac{\mathrm{d}z}{\sqrt{1-z^4}}\\ &=2\,K{(-1)}\\ &=\frac{\Gamma^2{\left(\frac14\right)}}{2\sqrt{2\pi}}\\ &=\frac{\pi^{3/2}}{2^{1/2}\Gamma^2{\left(\frac34\right)}}. \end{align}$$

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$$\int_0^1\frac{\mathrm dx}{\sqrt{1-x}\ \sqrt[4]x\ \sqrt[4]{2-x\,\sqrt3}}\stackrel?=\frac{2\,\sqrt2}{3\,\sqrt[8]3}\pi\tag1$$
The equality numerically holds up to at least $10^4$ decimal digits.
> Can we prove that the equality is exact?

An equivalent form of this conjecture is
$$I\left(\frac{\sqrt3}2;\ \frac14,\frac14\right)\stackrel?=\frac23,\tag2$$
where $I\left(z;\ a,b\right)$ is the [regularized beta function](http://mathworld.wolfram.com/RegularizedBetaFunction.html).

---
Even simpler case:
$$\int_0^1\frac{\mathrm dx}{\sqrt{1-x}\ \sqrt[6]{9-x}\ \sqrt[3]x}\stackrel?=\frac\pi{\sqrt3},\tag3$$
which is equivalent to
$$I\left(\frac19;\ \frac16,\frac13\right)\stackrel?=\frac12.\tag4$$

---
A [related question](https://math.stackexchange.com/questions/538564/conjecture-int-01-fracdx-sqrt3x-sqrt61-x-sqrt1-x-left-sqrt6).

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For $\alpha, \beta, \gamma \in (0,1)$ satisfying $\alpha+\beta+\gamma = 1$ and
$\mu \in \mathbb{C} \setminus [1,\infty)$, define

$$F_{\alpha\beta}(\mu) = \int_0^1\frac{dx}{x^\alpha(1-x)^\beta(1-\mu x)^\gamma} \quad\text{ and }\quad \Delta = \frac{\Gamma(1-\alpha)\Gamma(1-\beta)}{\Gamma(1+\gamma)}$$
When $|\mu| < 1$, we can rewrite the integral $F_{\alpha\beta}(\mu)$ as

$$\begin{align} F_{\alpha\beta}(\mu) = & \int_0^1 \frac{1}{x^\alpha(1-x)^{\beta}}\left(\sum_{n=0}^{\infty}\frac{(\gamma)_n}{n!}\mu^n x^n\right) dx = \sum_{n=0}^{\infty}\frac{(\gamma)_n}{n!}\frac{\Gamma(n+1-\alpha)\Gamma(1-\beta)}{\Gamma(n+1+\gamma)}\mu^n\\ = & \Delta\sum_{n=0}^{\infty}\frac{(\gamma)_n (1-\alpha)_n}{n!(\gamma+1)_n}\mu^n = \Delta\gamma \sum_{n=0}^{\infty}\frac{(1-\alpha)_n}{n!(\gamma+n)}\mu^n \end{align}$$
This implies
$$\mu^{-\gamma} \left(\mu\frac{\partial}{\partial \mu}\right) \mu^{\gamma} F_{\alpha\beta}(\mu) = \Delta\gamma \sum_{n=0}^{\infty}\frac{(1-\alpha)_n}{n!}\mu^n = \Delta\gamma\frac{1}{(1-\mu)^{1-\alpha}}$$
and hence
$$F_{\alpha\beta}(\mu) = \Delta\gamma \mu^{-\gamma} \int_0^\mu \frac{\nu^{\gamma-1}d\nu}{(1-\nu)^{1-\alpha}} = \Delta\gamma \int_0^1 \frac{t^{\gamma-1} dt}{(1-\mu t)^{1-\alpha}} = \Delta \int_0^1 \frac{dt}{(1 - \mu t^{1/\gamma})^{1-\alpha}}$$

Notice if we substitute $x$ by $y = 1-x$, we have

$$F_{\alpha\beta}(\mu) = \int_0^1 \frac{dy}{y^\beta(1-y)^\alpha(1-\mu - \mu y)^{\gamma}} = \frac{1}{(1-\mu)^\gamma} F_{\beta\alpha}(-\frac{\mu}{1-\mu})$$

Combine these two representations of $F_{\alpha\beta}(\mu)$ and let $\omega = \left(\frac{\mu}{1-\mu}\right)^{\gamma}$, we obtain

$$F_{\alpha\beta}(\mu) = \frac{\Delta}{(1-\mu)^{\gamma}}\int_0^1 \frac{dt}{( 1 + \omega^{1/\gamma} t^{1/\gamma})^{1-\beta}} = \frac{\Delta}{\mu^\gamma}\int_0^\omega \frac{dt}{(1 + t^{1/\gamma})^{1-\beta}}$$

Let $(\alpha,\beta,\gamma) = (\frac14,\frac12,\frac14)$ and $\mu = \frac{\sqrt{3}}{2}$, the identity we want to check becomes

$$\frac{\Gamma(\frac34)\Gamma(\frac12)}{\Gamma(\frac54) (\sqrt{3})^{1/4}}\int_0^\omega \frac{dt}{\sqrt{1+t^4}} \stackrel{?}{=} \frac{2\sqrt{2}}{3\sqrt[8]{3}} \pi\tag{*1}$$

Let $K(m)$ be the complete elliptic integral of the first kind associated with modulus $m$. i.e.

$$K(m) = \int_0^1 \frac{dx}{\sqrt{(1-x^2)(1-mx^2)}}$$
It is known that $\displaystyle K(\frac12) = \frac{8\pi^{3/2}}{\Gamma(-\frac14)^2}$. In term of $K(\frac12)$, it is easy to check $(*1)$ is equivalent to

$$\int_0^\omega \frac{dt}{\sqrt{1+t^4}} \stackrel{?}{=} \frac23 K(\frac12)\tag{*2}$$

To see whether this is the case, let $\varphi(u)$ be the inverse function of above integral.
More precisely, define $\varphi(u)$ by following relation:

$$u = \int_0^{\varphi(u)} \frac{dt}{\sqrt{1+t^4}}$$

Let $\psi(u)$ be $\frac{1}{\sqrt{2}}(\varphi(u) + \varphi(u)^{-1})$. It is easy to check/verify
$$\varphi'(u)^2 = 1 + \varphi(u)^4 \implies \psi'(u)^2 = 4 (1 - \psi(u)^2)(1 - \frac12 \psi(u)^2)$$

Compare the ODE of $\psi(u)$ with that of a Jacobi elliptic functions with modulus $m = \frac12$, we find

$$\psi(u) = \text{sn}(2u + \text{constant} | \frac12 )\tag{*3}$$

Since we are going to deal with elliptic functions/integrals with $m = \frac12$ only,
we will simplify our notations and drop all reference to modulus, i.e
$\text{sn}(u)$ now means $\text{sn}(u|m=\frac12)$ and $K$ means $K(m = \frac12)$.

Over the complex plane, it is known that $\text{sn}(u)$ is doubly periodic with
fundamental period $4 K$ and $2i K$. It has two poles at $i K$ and $(2 + i)K$ in the fundamental domain.
When $u = 0$, we want $\varphi(u) = 0$ and hence $\psi(u) = \infty$. So the constant
in $(*3)$ has to be one of the pole. For small and positive $u$, we want $\varphi(u)$ and hence $\psi(u)$ to be positive. This fixes the constant to $i K$. i.e.

$$\psi(u) = \text{sn}(2u + iK )$$

and the condition $(*2)$ becomes whether following equality is true or not.

$$\frac{1}{\sqrt{2}} (\omega + \omega^{-1}) \stackrel{?}{=} \text{sn}( \frac43 K + i K)\tag{*4}$$

Notice $3( \frac43 K + i K) = 4 K + 3 i K$ is a pole of $\text{sn}(u)$. if one repeat
apply the addition formula for $\text{sn}(u+v)$

$$\text{sn}(u+v) = \frac{\text{sn}(u)\text{cn}(v)\text{dn}(v)+\text{sn}(v)\text{cn}(u)\text{dn}(u)}{1-m\,\text{sn}(u)^2 \text{sn}(v)^2}$$

One find in order for $\text{sn}(3u)$ to blow up, $\text{sn}(u)$ will be a root of
following polynomial equation:
$$3 m^2 s^8-4 m^2 s^6-4 m s^6+6 m s^4-1 = 0$$
Substitute $m = \frac12$ and $s = \frac{1}{\sqrt{2}}(t+\frac{1}{t})$ into this, the equation $\omega$ need to satisfy is given by:

$$(t^8 - 6 t^4 - 3)(3 t^8 + 6 t^4 - 1 ) = 0$$

One can check that $\omega = \sqrt[4]{\frac{\sqrt{3}}{2-\sqrt{3}}}$ is indeed a root of this polynomial. As a result, the original equality is valid.

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Here is another approach for evaluating the integral (3).

Using **@achille hui**'s transformation, we can write

$$I=\int_0^1\frac{dx}{\sqrt[3]{x}\sqrt{1-x}\sqrt[6]{9-x}}= \frac{4\pi^2 2^{\frac{1}{3}}}{\Gamma\left(\frac{1}{3}\right)^3}\int_0^\frac{1}{\sqrt{2}}\frac{1}{\sqrt{1+x^6}}dx$$
Using the substitution $x=\sqrt{t}$, we get
$$I=\frac{\pi^2 2^{\frac{4}{3}}}{\Gamma\left(\frac{1}{3}\right)^3}\int_0^\frac{1}{2}\frac{1}{\sqrt{t+t^4}}dt$$
We can now transform this integral into a beta integral using the substitution $t=\frac{1-y}{2+y}$. This gives us

$$\begin{align*} I &= \frac{\pi^2 2^{\frac{4}{3}}}{\Gamma\left(\frac{1}{3}\right)^3} \int_0^1\frac{1}{\sqrt{1-y^3}}dy \\ &= \frac{\pi^2 2^{\frac{4}{3}}}{3\Gamma\left(\frac{1}{3}\right)^3}B\left(\frac{1}{3},\frac{1}{2} \right) \\ &= \frac{\pi^2 2^{\frac{4}{3}}}{3\Gamma\left(\frac{1}{3}\right)^3}\left( \frac{\sqrt{3}\Gamma\left(\frac{1}{3}\right)^3}{\pi 2^{\frac{4}{3}}}\right) \\ &= \frac{\pi}{\sqrt{3}} \end{align*}$$

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Let
$$\alpha=\sqrt{6}\ \sqrt{12+7\,\sqrt3}-3\,\sqrt3-6\,=\,\big(2+\sqrt{3}\big) \big(\sqrt{2} \sqrt[4]{27}-3\big)\,=\,\frac{3\sqrt{3}}{3+\sqrt2\ \sqrt[4]{27}}.\tag1$$
Note that $\alpha$ is the unique positive root of the polynomial equation
$$\alpha^4+24\,\alpha^3+18\,\alpha^2-27=0.\tag2$$
Now consider the following integral:
$$I=\int_0^1\frac{dx}{\sqrt[3]x\ \sqrt[6]{1-x}\ \sqrt{1-x\,\alpha^2}}.\tag3$$
I have a conjectured elementary value for it
$$I\stackrel?=\frac\pi9\Big(3+\sqrt2\ \sqrt[4]{27}\Big)=\color{blue}{\frac{\pi}{\sqrt{3}\,\alpha}}.\tag4$$

---
Actually, the integral $I$ can be evaluated in an exact form using _Mathematica_ or manually, using formula [DLMF 15.6.1](http://dlmf.nist.gov/15.6):
$$I=\frac{4\,\sqrt\pi}{\sqrt3}\cdot\frac{\Gamma\left(\frac56\right)}{\Gamma\left(\frac13\right)}\cdot{_2F_1}\left(\frac12,\frac23;\ \frac32;\ \alpha^2\right),\tag5$$
but I could not find a way to simplify this result to $(4)$.

So, my conjecture can be restated in a different form:
$${_2F_1}\left(\frac12,\frac23;\ \frac32;\ \alpha^2\right)\stackrel?=\frac{\sqrt2+\sqrt[4]3}{4\,\sqrt[4]{27}}\cdot\frac{\Gamma\left(\frac13\right)}{\Gamma\left(\frac56\right)}\cdot\sqrt\pi\tag6$$
or
$$\sum_{n=0}^\infty\frac{\Gamma\left(n+\frac23\right)}{(2\,n+1)\,\Gamma(n+1)}\alpha^{2\,n}\stackrel?=\frac{3+\sqrt2\,\sqrt[4]{27}}{18}\cdot\frac{\pi^{3/2}}{\Gamma\left(\frac56\right)}.\tag7$$

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The conjecture can also be given in terms of the [incomplete beta function](http://mathworld.wolfram.com/IncompleteBetaFunction.html):
$$B\left(\alpha^2;\ \frac12,\frac13\right)\stackrel?=\frac{\sqrt\pi}2\cdot\frac{\Gamma\left(\frac13\right)}{\Gamma\left(\frac56\right)}.\tag8$$

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> _Question:_ Is the conjecture indeed true?

_Note:_ It holds numerically up to at least $10^4$ decimal digits.

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__Conjecture 2__

Let
$${\small \begin{multline} \beta=\frac{21}4+\frac{9\,\sqrt5}4-\frac{15}8\sqrt{750-330\,\sqrt5}-\frac{33}8\sqrt{150-66\,\sqrt5} \\ + \frac12\sqrt{3\left(165+75\,\sqrt5-46\,\sqrt{750-330\,\sqrt5}-103\,\sqrt{150-66\,\sqrt5}\right)}. \end{multline}}\tag9$$
_Added later:_ We can simplify it to
$$\small\beta=\frac 3 4 \left(7+3 \sqrt 5-\sqrt[4] 5 \sqrt{66+30 \sqrt 5}\right)+\frac 1 2 \sqrt{495+225 \sqrt 5-3 \sqrt{6 \big(8545+3821 \sqrt 5\big)}}.\tag{$9'$}$$

I conjecture that:
$$\int_0^1\frac{dx}{\sqrt[3]x\ \sqrt[6]{1-x}\ \sqrt{1-x\,\beta^2}}\stackrel?=\color{blue}{\frac{2\,\pi}{5\,\sqrt3\,\beta}}.\tag{10}$$

I can imagine that if these conjectures are true, then there are generalizations for some other algebraic numbers.

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Let $t = 1 + y^3$, we can rewrite $B\left(\alpha^2; \frac12, \frac13 \right)$ as

$$\int_0^{\alpha^2} t^{-1/2} (1-t)^{-2/3} dt = \int_{-1}^{-\sqrt[3]{1-\alpha^2}} \frac{3y^2dy}{\sqrt{1+y^3} y^2} = 3 \int_{-1}^{-\sqrt[3]{1-\alpha^2}} \frac{dy}{\sqrt{1+y^3}}$$

Following the setup in my [answer](https://math.stackexchange.com/a/692107/59379) to a related question. Let
$\;\displaystyle\eta = \frac{\Gamma\left(\frac13\right)\Gamma\left(\frac16\right)}{\sqrt{3\pi}}\;$ and $\wp(z)$ be the [Weierstrass elliptic $\wp$ function](http://en.wikipedia.org/wiki/Weierstrass%27s_elliptic_function) with fundamental periods $1$ and $e^{i\pi/3}$. $\wp(z)$ is known to satisfy an ODE of the form
$$\wp'(z)^2 = 4 \wp(z)^3 - g_2 \wp(z) - g_3\quad\text{ where }\quad g_2 = 0 \;\text{ and }\;g_3 = \frac{\eta^6}{16}$$
If one perform variable substitution $\;\displaystyle y = -\frac{4}{\eta^2} \wp\left(\frac{iz}{\eta}\right)$, one has

$$\frac{dy}{\sqrt{1+y^3}} = -dz\quad\text{ and }\quad \begin{cases} y\left(\frac{\sqrt{3}\eta}{3}\right) = -\frac{4}{\eta^2}\wp\left(i\frac{\sqrt{3}}{3}\right) = 0\\ \\ y\left(\frac{\sqrt{3}\eta}{2}\right) = -\frac{4}{\eta^2}\wp\left(i\frac{\sqrt{3}}{2}\right) = -1 \end{cases}$$
Using this, we can express conjecture $(8)$ in terms of $y(\cdot)$ and/or $\wp(\cdot)$:
$$\begin{align} & B\left(\alpha^2; \frac12, \frac13 \right) \stackrel{?}{=} \frac{\sqrt{\pi}}{2}\frac{\Gamma\left(\frac13\right)}{\Gamma\left(\frac56\right)} = \frac{\sqrt{3}}{4}\eta\\ \iff & 3\left[y^{-1}(-1) - y^{-1}(-\sqrt[3]{1-\alpha^2})\right] \stackrel{?}{=} \frac{\sqrt{3}}{4}\eta\\ \iff & y^{-1}(-\sqrt[3]{1-\alpha^2}) \stackrel{?}{=} \frac{5\sqrt{3}}{12}\eta\\ \iff & \frac{4}{\eta^2}\wp\left(i\frac{5\sqrt{3}}{12}\right) \stackrel{?}{=} \sqrt[3]{1-\alpha^2} \end{align}$$
Let $u_0 = i\frac{\sqrt{3}}{3}$, $u_{-1} = i\frac{\sqrt{3}}{2}$ and $u = i\frac{5\sqrt{3}}{12} = \frac12(u_0 + u_{-1})$. Using the addition formula for $\wp$ function,
we have

$$\wp(2u) = \wp(u_0 + u_{-1}) = \frac14\left[\frac{\wp'(u_0)-\wp'(u_{-1})}{\wp(u_0)-\wp(u_{-1})}\right]^2 - \wp(u_0) - \wp(u_{-1})\\ =\frac14\left[\frac{-i\frac{\eta^3}{4} - 0}{0 - \frac{\eta^2}{4}}\right]^2 - 0 - \frac{\eta^2}{4} = -\frac{\eta^2}{2}$$
Using the duplication formula of $\wp$ function, we get
$$-\frac{\eta^2}{2} = \wp(2u) = \frac14\left(\frac{(6\wp(u)^2-\frac12 g_2)^2}{4\wp(u)^3-g_2\wp(u)-g_3}\right) -2\wp(u) = \frac{9\wp(u)^4}{4\wp(u)^3-\frac{\eta^4}{16}} - 2\wp(u)$$
Let $Y = \frac{4}{\eta^2}\wp(u)$ and $A^2 = 1 - Y^3$, above condition is equivalent to

$$\begin{align} & Y^4 + 8 Y^3 + 8 Y - 8 = 0\tag{*1a}\\ \iff & Y(8+Y^3) = 8(1-Y^3)\tag{*1b}\\ \implies & (9-A^2)^3(1-A^2) = 512A^6\tag{*1c}\\ \iff & (A^4-24A^3+18A^2-27)(A^4+24A^3+18A^2-27) = 0\tag{*1d} \end{align}$$

+ Since $\alpha$ is a root for one of the factors in $(*1d)$, $A = \alpha$ satisfies $(*1d)$ and hence $(*1c)$.
+ Since $0 < \alpha < 1$ implies $1 - \alpha^2 > 0$, $(*1c) \implies (*1b)$ in this particular case.
i.e. $Y = \sqrt[3]{1-\alpha^2}$ satisfies $(*1b)$ and hence $(*1a)$.
+ Since $u$ lies between $u_0$ and $u_{-1}$, $\frac{4}{\eta^2}\wp(u) > 0$. Using
the fact $(*1a)$ has only one positive root, we find $\frac{4}{\eta^2}\wp(u) = \sqrt[3]{1-\alpha^2}$. i.e. conjecture $(8)$ is true.

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$\def\Beta{B}\def\tfrac#1#2{{\textstyle\frac{#1}{#2}}}$
Perhaps this might be helpful to someone. The integral is equal to, as you note,
$$J(y) = \int_0^1 x^{-1/3}(1-x)^{-1/6}(1-xy^2)^{-1/2}\,dx = \frac{2\pi}{y\sqrt{3}} \frac{\Beta(y^2,\frac12,\frac13)}{\Beta(\frac12,\frac13)},$$
where $\Beta(z,a,b)$ is the incomplete beta function.
Consider the function
$$I(y^2,a,b) = \frac{\Beta(y^2,a,b)}{\Beta(a,b)},$$
and rewrite using [DLMF 8.17](http://dlmf.nist.gov/8.17) it as
$$I(y^2,\tfrac12,\tfrac13) = 1-2I(z,\tfrac13,\tfrac13), \qquad 4z(1-z) = 1-y^2.$$
Then the function
$$f(z) = I(z,{\textstyle\frac13,\frac13}) = 3z^{1/3}\frac{{}_2F_1(\tfrac13,\tfrac23;\tfrac43;z)}{\Beta(\frac13,\frac13)}$$
is the one for which the conjectures are equivalent to (choosing the right root $z$):
$$f(z) = \tfrac14 \quad\Leftrightarrow\quad z^4-14z^3+24z^2-14z+1=0,$$
$$f(z) = \tfrac25 \quad\Leftrightarrow\quad 1+17 z-107 z^2+164 z^3-155 z^4+164 z^5-107 z^6+17 z^7+z^8=0,$$

From my tests it appears (I don't know how to prove this)
that the function $z(w)$ which solves
the equation $f(z(w)) = w$ always has algebraic values when $w$
is a rational number. The original integral is
$\frac{2\pi}{y\sqrt{3}}(1-2f(z))$, which is algebraic times $\pi$ when $f(z)$
is rational and $z$ algebraic, so I think the question is really about solving $f(z)=w$.

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How to prove the following conjectured identity?
$$\int_0^\infty\frac{dx}{\sqrt[4]{7+\cosh x}}\stackrel{\color{#a0a0a0}?}=\frac{\sqrt[4]6}{3\sqrt\pi}\Gamma^2\big(\tfrac14\big)\tag1$$
It holds numerically with precision of at least $1000$ decimal digits.

Are there any other integers under the radical except $7$ and $1$ that result in a nice closed form?

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I will follow @user15302's idea. In [**this answer**](https://math.stackexchange.com/questions/1379472/closed-form-of-int-0-infty-frac1-lefta-cosh-x-right1-n-dx-for/1379605#1379605), I showed that

$$\int_{0}^{\infty} \frac{dx}{(a + \cosh x)^{s}} \, dx = \frac{1}{(a+1)^{s}} \int_{0}^{1} \frac{v^{s-1}}{\sqrt{(1-v)(1-bv)}} \, dv,$$

where $b = \frac{a-1}{a+1}$. Now let $I$ denote the Vladimir's integral and set $s = \frac{1}{4}$ and $a = 7$. Then we have $b = \frac{3}{4}$ and

$$I = 2^{-3/4} \int_{0}^{1} \frac{1}{v^{3/4}\sqrt{(1-v)(1-\frac{3}{4}v)}} \, dv.$$

The reason why the case $b = \frac{3}{4}$ is special is that, if we plug $v = \operatorname{sech}^2 t$ then we can utilize the triple angle formula to get the following surprisingly neat integral

$$I = 2^{5/4} \int_{0}^{\infty} \frac{\cosh t}{\sqrt{\cosh 3t}} \, dt.$$

Now using the substitution $x = e^{-6t}$, we easily find that

$$I = \frac{1}{3 \sqrt[4]{2}} \int_{0}^{1} \frac{x^{-11/12} + u^{-7/12}}{\sqrt{1+x}} \, dx = \frac{1}{3 \sqrt[4]{2}} \int_{0}^{\infty} \frac{dx}{x^{11/12}\sqrt{1+x}}.$$

The last integral can be easily calculated by the following formula

$$\int_{0}^{\infty} \frac{x^{a-1}}{(1+x)^{a+b}} \, dx = \beta(a, b) = \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}.$$

Therefore we obtain the following closed form

$$I = \frac{\Gamma(\frac{1}{12})\Gamma(\frac{5}{12})}{3 \sqrt[4]{2}\sqrt{\pi}}.$$

In order to verify that this is exactly the same as Vladimir's result, We utilize the [*Legendre multiplication formula*](https://en.wikipedia.org/wiki/Multiplication_theorem#Gamma_function-Legendre_function) and the [*reflection formula*](https://en.wikipedia.org/wiki/Reflection_formula) to find that

$$\Gamma(\tfrac{1}{12})\Gamma(\tfrac{5}{12}) = \frac{\Gamma(\tfrac{1}{12})\Gamma(\tfrac{5}{12})\Gamma(\tfrac{9}{12})}{\Gamma(\tfrac{3}{4})} = \frac{2 \pi \cdot 3^{1/4} \Gamma(\frac{1}{4})}{\Gamma(\tfrac{3}{4})} = 2^{1/2} 3^{1/4} \Gamma(\tfrac{1}{4})^2.$$

This completes the proof.

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By replacing $x$ with $4u$, then $\cosh u$ with $\frac{1}{t}$, we have:

$$I = \frac{1}{2^{3/4}}\int_{0}^{1}\frac{dt}{(1-t^2+t^4)^{1/4}(1-t^2)^{1/2}}=\frac{1}{2^{3/4}}\int_{0}^{1/2}\frac{dt}{(1-t(1-t))^{1/4}(t(1-t))^{1/2}}$$
Next, by replacing $t(1-t)$ with $v/4$,
$$I=\frac{1}{2^{11/4}}\int_{0}^{1}\frac{dv}{(1-v/4)^{1/4}(v(1-v))^{1/2}}$$
then, by setting $v=4-3z$,
$$I = \frac{3^{1/4}}{2^{9/4}}\int_{1}^{4/3}\frac{dz}{z^{1/4}((4-3z)(1+z))^{1/2}}=\frac{3^{1/4}}{2^{5/4}}\int_{1}^{2/\sqrt{3}}\sqrt{\frac{z}{(4-3z^2)(1+z^2)}}\,dz$$
that, at least, looks manageable. We also have:
$$I = \frac{1}{2^{1/4}}\int_{0}^{1}\frac{du}{(3u^4+u^2)^{1/4}(1-u^2)^{1/2}}\tag{1}$$
that Mathematica gladly evaluates to:
$$I = \frac{2^{1/4}\,\Gamma\left(\frac{1}{4}\right)^2}{3^{3/4}\sqrt{\pi}}.$$
Now we just need to understand *how*.

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I think this problem can be solved by invoking the theory of $j$-invariants for (hyper?)-elliptic curves, but I am not so confident in the topic to find the right change of variables that brings our integral into a complete elliptic integral. I think that [Noam Elkies][1] would solve this problem in a few seconds, so I am asking his help.

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**Update**. Found. Our claim was proven by Zucker and Joyce in [Special values of the hypergeometric series II][2], it is the result $(7\!\cdot\! 6)$. It is derived through standard hypergeometric manipulations, by starting with the elliptic modulus $k$ for which:
$$\frac{K'(k)}{K(k)}=3.$$
The modular function to be considered for regarding our integral as a period is so the [elliptic lambda function][3].

[1]: https://math.stackexchange.com/users/93983/noam-d-elkies
[2]: http://www.researchgate.net/publication/231927689_Special_values_of_the_hypergeometric_series_III
[3]: http://mathworld.wolfram.com/EllipticLambdaFunction.html

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Take the integral in the form $$I=\frac{1}{2^{1/4}}\int_{0}^{1}\frac{du}{\left(3u^{4}+u^{2}\right)^{1/4}\left(1-u^{2}\right)^{1/2}}=\frac{1}{2^{1/4}}\int_{0}^{1}\frac{du}{\left(3u^{2}+1\right)^{1/4}\left(1-u^{2}\right)^{1/2}\left(u^{2}\right)^{1/4}}$$ then put $u^{2}=s$ $$=\frac{1}{2^{5/4}}\int_{0}^{1}\frac{ds}{\left(3s+1\right)^{1/4}\left(1-s\right)^{1/2}s^{3/4}}$$ and now put $s=1-t$ $$=\frac{1}{2^{7/4}}\int_{0}^{1}\frac{dt}{\left(1-3t/4\right)^{1/4}\left(1-t\right)^{3/4}t^{1/2}}.$$ Now recalling the identity $$\,_{2}F_{1}\left(a,b;c;z\right)=\frac{\Gamma\left(c\right)}{\Gamma\left(b\right)\Gamma\left(c-b\right)}\int_{0}^{1}\frac{t^{b-1}\left(1-t\right)^{c-b-1}}{\left(1-tz\right)^{a}}dt$$ we have $$I=\frac{1}{2^{7/4}}\frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{1}{4}\right)}{\Gamma\left(\frac{3}{4}\right)}\,_{2}F_{1}\left(\frac{1}{4},\frac{1}{2};\frac{3}{4};\frac{3}{4}\right)$$ and in this case it is possible calculate the exact value of the hypergeometric function (see the update in the Jack D'Aurizio's answer for reference) $$\,_{2}F_{1}\left(\frac{1}{4},\frac{1}{2};\frac{3}{4};\frac{3}{4}\right)=\frac{2\sqrt{2}}{3^{3/4}}$$ and so $$I=\frac{\Gamma^{2}\left(\frac{1}{4}\right)}{6^{3/4}\sqrt{\pi}}.$$ The result is not equal to $\sqrt[4]{6}\Gamma^{2}\left(\frac{1}{4}\right)/\left(2\sqrt{\pi}\right)$ but I haven't found a mistake in my calculations.

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It is quite a well known fact that:
$$\int_0^{+\infty} \frac{\sin x}{x} \, dx = \frac{\pi}{2}$$
also the value of related series is very similiar:
$$\sum_{n = 1}^{+\infty} \frac{\sin n}{n} = \frac{\pi - 1}{2}$$
Combining these two identities and using ${\rm sinc}$ function we get:
$$\int_{-\infty}^{+\infty} {\rm sinc}\, x \, dx = \sum_{n = -\infty}^{+\infty} {\rm sinc}\, n = \pi$$
What is more interesting is the fact that the equality:
$$\int_{-\infty}^{+\infty} {\rm sinc}^k\, x \, dx = \sum_{n = -\infty}^{+\infty} {\rm sinc}^k\, n$$
holds for $k = 1,2,\ldots, 6$. There are some other nice identities with ${\rm sinc}$ where sum equals integral but moving on to other functions we have e.g.:
$$\sum_{n = -\infty}^{+\infty} \binom{\alpha}{n} e^{int} = \int_{-\infty}^{+\infty} \binom{\alpha}{n} e^{itx} \, dx = (1+e^{it})^\alpha, \; \alpha >-1$$ which is due to Pollard & Shisha.

And finally the identity which is related to the famous *Sophomore's Dream*:
$$\int_0^1 \frac{dx}{x^x} = \sum_{n = 1}^{+\infty} \frac{1}{n^n}$$
Unfortunately in this case the summation range is not even close to the interval of integration.

Do you know any other interesting identities which show that "sum = integral"?

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Several papers are dedicated to the subject of integrals of functions that equal the sum of the same function, primarily for estimation purposes.

[Boas and Pollard][1] (1973) has some interesting sum-integral equalities:

$$\pi/\alpha=\sum_{n=-\infty}^\infty \frac{\sin^2 (c+n)\alpha}{(c+n)^2}=\int_{-\infty}^\infty \frac{\sin^2 (c+n)\alpha}{(c+n)^2}\, \text{d}n$$

$$\pi\operatorname{sgn} a=\sum_{n=-\infty}^\infty \frac{\sin (n+c)\alpha}{n+c}=\int_{-\infty}^\infty \frac{\sin (n+c)\alpha}{n+c}\, \text{d}n$$

It also gives several general formulae for functions that suffice:

$$\sum_{n=-\infty}^\infty f(n)=\int_{-\infty}^\infty f(n) \, \text{d}n$$

mainly with Fourier analysis.

---
[This][2] paper gives an equality with the [Bessel J function][3]:

$$\int_{-\infty}^\infty \frac{J_y (at) J_y(bt)}{t}\, \text{d}t=\sum_{t=-\infty}^\infty \frac{J_y (at) J_y(bt)}{t}$$

and some more references:

> There have been a number of studies of this kind of sum-integral
> equality by various groups, for example, Krishnan & Bhatia in the
> 1940s (Bhatia & Krishnan 1948; Krishnan 1948a,b; Simon 2002) and Boas,
> Pollard & Shisha in the 1970s (Boas & Stutz 1971; Pollard & Shisha
> 1972; Boas & Pollard 1973).

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See also [Surprising Sinc Sums and Integrals][4] which has some other equalities. This paper also states that (paraphrasing)

*If $G$ is of bounded variation on $[−\delta, \delta]$, vanishes outside $(−α, α)$, is Lebesgue integrable over $(−α, α)$ with $0 < α < 2\pi$ and has a Fourier transform of $g$, then*

$$\sum_{n=\infty}^\infty g(n)=\int_{-\infty}^\infty g(x)\, \text{d}x+\sqrt{\frac{\pi}{2}}(G(0-)+G(0+))$$

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Ramanujan's [second lost notebook][5] contains some sums of functions that equal the integral of their functions (Chapter 14, entries 5(i), 5(ii), 16(i), 16(ii)).

---

If you want, even more references with examples are in the papers I have mentioned.

[1]: http://carma.newcastle.edu.au/jon/Preprints/Papers/Published-InPress/Sinc-sums/Related/boas.pdf
[2]: http://rsc.anu.edu.au/~pgill/papers/159_Bessel.pdf
[3]: http://en.wikipedia.org/wiki/Bessel_function#Bessel_functions_of_the_first_kind_:_J.CE.B1
[4]: http://web.cs.dal.ca/~jborwein/sinc-sums.pdf
[5]: http://www.scribd.com/doc/48940173/Ramanujan-Notebooks-II-pdf

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How one can prove that the infinite sum of this function equals its integral
$$\sum_{n=-\infty}^\infty\frac{\cos\pi\sqrt{n^2+1}}{3+4n^2}=\int_{-\infty}^\infty\frac{\cos\pi\sqrt{x^2+1}}{3+4x^2}dx\ ? \tag{1}$$

My analysis: Mathematica wasn't able to return any closed form for the integral or the sum. Then I checked this relation by numerical computations and it agreed to about 20 decimal places.

I know from this question [Sum equals integral][1] that the function $\text{sinc}\ x=\frac{\sin x}{x}$ has the same property
$$\int_{-\infty}^{+\infty} {\rm sinc}\, x \, dx = \sum_{n = -\infty}^{+\infty} {\rm sinc}\, n = \pi$$
I tried to find a closed form for the integral $(1)$ but couldn't.

Motivation: I was challenged by a friend to prove this relation. I'm curious how one can prove it?

Note: There has been a suggestion to straightforwardly apply Euler-MacLauren summation formula to prove this statement. Though I don't know why it can not be applied in this case, I checked numerically whether the sum equals the integral for the similar looking functions $f_1(x)=\frac{\cos\pi\sqrt{x^2+1}}{1+x^2}$ and $f_2(x)=\frac{\cos\pi\sqrt{x^2+1}}{2+x^2}$, but in both cases there was a difference of about 1% between the sum and the integral. In starck contrast to this, using the same algorithm for $\frac{\cos\pi\sqrt{x^2+1}}{3+4x^2}$ there wasn't any difference between the sum and the integral at least to 20 decimal places. So I think it is very unlikely that 1% error can be attributed to computational error.

[1]: https://math.stackexchange.com/questions/170747/sum-equals-integral

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For any positive $a$, define

$$f_a(x) =\frac{\cos\pi\sqrt{x^2+1}}{x^2+a^2}$$

What you have observed is caused by the equality
$$\sum_{n=-\infty}^\infty f_a(n) - \int_{-\infty}^\infty f_a(x) dx = \frac{2\pi}{a(e^{2\pi a} - 1)}\times \begin{cases} \cosh\pi\sqrt{a^2-1}, & a > 1\\ \\ \cos\pi\sqrt{1-a^2}, & a < 1 \end{cases} \tag{*1}$$
and the fact $$\cos\pi\sqrt{1-a^2} = \cos\frac{\pi}{2} = 0\quad\text{ when } a^2 = \frac34$$

To see why $(*1)$ is true, we use the fact $f_a(n)$ is an even function in $n$ to rewrite LHS of $(*1)$ as

$$2 \left[\sum_{n=0}^\infty f_a(n) - \left(\int_0^\infty f_a(x) dx + \frac12 f_a(0)\right)\right]$$

This is similar to what you will find in
the [Abel-Plana formula](https://en.wikipedia.org/wiki/Abel%E2%80%93Plana_formula)${}^{\color{blue}{[1]}}$,

> For any function $f(z)$ which is
>
> 1. continuous on $\Re z \ge 0$ and analytic on $\Re z > 0$
> 2. $f(z) \sim o(e^{2\pi|\Im z|} )$ as $\Im z \to \pm \infty$, uniformly with respect to $\Re z$.
> 3. $f(z) \sim O(e^{2\pi|\Im z|}/|z|^{1+\epsilon})$ as $\Re z \to +\infty$ ${}^{\color{blue}{[2]}}$.
>
> we have
>
$$\sum_{n=0}^\infty f(n) = \int_0^\infty f(x) dx + \frac12 f(0) + i \int_0^\infty \frac{f(it) - f(-it)}{e^{2\pi t}-1} dt\tag{*2}$$

However, $f_a(x)$ doesn't exactly satisfy the condition above. It has two
poles at $\pm a i$. After a little bit of tweaking of the contour used in the proof of the Abel-Plana formula, one find:

$$\text{LHS}(*1) = 2i \lim_{\epsilon\to 0^{+}} \int_0^\infty \frac{f_a(it+\epsilon) - f_a(-it+\epsilon)}{e^{2\pi t} - 1} dt$$

For $t \ne a$, since $f_a(z)$ is even, the two pieces in $f_a(it+\epsilon) - f_a(-it+\epsilon)$ cancels out as $\epsilon \to 0^{+}$.
For $t \approx a$, the two pieces can be combined to a integral of $\frac{f(it)}{e^{2\pi t}-1}$ over a circle centered at $a$.
As a result, RHS reduces to

$$(2i)(2\pi i)\text{Res}_{t = a}\left[\frac{\cos\pi\sqrt{1-t^2}}{(a^2 - t^2)(e^{2\pi t} - 1)}\right] = \frac{2\pi}{a(e^{2\pi a}-1)}\times\begin{cases} \cosh\pi\sqrt{a^2-1}, & a > 1\\ \\ \cos\pi\sqrt{1-a^2}, & a < 1 \end{cases}$$
Back to the special case $a^2 = \frac{3}{4}$ which corresponds to the equality in question.

When $a^2 = \frac{3}{4}$, the "pole" of $f_a(z)$ at $z = \pm a i$ become removable singularities. The original version of Abel-Plana formula in $(*2)$ applies. Since $f_a(x)$ is even, last integral in $(*2)$ vanishes and the equality follows. This explain why the sum equal to the integral for $\frac{\cos\pi\sqrt{x^2+1}}{3+4x^2}$ but not other similar looking integrand like $\frac{\cos\pi\sqrt{x^2+1}}{1+x^2}$ or $\frac{\cos\pi\sqrt{x^2+1}}{2+x^2}$.

**Notes**

+ $\color{blue}{[1]}$ For more details of Abel-Plana formula and its derivation, please refer to $\S 8.3$ of Frank W. J Olver's book: [Asymptotics and Special Functions](http://www.amazon.com/Asymptotics-Special-Functions-Computer-mathematics/dp/012525850X).
+ $\color{blue}{[2]}$ In order to convert the AP formula on finite sum in Olver's book to infinite sum here, I have added a condition $(3)$ for this particular problem. The whole purpose of that is to force following limits to zero.
$$\lim\limits_{b\to\infty} f(b) = 0\quad\text{ and }\quad\lim\limits_{b\to\infty}\int_0^\infty \frac{f(b+it)-f(b-it)}{e^{2\pi t} - 1}dt = 0$$
For other $f(z)$, if one can justifies these limits, we can forget condition $(3)$ and the AP formula remains valid.

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