Evaluation and generalisation of $\int_0^\infty\int_0^\infty\sin y\frac{\operatorname{gd}(xy)}{\cosh(xy)}\mathrm dx\mathrm dy=\frac{\pi^3}{16}$

Solution 1:

At first, I would separate the $x$ and $y$ variables. If we set $x=\frac{z}{y}$ we get

$$ \iint_{(0,+\infty)^2}\sin(y)\frac{\text{gd}(xy)}{\cosh(xy)}\,dx\,dy = \iint_{(0,+\infty)^2}\frac{\sin(y)}{y}\cdot\frac{\text{gd}(z)}{\cosh(z)}\,dy\,dz = \frac{\pi}{2}\int_{0}^{+\infty}\frac{\text{gd}(z)}{\cosh(z)}\,dz $$ and the involved problems boil down to the study of a one-dimensional integral.
Afterwards, it is practical to borrow a useful Lemma from this answer:

_{(Jack's Lemma about the integral representation of the $\zeta$ function in terms of the $\text{sech}$ function)} $$ \begin{eqnarray*}\zeta(s) &=& \left(1-\frac{2}{2^s}\right)^{-1}\sum_{n\geq 1}\frac{(-1)^{n+1}}{n^s}\\&=&\left(1-\frac{2}{2^s}\right)^{-1}\frac{1}{\Gamma(s)}\int_{0}^{+\infty}\frac{t^{s-1}}{e^t+1}\,dt\\(IBP)\quad&=&\left(1-\frac{2}{2^s}\right)^{-1}\frac{2^{s+1}}{4\,\Gamma(s+1)}\int_{0}^{+\infty}\frac{t^s}{\cosh^2(t)}\,dt\end{eqnarray*}$$ from which it follows that $$ g(s)=\int_{0}^{+\infty}\frac{t^s}{\cosh^2(t)}\,dt = \frac{(2^s-2)\,\Gamma(s+1)\,\zeta(s)}{2^{2s-1}}\tag{JL} $$

In the original case, through the substitution $z=\log(t)$, $$ \int_{0}^{+\infty}\frac{\text{gd}(z)}{\cosh(z)}\,dz=\int_{1}^{+\infty}\frac{4\arctan(t)-\pi}{t^2+1}\,dt = \int_{0}^{1}\frac{\pi-4\arctan(t)}{t^2+1}\,dt$$ the substitution $t=\tan\theta$ leads to an elementary integral. As an alternative: $$ \int_{0}^{+\infty}\int_{0}^{x}\frac{1}{\cosh(x)\cosh(t)}\,dt\,dx = \frac{1}{2}\left(\int_{0}^{+\infty}\frac{du}{\cosh(u)}\right)^2.$$

In the other cases, we may use the same ideas, integration by parts and $(JL)$ to derive a closed form.

Solution 2:

In this answer I'm not showing exactly what you aim at, but these are similar to the original problem.

Proposition: $$\displaystyle \int\limits_0^\infty \int\limits_0^\infty \sin y \dfrac{(gd(xy))^a}{\cosh(xy)}\;dx\;dy\;=\;\dfrac{1}{a+1}\left(\dfrac{\pi}{2}\right)^{a+2}$$

Proof: Take into account that $\dfrac{\partial}{\partial x}gd(xy)=\dfrac{y}{\cosh xy}$ and $gd(x)=2\arctan(e^x)-\dfrac{\pi}{2}$ ,

$\displaystyle \int\limits_0^\infty \int\limits_0^\infty \sin y \dfrac{(gd(xy))^a}{\cosh(xy)}\;dx\;dy\; \\ =\displaystyle \int\limits_0^\infty \dfrac{\sin y}{y} \left(\int\limits_0^\infty(gd(xy))^a\;d(gd(xy))\right)\;dy\;\\=\displaystyle \dfrac{\pi^{a+1}}{(a+1)2^{a+1}}\int\limits_0^\infty \dfrac{\sin y}{y} \; dy\\=\dfrac{1}{a+1}\left(\dfrac{\pi}{2}\right)^{a+2}$

Proposition: $$\displaystyle \int\limits_0^\infty \int\limits_0^\infty \sin^{2n+1}y \dfrac{(gd(xy))^a}{\cosh(xy)}\;dx\;dy\;=\;\dfrac{\pi^{a+2}}{2^{2n+a+2}(a+1)}\binom{2n}{n}$$

Proof: Evaluating the inner integral just like before,

$\displaystyle \int\limits_0^\infty \int\limits_0^\infty \sin^{2n+1}y \dfrac{(gd(xy))^a}{\cosh(xy)}\;dx\;dy\;\\=\displaystyle\dfrac{\pi^{a+1}}{2^{a+1}(a+1)}\underbrace{\int\limits_0^\infty \dfrac{\sin^{2n+1}y}{y}\; dy}_{I}$

The braced Integral can be evaluated as ,

$\displaystyle I=\int_0^\infty \frac{\sin^{2n+1}x}{x}\,dx = \int_0^\infty\int_0^\infty e^{-xy}\sin^{2n+1}x\,dx\,dy$

Now by IBP two times we can create a recurrence relation and thus evaluate,

$\displaystyle J_n = \int_0^\infty e^{-ax}\sin^{2n+1}x\,dx = \frac{(2n+1)!}{\prod_{r=1}^{n}(a^2 +(2r+1)^2)}$

Thus, $\displaystyle I=(2n+1)!\int_0^\infty \frac{dy}{\prod_{r=1}^{n}(y^2 +(2r+1)^2)}$

The expression $\displaystyle \frac{1}{\prod_{r=1}^{n}(y^2 +(2r+1)^2)} = \sum_{r=0}^n \frac{A_r}{y^2+(2r+1)^2}$

where the coefficients can be determined by the cover-up rule as $\displaystyle A_k = \frac{(-1)^k}{2^{2n}}\frac{2k+1}{(n-k)!(n+k+1)!}$

So, $\displaystyle I=(2n+1)!\sum_{k=0}^n \int_0^\infty \frac{A_k}{y^2+(2k+1)^2}\,dy = \frac{\pi}{2^{2n+1}} \sum_{k=0}^{n} (-1)^k \binom{2n+1}{n-k} = \frac{\pi}{2^{2n+1}}\binom{2n}{n}$

Combining We get the result.

I will make updates to the answer soon, if I get more varieties and even the original answer