Evaluation and generalisation of $\int_0^\infty\int_0^\infty\sin y\frac{\operatorname{gd}(xy)}{\cosh(xy)}\mathrm dx\mathrm dy=\frac{\pi^3}{16}$ I was thinking to combine integrals involving the Gudermannian function, see the definition in this Wikipedia, the hyperbolic cosine and a trick that I know from this site from a post high scored from the tag (integration) explaining how get 
$$\int_{0}^{\infty} \frac{\sin(x)}{x}\mathrm dx.$$
Then using limits to evaluate the limits of the integral, and calculating double integrals with techniques of integration and the properties of the Gudermannian function I believe that one could get some examples of closed-forms for integrals.
Claim. One could to justify $$\int_0^\infty\int_0^\infty\sin y\frac{\operatorname{gd}(xy)}{\cosh(xy)}\mathrm dx\mathrm dy=\frac{\pi^3}{16}.$$

Question. I would like to know how to study and provide a reasoning to get a nice generalization of previous integral (I am asking if it is feasible an explanation of how do you create a generealization, how do you explain you possibilities and exploit your knowledges to get a generalization, maybe is not the more general case but your reasoning and result are goods). But the main question is how get a generalization of previous Claim, see below my examples. Many thanks.

Example 1. Seems that it's possible to get a closed-form of the integral
$$\int_0^\infty\int_0^\infty\frac{\sin^\alpha y}{y^\beta}\frac{\operatorname{gd}(xy)}{\cosh(xy)}\mathrm dx\mathrm dy$$
 for some particular values $\alpha,\beta$ being these positive integers. I am saying that maybe it is possible that I can get simple examples with an online calculator.
Example 2. Seems that it's possible to get closed-form for integrals of the shape $$\int_0^\infty\int_0^\infty\sin y\frac{(\operatorname{gd}(xy))^\alpha}{\cosh^\beta(xy)}\mathrm dx\mathrm dy$$
for some particular values $\alpha,\beta$ being these positive integers. I am saying that maybe it is possible that I can get simple examples with an online calculator.
 A: 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.
A: 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
