Let $$\mathscr{I}(s) = \int_0^\infty {{x^{s - 1}}\left( {\frac{{\cos (\pi {x^2})}}{{\sinh \pi x}} - \frac{1}{{\pi x}}} \right)dx} $$
It suffices to prove, for $0<\Re(s)<1$, $$\tag{*}\mathscr{I}(s) + 2\frac{{\Gamma (s)}}{{{{(2\pi )}^s}}}\sin \frac{{\pi s}}{2}\mathscr{I}(1 - s) = 2\frac{{\Gamma (s)\zeta (s)}}{{{{(2\pi )}^s}}}$$
We need a nontrivial Fourier transform:
(Lemma 1) For $\xi\in \mathbb{R}\setminus \{0\}$, $$\int_{ - \infty }^\infty {(\frac{{{e^{i\pi {x^2}}}}}{{\sinh \pi x}} - \frac{1}{{\pi x}}){e^{ - 2\pi ix\xi }}dx} = i\frac{{{e^{ - i\pi {\xi ^2}}} - {e^{\pi \xi }}}}{{\sinh \pi \xi }} + 2i{\chi _{(0,\infty )}}(\xi )$$
where $\chi_A$ is the characteristic function of set $A$.
Proof: Let $C(r)$ be a contour along real axis, but with small indention above $r \in \mathbb{R}$, then $$\int_{C(0)} {\frac{{{e^{i\pi {z^2}}}{e^{ - 2\pi iz\xi }}}}{{\sinh \pi z}}dz} = {e^{ - i\pi {\xi ^2}}}\int_{C( - \xi )} {\frac{{{e^{i\pi {z^2}}}}}{{\sinh \pi (z + \xi )}}dz} $$
Let $F(z)=\dfrac{{{e^{i\pi {z^2}}}{e^{4\pi z}}}}{{\sinh \pi (z + \xi )\sinh 4\pi z}}$,
then $$F(z) - F(z + 4i) = \frac{{2{e^{i\pi {z^2}}}}}{{\sinh \pi (z + \xi )}}$$
therefore
$$\int_{C( - \xi )} {\frac{{2{e^{i\pi {z^2}}}}}{{\sinh \pi (z + \xi )}}dz} = \int_C {F(z)dz} = 2\pi i\color{red}{\frac{{1 - {e^{\pi \xi }}{e^{i\pi {\xi ^2}}}}}{{\pi \sinh \pi \xi }}}$$
where $C$ is the rectangular contour with vertices $\pm \infty, \pm \infty + 4i$, and has small indentions above $-\xi, -\xi+4i, 0, 4i$. Observe that $F$ has $20$ poles inside $C$, summing over residues at these points gives the red expression. Thus,
$$\begin{aligned}&\int_{ - \infty }^\infty {(\frac{{{e^{i\pi {x^2}}}}}{{\sinh \pi x}} - \frac{1}{{\pi x}}){e^{ - 2\pi ix\xi }}dx} = \int_{C(0)} {\frac{{{e^{i\pi {z^2}}}{e^{ - 2\pi iz\xi }}}}{{\sinh \pi z}}dz} - \int_{C(0)} {\frac{{{e^{ - 2\pi iz\xi }}}}{{\pi z}}dz} \\ &= i\frac{{{e^{ - i\pi {\xi ^2}}} - {e^{\pi \xi }}}}{{\sinh \pi \xi }} + 2i{\chi _{(0,\infty )}}(\xi )\end{aligned}$$
proving the lemma.
Let $0<\Re(s)<1$, we have $$\int_0^\infty {{x^{s - 1}}{e^{ - 2\pi ix\xi }}dx} = \frac{{\Gamma (s)}}{{{{(2\pi i)}^s}}}{\xi ^{ - s}}\qquad \Im(\xi)\leq 0$$
Plancherel theorem
$\int_\mathbb{R} f(x) \overline{g(x)} dx = \int_\mathbb{R} \hat{f}(\xi) \overline{\hat{g}(\xi)} d\xi$ produces
$$\begin{aligned}&\int_0^\infty {{x^{s - 1}}\left( {\frac{{{e^{-i\pi {x^2}}}}}{{\sinh \pi x}} - \frac{1}{{\pi x}}} \right)dx} = \frac{{\Gamma (s)}}{{{{(2\pi i)}^s}}}\int_{ - \infty }^\infty {{\xi ^{ - s}}\left[ { - i\frac{{{e^{i\pi {\xi ^2}}} - {e^{\pi \xi }}}}{{\sinh \pi \xi }} - 2i{\chi _{(0,\infty )}}(\xi )} \right]d\xi }
\\ &= - \frac{{\Gamma (s)}}{{{{(2\pi )}^s}}}{e^{ - \pi is/2}}i\int_0^\infty {{\xi ^{ - s}}\left[ {\frac{{{e^{i\pi {\xi ^2}}} - {e^{\pi \xi }}}}{{\sinh \pi \xi }} + 2} \right]d\xi } + \frac{{\Gamma (s)}}{{{{(2\pi )}^s}}}{e^{\pi is/2}}i\int_0^\infty {{\xi ^{ - s}}\frac{{{e^{i\pi {\xi ^2}}} - {e^{ - \pi \xi }}}}{{\sinh \pi \xi }}d\xi }\end{aligned}$$
Taking complex conjugation (i.e. replace $i$ by $-i$), then sum up with the original gives:
$$\begin{aligned}\mathscr{I}(s) &= - \frac{{\Gamma (s)}}{{{{(2\pi )}^s}}}\sin \frac{{\pi s}}{2}\int_0^\infty {{\xi ^{ - s}}\left[ {\frac{{{e^{ - i\pi {\xi ^2}}} - {e^{\pi \xi }}}}{{\sinh \pi \xi }} + 2 + \frac{{{e^{i\pi {\xi ^2}}} - {e^{ - \pi \xi }}}}{{\sinh \pi \xi }}} \right]d\xi } \\
& = - \frac{{\Gamma (s)}}{{{{(2\pi )}^s}}}\sin \frac{{\pi s}}{2} \left[ 2\mathscr{I}(1 - s) + 2\int_0^\infty {{x^{ - s}}\left( {\frac{1}{{\pi x}} - \coth \pi x + 1} \right)dx}\right]
\end{aligned}$$
the next lemma completes the proof of $(*)$.
(Lemma 2) For $0<\Re(s)<1$, $$\int_0^\infty {{x^{s - 1}}\left( {\frac{1}{{\pi x}} + 1 - \coth \pi x} \right)dx} = - \sec \frac{{\pi s}}{2}\zeta (1 - s)$$
Proof: A lucid method is via Mellin inversion. First, expand $\coth \pi x$ in power series of $e^{-\pi x}$, termwise integration yields
$$\int_0^\infty {{x^{s - 1}}\left( {1 - \coth \pi x} \right)dx} = - {2^{1 - s}}{\pi ^{ - s}}\Gamma (s)\zeta (s) = - \sec \frac{{\pi s}}{2}\zeta (1 - s) \qquad \Re(s)>1$$
where we used the functional equation of $\zeta$. Note that $\zeta$ has "moderate growth" in every vertical strip, so Mellin inversion is permissible
$$\frac{1}{{2\pi i}}\int_{\sigma - i\infty }^{\sigma + i\infty } {\left[ { - \sec \frac{{\pi s}}{2}\zeta (1 - s)} \right]{x^{ - s}}ds} = 1 - \coth \pi x \qquad \sigma>1$$
Now shift the path of integration, to make it has real part $0<\sigma'<1$, taking residue at $s=1$ into account, the LHS of above equation equals
$$ \frac{1}{{2\pi i}}\int_{\sigma ' - i\infty }^{\sigma ' + i\infty } {\left[ { - \sec \frac{{\pi s}}{2}\zeta (1 - s)} \right]{x^{ - s}}ds} - \frac{1}{{\pi x}} \qquad 0<\sigma'<1$$
apply Mellin inversion again proves the lemma.
Since the integrand is even, we have
$$\mathcal{I} \stackrel{def}{=}\int_{-\infty}^\infty \frac{dx}{\cosh^2(x-\frac{a}{x})} dx
= 2\int_0^\infty \frac{dx}{\cosh^2(x-\frac{a}{x})}dx$$
In one copy of the integral on RHS, change variables to $y = \frac{a}{x}$, one get
$$\int_0^\infty \frac{dx}{\cosh^2(x-\frac{a}{x})} =
\int_0^\infty \frac{1}{\cosh^2(\frac{a}{y} - y)} \frac{a}{y^2}dy$$
Renaming $y$ back to $x$ and add back to another copy of integral on RHS, we get
$$\mathcal{I} = \int_0^\infty \frac{1}{\cosh^2(x - \frac{a}{x})}\left(1 + \frac{a}{x^2}\right)dx
= \int_0^\infty \frac{1}{\cosh^2(x - \frac{a}{x})}\frac{d}{dx}\left(x - \frac{a}{x}\right) dx$$
Change variable to $z = x - \frac{a}{x}$, this becomes
$$\mathcal{I} = \int_{-\infty}^\infty \frac{dz}{\cosh^2(z)}$$
i.e. transform your integral $(3)$ to integral $(2)$, the one you already know.
Let's look at the more general identity $(1)$.
As you increases $x$ from $-\infty$ to $0$, $x - \frac{a}{x}$ increases from $-\infty$ to $\infty$ once. If one further increases $x$ from $0$ to $\infty$, $x - \frac{a}{x}$ increases from $-\infty$ to $\infty$ the second time.
For any $t \in \mathbb{R}$, let $x_1(t) < 0$, $x_2(t) > 0$ be the two roots of
$$x - \frac{a}{x} = t \equiv x^2 - tx - a = 0$$
This is a quadratic equation in $x$. By Vieta's formula, we have
$$x_1(t) + x_2(t) = t \implies x_1'(t) + x_2'(t) = 1$$
If you change variable to $t = x - \frac{a}{x}$ for both $(-\infty,0)$ and $(0,\infty)$, we obtain:
$$\begin{align}
\int_{-\infty}^\infty f(x - \frac{a}{x}) dx
&= \left(\int_{-\infty}^0 + \int_0^\infty\right) f(x - \frac{a}{x})dx\\
&= \int_{-\infty}^{\infty} f(t) x'_1(t) dt +
\int_{-\infty}^{\infty} f(t) x'_2(t) dt\\
&= \int_{-\infty}^\infty f(t) (x'_1(t) + x'_2(t))dt\\
&= \int_{-\infty}^\infty f(t)dt
\end{align}$$
This is the identity $(1)$ we seek.
Best Answer
Let $\mathcal{S}$ denote the sum of the following (convergent) infinite series:
$$\mathcal{S}:=4\sum_{n=0}^{\infty}\frac{(2n)!!}{(2n+1)!!}(2n+2)^{-2},\tag{1}$$
where here $n!!$ denotes the so-called double factorial of a number $n$.
(Note: My definition of $\mathcal{S}$ has an additional scalar factor of $4$ so as to simplify its expression in terms of the generalized hypergeometric function $_4F_3$.)
We'll make use of the following well-known integration formula for a subclass of Wallis' integrals (for proof see [wiki][1]):
$$\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\cos^{2n+1}{\left(\varphi\right)}=\frac{(2n)!!}{(2n+1)!!};~~~\small{n\in\mathbb{Z}_{\ge0}}.$$
Recall the definition of the [polylogarithm][2] as an infinite series. Given $s\in\mathbb{C}\land z\in\mathbb{C}\land|z|<1$, the polylogarithm $\operatorname{Li}_{s}{\left(z\right)}$ of order $s$ and argument $z$ is given by the (absolutely convergent) power series
$$\operatorname{Li}_{s}{\left(z\right)}=\sum_{n=1}^{\infty}\frac{z^{n}}{n^{s}}.$$
For positive integer order, the polylogarithm can be defined iteratively by
$$\operatorname{Li}_{1}{\left(z\right)}:=-\ln{\left(1-z\right)};~~~\small{z\in\left(-\infty,1\right)},$$
$$\operatorname{Li}_{n+1}{\left(z\right)}:=\int_{0}^{z}\mathrm{d}t\,\frac{\operatorname{Li}_{n}{\left(t\right)}}{t};~~~\small{n\in\mathbb{N}\land z\in\left(-\infty,1\right]}.$$
Another useful integral representation for $\operatorname{Li}_{n+1}{\left(z\right)}$, which can be obtained from the previous one by repeated integration by parts, is
$$\operatorname{Li}_{n+1}{\left(z\right)}=\frac{(-1)^{n}}{n!}\int_{0}^{1}\mathrm{d}t\,\frac{z\ln^{n}{\left(t\right)}}{1-zt};~~~\small{n\in\mathbb{N}\land z\in\left(-\infty,1\right]}.$$
An important auxiliary function pertaining to the polylogarithm is the so-called Nielsen generalized polylogarithm, defined for positive integer parameters via the integral representation
$$S_{n,p}{\left(z\right)}:=\frac{(-1)^{n+p-1}}{(n-1)!\,p!}\int_{0}^{1}\mathrm{d}t\,\frac{\ln^{n-1}{\left(t\right)}\ln^{p}{\left(1-zt\right)}}{t};~~~\small{\left(n,p\right)\in\mathbb{N}^{2}\land z\in\left(-\infty,1\right]}.$$
The following integration formula will be useful to have on hand later:
$$\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(1-t\right)}\ln{\left(1-zt\right)}}{t}=\operatorname{Li}_{3}{\left(z\right)}+S_{1,2}{\left(z\right)};~~~\small{z\in\left(-\infty,1\right]}.$$
Proof:
$$\begin{align} \int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(1-t\right)}\ln{\left(1-zt\right)}}{t} &=\int_{0}^{1}\mathrm{d}t\,\frac{\ln^{2}{\left(1-t\right)}+\ln^{2}{\left(1-zt\right)}-\left[\ln{\left(1-t\right)}-\ln{\left(1-zt\right)}\right]^{2}}{2t}\\ &=\int_{0}^{1}\mathrm{d}t\,\frac{\ln^{2}{\left(1-t\right)}}{2t}+\int_{0}^{1}\mathrm{d}t\,\frac{\ln^{2}{\left(1-zt\right)}}{2t}-\int_{0}^{1}\mathrm{d}t\,\frac{\ln^{2}{\left(\frac{1-t}{1-zt}\right)}}{2t}\\ &=\int_{0}^{1}\mathrm{d}t\,\frac{\ln^{2}{\left(1-t\right)}}{2t}+\int_{0}^{1}\mathrm{d}t\,\frac{\ln^{2}{\left(1-zt\right)}}{2t}\\ &~~~~~-\int_{0}^{1}\mathrm{d}u\,\frac{\left(1-z\right)}{\left(1-zu\right)^{2}}\cdot\frac{\ln^{2}{\left(u\right)}}{2\left(\frac{1-u}{1-zu}\right)};~~~\small{\left[t=\frac{1-u}{1-zu}\right]}\\ &=\int_{0}^{1}\mathrm{d}t\,\frac{\ln^{2}{\left(1-t\right)}}{2t}+\int_{0}^{1}\mathrm{d}t\,\frac{\ln^{2}{\left(1-zt\right)}}{2t}\\ &~~~~~-\int_{0}^{1}\mathrm{d}u\,\frac{\left(1-z\right)\ln^{2}{\left(u\right)}}{2\left(1-u\right)\left(1-zu\right)}\\ &=\int_{0}^{1}\mathrm{d}t\,\frac{\ln^{2}{\left(1-t\right)}}{2t}+\int_{0}^{1}\mathrm{d}t\,\frac{\ln^{2}{\left(1-zt\right)}}{2t}\\ &~~~~~-\int_{0}^{1}\mathrm{d}u\,\frac{\ln^{2}{\left(u\right)}}{2\left(1-u\right)}+\int_{0}^{1}\mathrm{d}u\,\frac{z\ln^{2}{\left(u\right)}}{2\left(1-zu\right)}\\ &=\frac12\int_{0}^{1}\mathrm{d}t\,\frac{z\ln^{2}{\left(t\right)}}{1-zt}+\frac12\int_{0}^{1}\mathrm{d}t\,\frac{\ln^{2}{\left(1-zt\right)}}{t}\\ &=\operatorname{Li}_{3}{\left(z\right)}+S_{1,2}{\left(z\right)}.\\ \end{align}$$
Using the technique of interchanging the order of summation and integration, we obtain an expression for the power series $\mathcal{S}$ as a definite integral.
$$\begin{align} \mathcal{S} &=4\sum_{n=0}^{\infty}\frac{(2n)!!}{(2n+1)!!}(2n+2)^{-2}\\ &=\sum_{n=0}^{\infty}\frac{1}{(n+1)^{2}}\cdot\frac{(2n)!!}{(2n+1)!!}\\ &=\sum_{n=0}^{\infty}\frac{1}{(n+1)^{2}}\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\cos^{2n+1}{\left(\varphi\right)}\\ &=\sum_{n=0}^{\infty}\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\frac{\cos^{2n+1}{\left(\varphi\right)}}{(n+1)^{2}}\\ &=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\sum_{n=0}^{\infty}\frac{\cos^{2n+1}{\left(\varphi\right)}}{(n+1)^{2}}\\ &=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\sum_{n=1}^{\infty}\frac{\cos^{2n-1}{\left(\varphi\right)}}{n^{2}}\\ &=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\frac{1}{\cos{\left(\varphi\right)}}\sum_{n=1}^{\infty}\frac{\left[\cos^{2}{\left(\varphi\right)}\right]^{n}}{n^{2}}\\ &=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\sec{\left(\varphi\right)}\operatorname{Li}_{2}{\left(\cos^{2}{\left(\varphi\right)}\right)}\\ &=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\frac{\cos{\left(\varphi\right)}\operatorname{Li}_{2}{\left(1-\sin^{2}{\left(\varphi\right)}\right)}}{1-\sin^{2}{\left(\varphi\right)}}\\ &=\int_{0}^{1}\mathrm{d}x\,\frac{\operatorname{Li}_{2}{\left(1-x^{2}\right)}}{1-x^{2}};~~~\small{\left[\varphi=\arcsin{\left(x\right)}\right]}.\\ \end{align}$$
Then,
$$\begin{align} \mathcal{S} &=\int_{0}^{1}\mathrm{d}x\,\frac{\operatorname{Li}_{2}{\left(1-x^{2}\right)}}{1-x^{2}}\\ &=-\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(\frac{1+x}{1-x}\right)}}{2}\cdot\frac{2x\ln{\left(x^{2}\right)}}{1-x^{2}};~~~\small{I.B.P.s}\\ &=\int_{0}^{1}\mathrm{d}x\,\frac{2x\ln{\left(x\right)}\ln{\left(\frac{1-x}{1+x}\right)}}{1-x^{2}}\\ &=\int_{0}^{1}\mathrm{d}x\,\frac{2x\ln{\left(x\right)}\ln{\left(1-x^{2}\right)}}{1-x^{2}}-\int_{0}^{1}\mathrm{d}x\,\frac{2x\ln{\left(x\right)}\ln{\left((1+x)^2\right)}}{1-x^{2}}\\ &=\frac12\int_{0}^{1}\mathrm{d}x\,\frac{2x\ln{\left(x^{2}\right)}\ln{\left(1-x^{2}\right)}}{1-x^{2}}-\int_{0}^{1}\mathrm{d}x\,\frac{4x\ln{\left(x\right)}\ln{\left(1+x\right)}}{1-x^{2}}\\ &=\frac12\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(y\right)}\ln{\left(1-y\right)}}{1-y};~~~\small{\left[x^{2}=y\right]}\\ &~~~~~+\int_{0}^{1}\mathrm{d}x\,\frac{2\ln{\left(x\right)}\ln{\left(1+x\right)}}{1+x}-\int_{0}^{1}\mathrm{d}x\,\frac{2\ln{\left(x\right)}\ln{\left(1+x\right)}}{1-x}\\ &=\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(t\right)}\ln{\left(1-t\right)}}{2t};~~~\small{\left[y=1-t\right]}\\ &~~~~~-\int_{0}^{1}\mathrm{d}x\,\frac{\ln^{2}{\left(1+x\right)}}{x};~~~\small{I.B.P.s}\\ &~~~~~-\int_{0}^{1}\mathrm{d}t\,\frac{2\ln{\left(1-t\right)}\ln{\left(2-t\right)}}{t};~~~\small{\left[x=1-t\right]}\\ &=\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(t\right)}\ln{\left(1-t\right)}}{2t}-\int_{0}^{1}\mathrm{d}t\,\frac{\ln^{2}{\left(1+t\right)}}{t}\\ &~~~~~-\int_{0}^{1}\mathrm{d}t\,\frac{2\ln{\left(1-t\right)}\ln{\left(2\right)}}{t}-\int_{0}^{1}\mathrm{d}t\,\frac{2\ln{\left(1-t\right)}\ln{\left(1-\frac12t\right)}}{t}\\ &=\frac12\,S_{2,1}{\left(1\right)}-2S_{1,2}{\left(-1\right)}\\ &~~~~~+2\ln{\left(2\right)}\operatorname{Li}_{2}{\left(1\right)}-2\left[\operatorname{Li}_{3}{\left(\frac12\right)}+S_{1,2}{\left(\frac12\right)}\right]\\ &=3\ln{\left(2\right)}\,\zeta{\left(2\right)}-\frac74\,\zeta{\left(3\right)}.\blacksquare\\ \end{align}$$