Help with $\int _0^{\infty }\frac{\sinh \left(x\right)}{x\cosh ^2\left(x\right)}\:\mathrm{d}x$ I want to know how to prove that
$$\int _0^{+\infty }\frac{\sinh \left(x\right)}{x\cosh ^2\left(x\right)}\:\mathrm{d}x=\frac{4G}{\pi }$$
Here $G$ denotes Catalan's constant, I obtained such result with the help of mathematica.
I also found that the integral equals a certain infinite series
$$\int _0^{+\infty }\frac{\sinh \left(x\right)}{x\cosh ^2\left(x\right)}\:\mathrm{d}x=\sum _{n=0}^{+\infty }\frac{\binom{2n}{n}^2}{16^n\left(2n+1\right)}=\frac{4G}{\pi }$$
which can also be found in this link.
So I've $2$ questions
$1)$$¿$How can we transform the integral into the mentioned series?
$2)$$¿$Is there a simple way to evaluate the main integral without resorting to series expansion?
What I did for question $\#2$ is to employ the substitution $x=\ln\left(t\right)$
$$\int _0^{+\infty }\frac{\sinh \left(x\right)}{x\cosh ^2\left(x\right)}\:\mathrm{d}x=-2\int _1^{\infty }\frac{1-t^2}{\ln \left(t\right)\left(1+t^2\right)^2}\:\mathrm{d}t$$
But I'm not sure how to proceed.
 A: For $\Re(s)>0$, the Dirichlet beta function is defined by the infinite series $$\beta(s) = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)^{s}}. $$
The Dirichlet beta function can be extended to rest of the complex plane by its functional equation, from which we see that $\beta(s)$ has zeroes at the negative odd integers.
An integral representation of the Dirichlet beta function is $$\beta(s) = \frac{1}{\Gamma(s)}\int_{0}^{\infty} \frac{x^{s-1}e^{-x}}{1+e^{-2x}} \, \mathrm dx = \frac{1}{2\Gamma(s)} \int_{0}^{\infty} \frac{x^{s-1}}{\cosh (x)} \, \mathrm dx \, , \quad \Re(s) >0.$$
Integrating by parts, we get $$\beta (s) = \frac{1}{2 \Gamma(s+1)} \int_{0}^{\infty} \frac{x^{s} \sinh (x) }{\cosh^{2} (x)} \, \mathrm dx . \tag{1}$$
Since the Dirichlet beta function and the reciprocal gamma function are entire functions, and since the Mellin transform defines an analytic function where it converges absolutely, the identity theorem says that equation $(1)$ holds for $\Re(s) >-2$.
Now if we multiply both sides of $(1)$ by $2 \Gamma(s+1)$ and let $s \to -1$, we get $$\int_{0}^{\infty} \frac{\sinh (x)}{x \cosh^{2} (x)} \, \mathrm dx = \lim_{s \to -1} 2 \beta(s) \Gamma(s+1) = 2 \lim_{s \to -1} \beta(s) \left(\frac{1}{s+1} + \mathcal{O}(1) \right) = 2\beta'(-1).$$
But from differentiating both sides of the functional equation, we see that $$-\beta'(-1) =  -\frac{2}{\pi} \beta(2) = -\frac{2}{\pi} \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)^{2}} = -\frac{2 G}{\pi}.$$
Therefore, $$\int_{0}^{\infty} \frac{\sinh (x)}{x \cosh^{2} (x)} \, \mathrm dx = \frac{4 G}{\pi}. $$
A: Enforcing the substitution $x\mapsto \frac12\log(x)$, we find that
$$\begin{align}
\int_{0}^\infty \frac{\sinh(x)}{x\cosh^2(x)}\,dx&=\int_0^\infty \color{blue}{\left(\frac{x-1}{\log(x)}\right)}\frac1{x^{1/2}(x+1)^2}\,dx\\\\
&=\int_0^\infty \color{blue}{\left(\int_0^1 x^t\,dt\right)} \frac{1}{x^{1/2}(x+1)^2}\,dx\\\\
&=\int_0^1\int_0^\infty  \frac{x^{t-1/2}}{(x+1)^2}\,dx\,dt\tag1\\\\
&=\int_0^1\frac{\pi(t-1/2)}{\sin(\pi (t-1/2))}\,dt\tag2\\\\
&=\frac1\pi\int_{-\pi/2}^{\pi/2} \frac{t}{\sin( t)}\,dt\tag3\\\\
&=\frac{4G}{\pi}\tag4
\end{align}$$
as was to be shown!

NOTES:
In arriving at $(1)$, we applied the Fubini-Tonelli Theorem.
To go from $(1)$ to $(2)$, we evaluated the inner integral, $\int_0^\infty \frac{x^{t-1/2}}{(1+t)^2}\,dt$.  To do so, one can use complex analysis and integrate $\frac{z^{t-1/2}}{(1+z)^2}$ over the classical keyhole contour and apply the reside theorem.  Alternatively one can use real analysis only.  To proceed accordingly, we enforce the substitution $x\mapsto \tan^2(x)$ to find that
$$\begin{align}
\int_0^\infty \frac{x^{t-1/2}}{(1+t)^2}\,dt&=2\int_0^{\pi/2}\sin^{2t}(x)\cos^{2-2t}(x)\,dx\\\\
&=B(1/2+t,3/2-t)\tag{N1}
\end{align}$$
Then, applying the identity $B(1+x,y)=\frac{x}{x+y}B(x,y)$ (with $x=t-1/2$) and the reflection formula for the Gamma function to $(N1)$ reveals
$$\begin{align}
\int_0^\infty \frac{x^{t-1/2}}{(1+t)^2}\,dt&=(t-1/2)B(t-1/2,1-(t-1/2))\\\\
&=(t-1/2)\Gamma(t-1/2)\Gamma(1-(t-1/2))\\\\
&\frac{\pi(t-1/2)}{\sin(\pi(t-1/2))}
\end{align}$$
In going from $(2)$ to $(3)$, we enforced the substitution $\pi(t-1/2)\mapsto t$.
Finally, in arriving at $(4)$,  we relied on the well-known representation for Catalan's Constant $G$, as given HERE.
A: You can integrate the function:
$$
f(z):=\frac{\psi^{(0)}\left [ -iz/(2\pi)\right ]\sinh z}{\cosh^2z}
$$
Along a rectangular contour with vertices
$(\pm\infty,\pm\infty+2\pi i)$
Then you can get:
$$
-2\pi i\int_{-\infty}^{\infty} \frac{\sinh z}{z\cosh^2z}\text{d}z
= 2\pi i\sum_{z_k=\pi i/2,3\pi i/2}\text{Res}(f(z),z_k)
$$
It equivalent to say:
$$
\int_{0}^{\infty} \frac{\sinh z}{z\cosh^2z} \text{d}z
=-\frac{1}{2} \left [ -\frac{1}{2\pi}\psi^{(1)} \left ( \frac{1}{4}  \right ) 
+ \frac{1}{2\pi}\psi^{(1)} \left ( \frac{3}{4}  \right )
 \right ]
$$
And we have some special values(trival)
$$
\begin{aligned}
&\psi^{(1)} \left ( \frac{1}{4}  \right )=\pi^2+8G\\
& \psi^{(1)} \left ( \frac{3}{4}  \right )=\pi^2-8G
\end{aligned}
$$
Finally,
$$
\int_{0}^{\infty} \frac{\sinh z}{z\cosh^2z} \text{d}z
=\frac{1}{2} \cdot\frac{1}{2\pi} \cdot16G=\frac{4G}{\pi}
$$
Completed the proof.

Here I write some other results,they're all trivial.
$$\begin{aligned}
&\int_{0}^{\infty} \frac{\sinh z}{z\cosh^4z} \text{d}z
=\frac{2}{3\pi}G +\frac{16}{\pi^3} \beta(4)\\
&\int_{0}^{\infty} \frac{\sinh^3 z}{z\cosh^4z} \text{d}z
=\frac{10}{3\pi}G -\frac{16}{\pi^3} \beta(4)\\
&\int_{0}^{1} x\ln\left ( \frac{2-x^2}{1-x^2}  \right ) 
\arctan\left ( \frac{x}{\sqrt{2-2x^2} }  \right ) \text{d}x
=\frac{\pi}{2} (\sqrt{2}-1 )\ln(1+\sqrt{2} )
\end{aligned}$$
A: Let $I(a)=\int _0^{\infty }\frac{\sinh (ax)}{x\cosh ^2x}{d}x$. Then
$$I’(a)= \int _{-\infty}^{\infty }\frac{\cosh (ax)}{2\cosh ^2x}{d}x
\overset{t=e^{2x}}=\int_0^\infty \frac{t^{-\frac a2}+t^{\frac a2}}{2(1+t)^2}dt
=\frac{\pi a}{2\sin\frac{\pi a}2}
$$
and
$$\int _0^{\infty }\frac{\sinh x}{x\cosh ^2x}{d}x
=\int_0^1 I’(a)da
\overset{s=\tan\frac {\pi a}4} =\frac 4\pi \int_0^{1}\frac {\tan^{-1}s}{s}ds
=\frac{4}{\pi}G
\\
$$
A: I present another approach, utilising a well-known property of the Laplace Transform i.e.
$$\int_{0}^{+\infty} f\left(x\right) g\left(x\right) \, dx = \int_{0}^{+\infty} \left(\mathcal{L} f\right)\left(y\right)\left(\mathcal{L}^{-1} g\right)\left( y\right) \, dy$$
Letting $f\left(x\right) = \tanh \left(x\right) \text{sech} \left(x\right)$ and $g(x) = \frac{1}{x}$:
$$\left( \mathcal{L} f \right) (y) = 1 + \frac{1}{2} y \left( \psi^{(0)} \left(\frac{1+y}{4}\right)- \psi^{(0)} \left(\frac{3+y}{4}\right)\right)$$
$$\left( \mathcal{L}^{-1} g \right) \left(y \right) = 1$$
Where $\psi$ represents the polygamma function.
$$\implies I = \int_{0}^{\infty} \left(1 + \frac{1}{2} y \left( \psi^{(0)} \left(\frac{1+y}{4}\right)- \psi^{(0)} \left(\frac{3+y}{4}\right)\right)\right) \, dy$$
I will now proceed to integrate indefinitely, then take limits at the end.
$$\int 1 \, dy + \frac{1}{2} \int y \, \psi^{(0)} \left(\frac{1+y}{4}\right) \, dy - \frac{1}{2} \int y \, \psi^{(0)} \left(\frac{3+y}{4}\right) \, dy$$
We proceed with integrating by parts.
$$=y + \frac{1}{2} \left( 4y \ln \Gamma \left( \frac{1+y}{4} \right) - 4 \int \ln \Gamma \left( \frac{1+y}{4} \right) \, dy \right)-\frac{1}{2} \left( 4y \ln \Gamma \left( \frac{3+y}{4} \right) - 4 \int \ln \Gamma \left( \frac{3+y}{4} \right) \, dy \right)$$
$$=y + 2 y \ln \Gamma \left( \frac{1+y}{4}\right) - 8 \psi^{(-2)} \left(\frac{1+y}{4}\right) - 2y \ln \Gamma \left( \frac{3+y}{4} \right) + 8 \psi^{(-2)} \left( \frac{3+y}{4} \right)$$
Taking limits to $\infty$ and $0$, and then subtracting gives us:
$$2\ln (2\pi) + 8 \psi^{(-2)} \left( \frac{1}{4} \right)-8 \psi^{(-2)} \left( \frac{3}{4} \right)$$
$$=\boxed{\frac{4G}{\pi}}$$ (the equality was checked with WolframAlpha) as required.
A: We can use contour integration, but we have to be a little careful with it.
Begin by noting that the integrand is an even function, so we shall evaluate
$I=\int_{-\infty}^{+\infty}\dfrac{\sinh x}{x\cosh^2x}~dx$
and take half of this result. in this way the contour accesses infinity at both ends of the desired integration range, which facilitates using contour integration.
A semicircular contour that is often used with this doubly infinite integration range is problematic because the integrand will not be $o(1/z)$ over the entire semicircle. We violate the small-$o$ relation at the imaginary axis where there are infinitely many singularities.
We therefore design a more refined contour: a rectangle with corners at $-N, +N, +N+(2\pi N)i, -N+(2\pi N)i$. Traversing the rectangle counterclockwise and applying the Residue Theorem then gives
$I(1) + I(2) + I(3) + I(4)=2\pi i\sum _{k=0}^{2N} R[(2k+1)\pi i/2]$
where the integration path for $I(1)$ runs from $-N$ to $+N$, the path for $I(2)$ runs from $+N$ to $+N+(2\pi N)i$, and so on around the remaining sides of the rectangle for $I(3)$ and $I(4)$. Then the required integral is the limit of $I(1)$ as $N\to\infty$, which matches with the sum of residues provided that $I(2),I(3),I(4)$ go to zero in this limit.
To verify the zero limit for $I(2)$ render (limits represent the asymptotic behavior as $N\to\infty$):
$|\sinh(N+iy)|=\sqrt{\sinh^2N+\sin^2y}\to e^{N/2}$
$|\cosh(N+iy)|=\sqrt{\cosh^2N-\sin^2y}\to e^{N/2}$
Thus since the length of the side is $2\pi N$, the Triangle Inequality guarantees
$|I(2)|\le\dfrac{2\pi\sqrt{\sinh^2N+\sin^2y}}{(\cosh^2N-\sin^2y)}\to 2\pi e^{-N/2}\to0$
A similar analysis eliminates $I(4)$ as $N\to\infty$.
For $I(3)$ we use the fact that the hyperbolic functions are periodic, thus for the selected values of $y$ they match the real-argument values at $x$. Thereby
$|\dfrac{\sinh z}{z\cosh^2 z}|=|\dfrac{\sinh x}{z\cosh^2 x}|<\dfrac{1}{y\cosh x}<[1/(2\pi N)]e^{-|x|/2}$
$|I(3)|<[1/(2\pi N)]\int_{-N}^Ne^{-|x|/2}dx\to0$ (the integral is bounded as $N\to\infty$.)
So we have
$I=\lim_{N\to\infty}(I(1))=2\pi i\sum _{k=0}^\infty R[(2k+1)\pi i/2]$
and it remains to find the residues.
To find these residues $R[(2k+1)\pi i/2]$, it is convenient to define a difference parameter $\delta$ at each of these singularities. Since the singularities are second-order poles, we need to carry the Laurent series expansion to two terms.
Therefore, at each singularity $z=(2k+1)\pi i/2$, we render the following Laurent series:
$\sinh z = (-1)^ki+0\delta+O(\delta^2)$
$\dfrac{1}{\cosh^2 z} = 1/\delta^2+0/\delta+O(1)$
$(1/z)=\dfrac{-2i}{(2k+1)\pi}-\dfrac{4}{(2k+1)^2\pi^2}\delta+O(\delta^2)$
And upon multiplying theses:
$\dfrac{\sinh z}{z\cosh^2z}=\dfrac{2(-1)^k(2k+1)}{\pi\delta^2}\color{blue}{-\dfrac{4(-1)^ki}{(2k+1)^2\pi^2\delta}}+O(1)$
from which we read off the residues
$R[(2k+1)\pi i/2]=\dfrac{-4(-1)^ki}{(2k+1)^2\pi^2}$
So
$I=\int_{-\infty}^{+\infty}\dfrac{\sinh x}{x\cosh^2x}~dx=(8/\pi)\color{blue}{\sum _{k=0}^\infty \dfrac{(-1)^k}{(2k+1)^2}}$
where the blue series directly defines the Catalan constant $G$.  Thus $I=8G/\pi$ and the halved value in the original problem becomes
$\int_0^\infty\dfrac{\sinh x}{x\cosh^2x}~dx=4G/\pi.$
