value of $\int_{-\infty}^{\infty}\arcsin\frac1{\cosh x}\,dx$ I want to know the value of
$$I=\int_{-\infty}^{\infty}\arcsin\frac1{\cosh x}\,dx$$
The Symbolab integral calculator says that the integral diverges, but when one graphs it obvious that it converges. So what is the value? 
I was thinking that I might try Feynman integration, but I can't think of the right substitution.
Alert! Alert! I've found an antiderivative!
From the answer provided by @user10354138, we can reach 
$$\int\arcsin\frac1{\cosh x}dx=i\operatorname{Li}_2(i\phi)-i\operatorname{Li}_2(-i\phi)+C$$
Where $$\phi=\tan\bigg(\frac12\arcsin\frac1{\cosh x}\bigg)$$
And $$\operatorname{Li}_2(z)=\sum_{n\geq1}\frac{z^n}{n^2}$$
is the Di-logarithm.
 A: Yet another alternative approach: once shown that
$$ \int_{\mathbb{R}}\frac{dx}{\cosh(x)^{2k+1}} = \frac{\pi \binom{2k}{k}}{4^k}\tag{1}$$
and recalled that
$$ \arcsin z = \sum_{k\geq 0}\frac{\binom{2k}{k}}{4^k(2k+1)}z^{2k+1} \tag{2} $$
we have the following identity:
$$ \int_{\mathbb{R}}\arcsin\frac{1}{\cosh x}\,dx = \pi\sum_{k\geq 0}\frac{1}{2k+1}\left[\frac{1}{4^k}\binom{2k}{k}\right]^2.\tag{3} $$
Now we may invoke a function whose Maclaurin series involves squared central binomial coefficients, namely the complete elliptic integral of the first kind $K(x)$, here denoted according to Mathematica's notation (the argument of $K$ is the elliptic modulus):
$$ \sum_{k\geq 0}\left[\frac{1}{4^k}\binom{2k}{k}\right]^2 x^{2k}=\frac{2}{\pi}K(x^2)\tag{4} $$
leading to:
$$ \int_{\mathbb{R}}\arcsin\frac{1}{\cosh x}\,dx = 2\int_{0}^{1} K(x^2)\,dx = \int_{0}^{1}\frac{K(x)}{\sqrt{x}}\,dx.\tag{5} $$
At last, we recall that both $K(x)$ and $\frac{1}{\sqrt{x}}$ have fairly simple Fourier-Legendre series expansions:
$$ K(x)=2\sum_{n\geq 0}\frac{P_n(2x-1)}{2n+1},\qquad \frac{1}{\sqrt{x}}=2\sum_{n\geq 0}(-1)^n P_n(2x-1) $$
and by the orthogonality of shifted Legendre polynomials

$$ \int_{\mathbb{R}}\arcsin\frac{1}{\cosh x}\,dx =\int_{0}^{1}\frac{K(x)}{\sqrt{x}}\,dx = 4\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)^2} = 4G.\tag{6} $$

A: Alternatively, you can integrate by parts:
\begin{align}
\int \limits_{-\infty}^\infty \arcsin(\operatorname{sech}(x)) \, \mathrm{d} x &= 2 \int \limits_0^\infty \arcsin(\operatorname{sech}(x)) \, \mathrm{d} x \\
&= 2x \arcsin(\operatorname{sech}(x)) \Bigg \rvert_{x=0}^{x=\infty} - 2 \int \limits_0^\infty x \frac{- \sinh(x) \operatorname{sech}^2(x)}{\sqrt{1-\operatorname{sech}^2(x)}} \, \mathrm{d} x \\
&= 2 \int \limits_0^\infty \frac{x}{\cosh(x)} \, \mathrm{d} x = 4 \sum \limits_{n=0}^\infty (-1)^n \int \limits_0^\infty x \, \mathrm{e}^{-(2n+1) x} \, \mathrm{d} x \\
&= 4 \Gamma(2) \sum \limits_{n=0}^\infty \frac{(-1)^n}{(2n+1)^2} = 4 \mathrm{G} \, .
\end{align}
A: Wolfy says it is 4 times the Catalan's constant.
One (not optimal) way to derive this is
$$\def\sech{\operatorname{sech}}
\begin{align*}
\int_{-\infty}^\infty\arcsin\sech x\,\mathrm{d}x&=2\int_0^\infty\arcsin\sech x\,\mathrm{d}x\\
&=2\int_0^1\frac{\arcsin u\,\mathrm{d}u}{u\sqrt{1-u^2}}\quad(u=\sech x)\\
&=2\int_0^{\pi/2}\frac{\theta\,\mathrm{d}\theta}{\sin\theta}\quad(u=\sin\theta)\\
&=2\int_0^1\frac{2\tan^{-1}t\,\frac{2\,\mathrm{d}t}{1+t^2}}{\frac{2t}{1+t^2}}\quad(t=\tan\tfrac12\theta)\\
&=4\int_0^1\frac{\tan^{-1}t}{t}\,\mathrm{d}t\\
&=4\int_0^1\sum_{n=0}^\infty\frac{(-1)^n}{2n+1}t^{2n}\,\mathrm{d}t\\
&=4\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^2}=4G\\
\end{align*}
$$
