TITLE: value of $\int_{-\infty}^{\infty}\arcsin\frac1{\cosh x}\,dx$
QUESTION [5 upvotes]: 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.

REPLY [7 votes]: 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

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