Krdytkyu

I want to know the value of
$$I=int_{-infty}^{infty}arcsinfrac1{cosh x},dx$$
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
$$intarcsinfrac1{cosh x}dx=ioperatorname{Li}_2(iphi)-ioperatorname{Li}_2(-iphi)+C$$
Where $$phi=tanbigg(frac12arcsinfrac1{cosh x}bigg)$$
And $$operatorname{Li}_2(z)=sum_{ngeq1}frac{z^n}{n^2}$$
is the Di-logarithm.

Wolfy says it is 4 times the Catalan's constant.

One (not optimal) way to derive this is
$$defsech{operatorname{sech}}
begin{align*}
int_{-infty}^inftyarcsinsech x,mathrm{d}x&=2int_0^inftyarcsinsech x,mathrm{d}x\
&=2int_0^1frac{arcsin u,mathrm{d}u}{usqrt{1-u^2}}quad(u=sech x)\
&=2int_0^{pi/2}frac{theta,mathrm{d}theta}{sintheta}quad(u=sintheta)\
&=2int_0^1frac{2tan^{-1}t,frac{2,mathrm{d}t}{1+t^2}}{frac{2t}{1+t^2}}quad(t=tantfrac12theta)\
&=4int_0^1frac{tan^{-1}t}{t},mathrm{d}t\
&=4int_0^1sum_{n=0}^inftyfrac{(-1)^n}{2n+1}t^{2n},mathrm{d}t\
&=4sum_{n=0}^inftyfrac{(-1)^n}{(2n+1)^2}=4G\
end{align*}
$$

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}

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
$$ arcsin z = sum_{kgeq 0}frac{binom{2k}{k}}{4^k(2k+1)}z^{2k+1} tag{2} $$
we have the following identity:
$$ int_{mathbb{R}}arcsinfrac{1}{cosh x},dx = pisum_{kgeq 0}frac{1}{2k+1}left[frac{1}{4^k}binom{2k}{k}right]^2.tag{3} $$
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):
$$ sum_{kgeq 0}left[frac{1}{4^k}binom{2k}{k}right]^2 x^{2k}=frac{2}{pi}K(x^2)tag{4} $$
leading to:
$$ int_{mathbb{R}}arcsinfrac{1}{cosh x},dx = 2int_{0}^{1} K(x^2),dx = int_{0}^{1}frac{K(x)}{sqrt{x}},dx.tag{5} $$
At last, we recall that both $K(x)$ and $frac{1}{sqrt{x}}$ have fairly simple Fourier-Legendre series expansions:
$$ K(x)=2sum_{ngeq 0}frac{P_n(2x-1)}{2n+1},qquad frac{1}{sqrt{x}}=2sum_{ngeq 0}(-1)^n P_n(2x-1) $$
and by the orthogonality of shifted Legendre polynomials

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