# 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.

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.

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*}

• Dude that's dope! Thank you! – clathratus Oct 31 '18 at 3:31
• This substitution simplifies to $t=e^{-x}$. – J.G. Oct 31 '18 at 8:44
• how do you get from $\frac{\theta d\theta}{\sin\theta}$ to $$\frac{2\arctan(t)\frac{2dt}{1+t^2}}{\frac{2t}{1+t^2}}$$ With the substitution $t=\tan\frac\theta2$? – clathratus Nov 1 '18 at 23:06
• I keep getting $$\frac{\theta d\theta}{\sin\theta}=\frac{2\arctan t}{t\sqrt{t^2+1}}dt$$ – clathratus Nov 1 '18 at 23:10
• @clathratus This is the universal trigonometric substitution – user10354138 Nov 1 '18 at 23:25

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}

• I'm a fan of the alternative approach. Thank you! – clathratus Oct 31 '18 at 14:59

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

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

• This is really cool. I'll have to look at this one a little more in order to understand it better... But this is great. Thanks! – clathratus Oct 31 '18 at 18:37
• You seem to link any special function with any other special function. The link between elliptic integral and Legendre function is pretty smart. +1 – Paramanand Singh Nov 1 '18 at 16:41
• How do we show $$\int_{\Bbb R}\frac{\mathrm dx}{\cosh(x)^{2k+1}}=\frac{\pi{2k\choose k}}{4^k}$$ It looks a lot like a Beta integral – clathratus Dec 21 '18 at 0:56
• @clathratus: it is, indeed, it is enough to set $\frac{1}{\cosh(x)}=u$. – Jack D'Aurizio Dec 21 '18 at 10:48