Feynman's trick VI

Interesting integral involving the integral representation of $\sec(x)$

We prove the following result using Feynman's trick and the integral respresentation of $\sec(x)$ \[\int_{0}^{\frac{\pi}{2}} \operatorname{arctanh}^2(\sin(x)) dx = \frac{\pi^3}{8}\]
Proof
\begin{align*} S=\int_{0}^{\frac{\pi}{2}} \operatorname{arctanh}^2(\sin(x)) dx = & \int_{0}^{1} \frac{\operatorname{arctanh}^2 w}{\sqrt{1-w^2}} dw \quad (w \mapsto \sin(x))\\ =& \frac{1}{4} \int_{0}^{1} \frac{\ln^2\left(\frac{1+w}{1-w}\right)}{\sqrt{1-w^2}} dw \\ =& \frac{1}{4} \int_{1}^{\infty} \frac{\ln^2(r) }{\sqrt{r}(1+r)} dr \quad \left(r \mapsto \frac{1+w}{1-w}\right)\\ =& \frac{1}{4} \int_{0}^{1} \frac{\ln^2(s) }{\sqrt{s}(1+s)} ds \quad (s \mapsto \frac{1}{r})\\ \end{align*} \[ \therefore S = \frac{1}{4} \int_{0}^{\infty} \frac{\ln^2(r) }{\sqrt{r}(1+r)} dr - \underbrace{\frac{1}{4} \int_{0}^{1} \frac{\ln^2(r) }{\sqrt{r}(1+r)} dr}_{S} = \underbrace{\frac{1}{4} \int_{1}^{\infty} \frac{\ln^2(r) }{\sqrt{r}(1+r)} dr}_{S}\] \begin{align*} \therefore 2S =& \frac{1}{4} \int_{0}^{\infty} \frac{\ln^2(r) }{\sqrt{r}(1+r)} dr\\ =& \frac{1}{4} \int_{0}^{\infty} \frac{ \frac{d^2}{dt^2}\Big|_{t=0+} r^t }{\sqrt{r}(1+r)} dr\\ =& \frac{1}{4} \frac{d^2}{dt^2}\Big|_{t=0+} \int_{0}^{\infty} \frac{ r^{t-\frac{1}{2}} }{(1+r)} dr\\ \end{align*} Recall the integral representation of $\sec(x)$ (Proof in the appendix at the end): \[\sec(x) = \frac{2}{\pi} \int_{0}^{\infty} \frac{y^{\frac{2x}{\pi}}}{y^2+1} dy \; \underset{v=y^2}{\Longrightarrow}\; \pi \sec(\pi x) = \int_{0}^{\infty} \frac{v^{x-\frac{1}{2}}}{v+1}dv \] Therefore \begin{align*} S = \frac{1}{8} \frac{d^2}{dt^2}\Big|_{t=0+} \int_{0}^{\infty} \frac{ r^{t-\frac{1}{2}} }{(1+r)} dr =& \frac{1}{8} \frac{d^2}{dt^2}\Big|_{t=0+} \pi \sec(\pi t)\\ =& \lim_{t \to 0+} \frac{\pi^3}{8}\sec(\pi t)\left[\tan^2(\pi t) + \sec^2(\pi t)\right]\\ =& \frac{\pi^3}{8} \end{align*} \[\boxed{\int_{0}^{\frac{\pi}{2}} \operatorname{arctanh}^2(\sin(x)) dx = \frac{\pi^3}{8}}\]

Appendix: Integral representation of $\sec x$

How do we know that $\displaystyle \sec x= \frac{2}{\pi}\int_{0}^{\infty}\frac{y^\frac{2x}{\pi}}{1+y^2}dy $?

The proof can be obtained trough contour integration in a branch cut but here we will use a more straigthforward approach with the help of the Ramanujan's master theorem

Let $|a|<1$
Recall \[\frac{1}{1+y^2} = \int_{0}^{1} t^{y^2} dt \] Therefore \begin{align*} \int_{0}^{\infty}\frac{y^a}{1+y^2}dy =& \int_{0}^{\infty}\int_{0}^{1} y^a t^{y^2} dtdy\\ =&\int_{0}^{1} \int_{0}^{\infty} y^a t^{y^2} dydt\\ =& \frac{1}{2}\int_{0}^{1} \int_{0}^{\infty} w^{\frac{a}{2}-\frac{1}{2}} t^{w} dwdt \quad (w \mapsto y^2) \end{align*} The integral $\displaystyle \int_{0}^{\infty} w^{\frac{a}{2}-\frac{1}{2}} t^{w} dw$ is the Mellin transform of $\displaystyle t^w$ at $\displaystyle \frac{a+1}{2}$
Recall that \[t^w = e^{w\ln(t)} = \sum_{n=0}^{\infty} \frac{\ln^n(t)w^n}{n!} = \sum_{n=0}^{\infty} \frac{(-\ln(t))^n(-w^n)}{n!} \] Therefore, by Ramanujan's master theorem \[ \int_{0}^{\infty} w^{\frac{a}{2}-\frac{1}{2}} t^{w} dw = \Gamma\left(\frac{a+1}{2}\right)\left[-\ln(t)\right]^{-\frac{a+1}{2}} \] Therefore \begin{align*} \frac{1}{2}\int_{0}^{1} \int_{0}^{\infty} w^{\frac{a}{2}-\frac{1}{2}} t^{w} dwdt =& \frac{1}{2}\int_{0}^{1} \Gamma\left(\frac{a+1}{2}\right)\left[-\ln(t)\right]^{-\frac{a+1}{2}} dt\\ =& \frac{1}{2}\Gamma\left(\frac{a+1}{2}\right)\int_{0}^{1}\left[-\ln(t)\right]^{-\frac{a+1}{2}} dt\\ =& \frac{1}{2}\Gamma\left(\frac{a+1}{2}\right)\int_{0}^{1}v^{-\frac{a+1}{2}}e^{-v} dv \quad (v\mapsto -ln(t))\\ =& \frac{1}{2}\Gamma\left(\frac{a+1}{2}\right)\Gamma\left(\frac{1-a}{2}\right) \end{align*} Finally, recall the Euler's reflection formula \[\Gamma(z)\Gamma(1-z)= \frac{\pi}{\sin \pi z} = \pi \csc \pi z \] Therefore \[\int_{0}^{\infty}\frac{y^a}{1+y^2}dy = \frac{1}{2}\Gamma\left(\frac{a+1}{2}\right)\Gamma\left(\frac{1-a}{2}\right) = \frac{\pi}{2} \csc\left(\pi \frac{1+a}{2}\right)=\frac{\pi}{2} \sec\left(\pi \frac{a}{2}\right)\] If we let $\displaystyle x = \pi \frac{a}{2}$ then $\displaystyle a=\frac{2x}{\pi}$. Then \[\boxed{\frac{2}{\pi}\int_{0}^{\infty}\frac{y^\frac{2x}{\pi}}{1+y^2}dy = \sec x}\]