When I came across the integral
$$J=\int_0^{\infty} \frac{\operatorname{artanh}\left(\frac{1}{\sqrt{1+x^2}}\right)}{\sqrt{1+x^2}} d x=\frac{\pi^2}{4} $$
whose answer is surprisingly decent, I, as usual, put $x=\tan \theta$ and transform the integral into
$$
\begin{aligned}
J & =\int_0^{\frac{\pi}{2}} \frac{\operatorname{artanh}\left(\frac{1}{\sec \theta}\right)}{\sec \theta} \sec ^2 \theta d \theta \\
& =\int_0^{\frac{\pi}{2}} \sec \theta \operatorname{artanh}\left(\frac{1}{\sec \theta}\right) d \theta
\end{aligned}
$$
Feynman’s trick reminds me to deal with its parametrized integral
$$
J(a)= \int_0^{\infty} \frac{\operatorname{artanh}\left(\frac{a}{\sqrt{1+x^2}}\right)}{\sqrt{1+x^2}} d x =\int_0^{\frac{\pi}{2}} \sec \theta \operatorname{artanh} \left(\frac{a}{\sec \theta}\right) d \theta
$$
with $|a|<1$ and $J(0)=0$.
Consequently, differentiation under integral make our life easier as
$$
\begin{aligned}
J^{\prime}(a)&=\int_0^{\frac{\pi}{2}} \frac{1}{1-\frac{a^2}{\sec ^2 \theta}} d \theta \\& =\int_0^{\frac{\pi}{2}} \frac{\sec ^2 \theta}{\sec ^2 \theta-a^2} d \theta \\
& =\int_0^{\frac{\pi}{2}} \frac{d(\tan \theta)}{\left(1-a^2\right)+\tan ^2 \theta} \\
& =\frac{1}{\sqrt{1-a^2}}\left[\operatorname{artan}\left(\frac{\tan \theta}{\sqrt{1-a^2}}\right)\right]_0^{\frac{\pi}{2}} \\
& =\frac{\pi}{2 \sqrt{1-a^2}} \\
&
\end{aligned}
$$
Integrating back with $J(0)=0$ yields
$$
\boxed{J(a)= \int_0^{\infty} \frac{\operatorname{artanh}\left(\frac{a}{\sqrt{1+x^2}}\right)}{\sqrt{1+x^2}} d x=\frac{\pi}{2} \int \frac{1}{\sqrt{1-a^2}} d a=\frac{\pi}{2} \operatorname{arcsin}a \,} \tag*{(1)}
$$
In particular, let $a$ approach to $1$, we get
$$
J= \lim _{a \rightarrow 1} J(a)= \lim _{a \rightarrow 1}\frac{\pi}{2} \operatorname{arcsin} 1=\frac{\pi^2}{4}
$$
Inspired by $J(a)$, I believe that whenever $\left|\frac ab \right| <1$, we can further evaluate
$$
\int_0^{\infty} \frac{\operatorname{artanh}\left(\frac{a}{\sqrt{b^2+x^2}}\right)}{\sqrt{b^2+x^2}} d x
$$
by simply letting $x\mapsto bx$ which transforms
$$
\boxed{\int_0^{\infty} \frac{\operatorname{artanh}\left(\frac{a}{\sqrt{b^2+x^2}}\right)}{\sqrt{b^2+x^2}} d x=\int_0^{\infty} \frac{\operatorname{artanh}\left(\frac{\frac{a}{b}}{\sqrt{1+x^2}}\right)}{\sqrt{1+x^2}} d x=J\left(\frac{a}{b}\right )= \frac{\pi}{2} \operatorname{arcsin}\left(\frac{a}{b}\right) \,}
$$
Latest Edit
:After submitting the question, I found that there is a simpler version with Feynman’s trick to share with you.
Considering $b$ is a constant and let
$$\displaystyle J(a)=\int_0^{\infty} \frac{\operatorname{artanh}\left(\frac{a}{\sqrt{b^2+x^2}}\right)}{\sqrt{b^2+x^2}} d x\tag*{} $$
where $J(0)=0$.
Differentiating $J(a)$ w.r.t. $a$ yields
$$\displaystyle \begin{aligned}J^{\prime}(a) & =\int_0^{\infty} \frac{1}{\left(1-\frac{a^2}{b^2+x^2}\right)\left(b^2+x^2\right)} d x \\& =\int_0^{\infty} \frac{1}{b^2-a^2+x^2} d x \\& =\frac{1}{\sqrt{b^2-a^2}}\left[\tan ^{-1}\left(\frac{x}{\sqrt{b^2-a^2}}\right)\right]_0^{\infty} \\& =\frac{\pi}{2 \sqrt{b^2-a^2}}\end{aligned}\tag*{} $$
Integrating back w.r.t. $a$ with $J(0)=0$ yields
$$\displaystyle \boxed{ \begin{aligned}J(a) & =\frac{\pi}{2} \int \frac{d a}{\sqrt{b^2-a^2}} =\frac{\pi}{2} \arcsin \left(\frac{a}{b}\right)\end{aligned}}\tag*{} $$
Can it be done without Feynman’s trick?
Your comments and alternative methods are highly appreciated.
Best Answer
Substituting $x = \operatorname{csch} t$ and noting that $ \frac{1}{\sqrt{x^2+1}} = \tanh t ,$ the integral reduces to
\begin{align*} J &= \int_{0}^{\infty} \frac{t}{\sinh t} \, \mathrm{d} t \\ &= 2 \sum_{n=0}^{\infty} \int_{0}^{\infty} t e^{-(2n+1)t} \, \mathrm{d}t \\ &= 2 \sum_{n=0}^{\infty} \frac{1}{(2n+1)^2} \\ &= \frac{\pi^2}{4}. \end{align*}
Also, for $|\alpha| < 1$, OP's substitution shows that
\begin{align*} J(\alpha) &= \int_{0}^{\frac{\pi}{2}} \frac{\operatorname{artanh}(\alpha \sin\theta)}{\sin \theta} \, \mathrm{d} \theta \\ &= \sum_{n=0}^{\infty} \frac{\alpha^{2n+1}}{2n+1} \int_{0}^{\frac{\pi}{2}} \sin^{2n}\theta \, \mathrm{d} \theta \\ &= \sum_{n=0}^{\infty} \frac{\alpha^{2n+1}}{2n+1} \cdot (-1)^n \frac{\pi}{2} \binom{-1/2}{n} \\ &= \frac{\pi}{2} \int_{0}^{\alpha} \frac{\mathrm{d}t}{\sqrt{1 - t^2}} \\ &= \frac{\pi}{2} \arcsin \alpha. \end{align*}