Gradshteyn & Ryzhik $3.535$ states that
$$\int_0^1 \frac{x \operatorname{csch}(a x)}{\sqrt{\cosh (2 a)-\cosh (2 a x)}} \mathrm dx=\frac{\pi \sin ^{-1}(\tanh (a))}{2 \sqrt{2} a^2 \sinh (a)}$$
Holds for $a>0$. I found this result neat enough but can't prove it so far. Any help is appreciated!
Best Answer
Let $f \colon (0,\infty) \to (0,\infty)$, \begin{align} f(a) &= \int \limits_0^1 \frac{x \operatorname{csch}(a x)}{\sqrt{\cosh(2a) - \cosh(2ax)}} \, \mathrm{d} x = \int \limits_0^1 \frac{x \operatorname{csch}(a x)}{\sqrt{2 \left[\frac{1}{1-\tanh^2(a)} - \frac{1}{1-\tanh^2(ax)}\right]}} \, \mathrm{d} x \\ &=\frac{1}{\sqrt{2} \sinh(a)} \int \limits_0^1 \frac{x}{\tanh(ax) \sqrt{1 - \frac{\tanh^2(ax)}{\tanh^2(a)}} \cosh^2(ax)} \, \mathrm{d} x \\ &\!\!\!\!\!\!\!\stackrel{u = \frac{\tanh(ax)}{\tanh(a)}}{=} \frac{1}{\sqrt{2} a^2 \sinh(a)} \int \limits_0^1 \frac{\operatorname{artanh}(\tanh(a) u)}{u \sqrt{1-u^2}} \, \mathrm{d} u \equiv \frac{g(\tanh(a))}{\sqrt{2} a^2 \sinh(a)} \, . \end{align} Here we have introduced $g \colon [0,1] \to [0,\infty)$, $$ g(b) = \int \limits_0^1 \frac{\operatorname{artanh}(b u)}{u \sqrt{1-u^2}} \, \mathrm{d} u \, . $$ We can differentiate under the integral sign to obtain \begin{align} g'(b) &= \int \limits_0^1 \frac{\mathrm{d}u}{\sqrt{1-u^2}(1-b^2 u^2)} \stackrel{u = \frac{1}{\sqrt{1+v^2}}}{=} \int \limits_0^\infty \frac{\mathrm{d} v}{1 - b^2 + v^2} = \frac{\pi}{2 \sqrt{1-b^2}} \end{align} for $b \in (0,1)$, which implies $$ g(b) = g(0) + \int \limits_0^b g'(c) \mathrm{d} c = 0 + \frac{\pi}{2} \int \limits_0^b \frac{\mathrm{d} c}{\sqrt{1-c^2}} = \frac{\pi}{2} \arcsin(b) $$ for $b \in [0,1]$. Therefore, $$ f(a) = \frac{g(\tanh(a))}{\sqrt{2} a^2 \sinh(a)} = \frac{\pi \arcsin(\tanh(a))}{2 \sqrt{2} a^2 \sinh(a)}$$ holds for $a > 0$.