Evaluate $\int_{0}^{1} \frac{\operatorname{arsinh}x}{x\sqrt{1-x^4}} \,\mathrm{d}x$

Question

The problem is to show $$\int_{0}^{1} \frac{\operatorname{arsinh}x}{x\sqrt{1-x^4}} \,\mathrm{d}x = \frac{\pi^2}{8}$$ which is a typical integral emerges in many specific value, yet I have no more insight how to prove this one directly (any possible method other than the example below is acceptable).

Here is an example when the integral pops out in Legendre-chi function $\chi_2$, like (1) in this post, we have $$ \int_{0}^{\pi/2} {\arctan(\sin x) \,\mathrm{d}x} = 2\chi_2(\sqrt2-1) = \frac{\pi^2}{8} - \frac{\ln^{2}(\sqrt{2}+1)}{2} $$ which is a very typical integral, like in this post, from where we already know that $$ \int_{0}^{\pi/2} {\arctan(\sin x) \,\mathrm{d}x} = \int_{0}^{1} {\frac{\arctan x}{\sqrt{1-x^2}} \,\mathrm{d}x} = \int_{0}^{1} {\frac{\operatorname{arsinh}x}{x\sqrt{1+x^{2}}} \,\mathrm{d}x} $$ or $$ \int_{0}^{\pi/4} {\arcsin(\tan x) \,\mathrm{d}x} = \frac{\pi^2}{8} - \int_{0}^{\pi/2} {\arctan(\sin x) \,\mathrm{d}x} = \frac{\ln^{2}(\sqrt{2}+1)}{2} $$ Now, for $a<1$ consider the following parameterization $$ I(a) = \int_{0}^{\pi/4} {\arcsin(a\tan x) \,\mathrm{d}x} $$ let $u=\tan x$ and $y^2=(1-a^2u^2)/(1+u^2)$ in its derivative $$ \begin{aligned} I'(a) &= \int_{0}^{\pi/4} {\frac{\tan x}{\sqrt{1-a^2\tan^2x}} \,\mathrm{d}x} = \int_{0}^{1} {\frac{u}{(1+u^2)\sqrt{1-a^2u^2}} \,\mathrm{d}u} \\ &= \frac1{\sqrt{1+a^2}}\int_{\sqrt{(1-a^2)/2}}^{1} {\frac{\mathrm{d}y}{\sqrt{a^2+y^2}}} = \frac1{\sqrt{1+a^2}}\operatorname{arsinh}\left(\frac{y}{a}\right)\biggr|_{y=\sqrt{(1-a^2)/2}}^{1} \\ &= \frac{1}{\sqrt{1+a^{2}}} \operatorname{arsinh}\frac1{a} - \frac{1}{\sqrt{1+a^{2}}} \operatorname{arsinh}\sqrt{\frac{1-a^2}{2a^2}} \end{aligned} $$ Integrtation by parts gives $$ \begin{aligned} \int_{0}^{\pi/4} {\arcsin(\tan x) \,\mathrm{d}x} &= \int_{0}^{1} \frac{1}{\sqrt{1+a^{2}}} \operatorname{arsinh}\frac1{a} \,\mathrm{d}x - \int_{0}^{1} \frac{1}{\sqrt{1+a^{2}}} \operatorname{arsinh}\sqrt{\frac{1-a^2}{2a^2}} \,\mathrm{d}x \\ &= \operatorname{arsinh}^2(1) + \int_{0}^{1} {\frac{\operatorname{arsinh}a}{a\sqrt{1+a^{2}}}\,\mathrm{d}a} - \int_{0}^{1} \frac{\operatorname{arsinh}a}{a\sqrt{1-a^4}} \,\mathrm{d}a \end{aligned} $$ Combining all the special case of $\chi_2(\sqrt2-1)$ from above, we will find the required integral. And there are other equivalent form of required integral, for example $$ \int_{0}^{1} \frac{\operatorname{arsinh}x}{x\sqrt{1-x^4}} \,\mathrm{d}x = \frac1{2} \int_{0}^{1} \frac{1}{\sqrt{1+x^2}} \operatorname{arcosh}\frac1{x^2} \,\mathrm{d}x = \frac1{4} \int_{0}^{1} \frac{1}{\sqrt{x(1+x)}} \operatorname{arcosh}\frac1{x} \,\mathrm{d}x $$ Yet, I still can not find any convenience for further calculation.

Thanks in advance for any help.

Answer 1

Utilize $$\int_{0}^{\pi/2}\frac{\sin t}{1+x^2\sin^2 t}dt = \frac{\text{arcsinh}{\>x}}{x\sqrt{1+x^2}} $$ to integrate \begin{align} &\int_{0}^{1} \frac{\operatorname{arsinh}x}{x\sqrt{1-x^4}} {d}x \\ =&\int_0^1 \int_0^{\pi/2}\frac{\sin t}{\sqrt{1-x^2}(1+x^2\sin^2 t)}dt\ dx \\ =& \ \frac\pi2 \int_0^{\pi/2}\frac{\sin t}{\sqrt{1+\sin^2t}}dt = \frac{\pi^2}{8} \end{align}

Answer 2

$$\int_0^1 \frac{\operatorname{arsinh}x}{x\sqrt{1-x^4}}dx\overset{IBP}=\frac12\int_0^1 \frac{\operatorname{arctanh}\left(\sqrt{1-x^2}\sqrt{1+x^2}\right)}{\sqrt{1+x^2}}dx$$

$$=\frac12\int_0^1 \int_0^\sqrt{1-x^2}\frac{1}{1-(1+x^2)y^2}dydx=\frac12\int_0^1 \int_0^\sqrt{1-y^2}\frac{1}{1-y^2(1+x^2)}dxdy$$

$$=\frac12\int_0^1 \frac{\operatorname{arctanh} y}{y\sqrt{1-y^2}}dy\overset{y\to \frac{1-y}{1+y}}=-\frac14 \int_0^1 \frac{\ln y}{\sqrt y (1-y)}dy\overset{\sqrt y \to y}=-\int_0^1 \frac{\ln y}{1-y^2}dy$$

$$=-\sum_{n=0}^\infty \int_0^1 y^{2n}\ln y\, dy=\sum_{n=0}^\infty \frac{1}{(2n+1)^2}=\frac34\sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{8}$$

Answer 3

Feynman’s Trick

Noting that $$ I=\int_0^1 \frac{\sinh ^{-1} x}{x \sqrt{1-x^4}} d x=\int_0^1 \frac{\tanh ^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right)}{x \sqrt{1-x^4}} d x $$ Let’s consider the parametrised integral $$ I(a)=\int_0^1 \frac{\tanh ^{-1} \frac{x a}{\sqrt{1+x^2}}}{x \sqrt{1-x^4}} d x $$ Differentiating $I(a)$ w.r.t. $a$ yields $$ \begin{aligned} I^{\prime}(a) & =\int_0^1 \frac{1}{1-\frac{x^2 x^2}{1+x^2}} \frac{d x}{\left(1+x^2\right) \sqrt{1-x^2}} \\ & =\int_0^1 \frac{1}{\left(1-a^2\right) x^2+1} \frac{d x}{\sqrt{1-x^2} } \\ & =\frac{\pi}{2 \sqrt{2-a^2}} \quad (\textrm{ via }x\mapsto \sin x) \end{aligned} $$ Integrating $I’(a)$ from $a=0$ to $1$ brings us $$ \boxed{I=\frac{\pi}{2} \int_0^1 \frac{d a}{\sqrt{2-a^2}}=\frac{\pi}{2} \cdot \frac{\pi}{4}=\frac{\pi^2}{8}} $$

Answer 4

With a somewhat roundabout exchange of variables and integration by parts, we can obtain the following equivalent forms:

$$\newcommand{\arsinh}{\operatorname{arsinh}} \newcommand{\artanh}{\operatorname{artanh}} \begin{align*} I &= \int_0^1 \frac{\arsinh x}{x \sqrt{1-x^4}} \, dx \\ &= \int_0^1 \frac{\artanh \frac x{\sqrt{1+x^2}}}{x \sqrt{1-x^4}} \, dx \\ &= \int_0^\tfrac1{\sqrt2} \frac{\artanh y}{y\sqrt{1-2y^2}} \, dy & y=\frac x{\sqrt{1+x^2}} \\ &= \int_0^\tfrac1{\sqrt2} \frac{\artanh \sqrt{1-2y^2}}{1-y^2} \, dy & \rm IBP \\ &= \sqrt2 \int_0^1 \frac{z \artanh z}{\sqrt{1-z^2} (1+z^2)} \, dz & z=\sqrt{1-2y^2} \\ &= \sqrt2 \int_0^\tfrac\pi2 \frac{\sin w \log\frac{1+\sin w}{\cos w}}{1+\sin^2w} \, dw & w=\arcsin z \\ &= \sqrt2 \int_1^\infty \frac{v^2-1}{v^4+1} \log v \, dv & v=\sec w+\tan w \\ &= \sqrt2 \int_0^1 \frac{u^2-1}{u^4+1} \log u \, du & u=\frac1v \\ \end{align*}$$

the last of which can be evaluated to trigammas by exploiting Taylor series, integrating by parts once more, and recalling the trigamma reflection formula.

$$\begin{align*} I &= \sqrt2 \sum_{n\ge0} (-1)^n \int_0^1 \left(u^{4n+2} - u^{4n}\right) \log u \, du \\ &= \sqrt2 \sum_{n\ge0} (-1)^n \left[\frac1{(4n+1)^2} - \frac1{(4n+3)^2}\right] \\ &= \frac1{32\sqrt2} \sum_{n\ge0} \left[\frac1{\left(n+\frac18\right)^2} - \frac1{\left(n+\frac38\right)^2} - \frac1{\left(n+\frac58\right)^2} + \frac1{\left(n+\frac78\right)^2}\right] \\ &= \frac{\left[\psi'\left(\frac18\right) + \psi'\left(\frac78\right)\right] - \left[\psi'\left(\frac38\right) + \psi'\left(\frac58\right)\right]}{32\sqrt2} \\ &= \frac{\pi^2}{32\sqrt2} \left(\csc^2\frac\pi8 - \csc^2\frac{3\pi}8\right) = \boxed{\frac{\pi^2}8} \end{align*}$$

since $\sin\dfrac\pi8=\dfrac{\sqrt{2-\sqrt2}}2$.