Is it very hard to evaluate $\int_0^1 \frac{\operatorname{arcsinh} x \arccos x}{\sqrt{x^2+1}} d x$ without using series representation?

Question

The exact value of the integral $$ I=\int_0^1 \frac{\operatorname{arcsinh} x \arccos x}{\sqrt{x^2+1}} d x= \frac{\pi^3}{96} $$ prompted me to try all methods I know, but in vain. After few days, integration by parts came to my mind that $$ \begin{aligned} I & =\int_0^1\frac{\operatorname{arcsinh} x \arccos x}{\sqrt{x^2+1}} d x \\ & =\frac{1}{2} \int_0^1 \arccos x d\left(\operatorname{arcsinh}^2 x\right) \\ & =\frac{1}{2} \int_0^1 \frac{\operatorname{arcsinh}{ }^2 x}{\sqrt{1-x^2}} d x , \end{aligned} $$ which lead me to use the usual substitution $x=\sin \theta$.
$$ I=\frac{1}{2} \int_0^ {\frac \pi 2} \operatorname{arcsinh}^2(\sin \theta) d \theta, $$ by which I was stuck for couples of hours. I was forced to use series representation of $$\operatorname{arcsinh}^2 x= \sum_{k=0}^{\infty} \frac{(-4)^k(k!)^2}{(1+k)(1+2 k)!} x ^{2+2 k} .$$ Then $$ \begin{aligned} I & =\frac{1}{2} \int_0^ {\frac \pi 2} \sum_{k=0}^{\infty} \frac{(-4)^k(k!)^2}{(1+k)(1+2 k)!} \sin ^{2+2 k} \theta d \theta\\ & =\frac{1}{4} \sum_{k=0}^{\infty} \frac{(-4)^k(k!)^2}{(1+k)(1+2 k)!} B\left(k+\frac{3}{2}, \frac{1}{2}\right) \\ & =\frac{1}{4} \sum_{k=0}^{\infty} \frac{(-4)^k(k!)^2}{(1+k)(1+2 k)!} \cdot \frac{\Gamma\left(k+\frac{3}{2}\right) \Gamma\left(\frac{1}{2}\right)}{\Gamma(k+2)} \\ & =\frac{1}{4} \sum_{k=0}^{\infty} \frac{(-4)^k(k!)^2}{(1+k)(1+2 k)!} \frac{\left(k+\frac{1}{2}\right)\left(k-\frac{1}{2}\right) \cdots \frac{1}{2!} \Gamma^2\left(\frac{1}{2}\right)}{(k+1)!} \\ & =\frac{\pi}{4} \sum_{k=0}^{\infty} \frac{(-1)^k 2^{2 k}(k!)^2}{(1+k)(1+2 k)!} \cdot \frac{(2 k+1)(2 k-1) \cdots 1}{(k+1)!2^{k+1}}\\& =\frac{\pi}{4} \sum_{k=0}^{\infty} \frac{(-1)^k 2^{2 k}(k!)^2}{(1+k)(k+1)!2^{k+1}(2 k)(2(k-1)) \cdots \cdot 2}\\&= \frac{\pi}{8} \sum_{k=0}^{\infty} \frac{(-1)^k}{(k+1)^2}\\&=\frac{\pi^3}{96} \end{aligned} $$

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My Question: Can we evaluate the integral without series representation?

Your comments and alternatives are highly appreciated.

Answer 1

Integrate by parts, followed by $x=\cos t$

\begin{aligned} I & =\int_0^1\frac{\sinh^{-1} x \cos^{-1} x}{\sqrt{x^2+1}} d x =\frac{1}{2} \int_0^{\pi/2} [\sinh^{-1}\cos t]^2\ d t \end{aligned} Then, let $K(a)=\frac12\int_0^{\pi/2}{[\sinh^{-1}(a\cos t)]^2}dt$ \begin{align} K’(a) =&\int_0^{\pi/2}\frac{\sinh^{-1}(a\cos t) \cos t }{\sqrt{1+a^2\cos^2t}} dt\\ =&\ \frac1a \int_0^a \bigg[\int_0^{\pi/2}\frac{y \sinh^{-1}(y\cos t) \cos t}{\sqrt{1+y^2\cos^2t}}\ dt\bigg]’_y dy\\ =& \ \frac1a \int_0^a \int_0^{\pi/2} \bigg[\frac{\sinh^{-1}(y\cos t)\cos t}{(1+y^2 \cos^2 t)^{3/2}}+\frac{y \cos^2 t}{1+y^2 \cos^2 t} \bigg]dt dy\\ =& \ \frac1a \int_0^a \int_0^{\pi/2} \bigg[\frac{\sinh^{-1}(y\cos t)}{1+y^2}\ \overset{ibp}{d_t}\bigg(\frac{\sin t}{\sqrt{1+y^2\cos^2t}}\bigg)\\ & \hspace{30mm}+\frac{y\cos^2 t}{1+y^2\cos^2t} dt\bigg] dy\\ =&\ \frac1a \int_0^a\int_0^{\pi/2}\frac {y}{1+y^2}dt dy= \frac\pi{4a}{\ln(1+a^2)} \end{align} Thus \begin{align} I=K(1) =\int_0^1 K’(a)da= \frac\pi4\int_0^1 \frac{\ln(1+a^2)}ada=\frac{\pi^3}{96} \end{align}

Answer 2

Although it may not be exactly what OP wants, here is a solution that only uses the power series for $\log(1+x)$. I humbly suspect that this approach may possibly be adapted to avoid series at all.

Starting from @Quanto's observation,

$$ \frac{\operatorname{arsinh} x}{\sqrt{1+x^2}} = \int_{0}^{\frac{\pi}{2}} \frac{x \sin \alpha}{1+x^2 \sin^2\alpha} \, \mathrm{d}\alpha, $$

integrating both sides with respect to $x$ yields

$$ \operatorname{arsinh}^2 x = \int_{0}^{\frac{\pi}{2}} \frac{\log(1+x^2 \sin^2\alpha)}{\sin\alpha} \, \mathrm{d}\alpha. $$

Consequently,

$$ \begin{align*} I &= \frac{1}{2}\int_{0}^{\frac{\pi}{2}} \int_{0}^{\frac{\pi}{2}} \frac{\log(1 +\sin^2\alpha \sin^2\beta)}{\sin\alpha} \, \mathrm{d}\alpha \mathrm{d}\beta \\ &= \operatorname{Re} \int_{0}^{\frac{\pi}{2}} \int_{0}^{\frac{\pi}{2}} \frac{\log(1 + i \sin\alpha \sin\beta)}{\sin\alpha} \, \mathrm{d}\alpha \mathrm{d}\beta. \end{align*} $$

Now consider the function $I(x)$ defined as:

$$ \begin{align*} I(x) &\triangleq \int_{0}^{\frac{\pi}{2}} \int_{0}^{\frac{\pi}{2}} \frac{\log(1 + x \sin\alpha \sin\beta)}{\sin\alpha} \, \mathrm{d}\alpha \mathrm{d}\beta \\ &= \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} x^n \int_{0}^{\frac{\pi}{2}} \int_{0}^{\frac{\pi}{2}} (\sin\alpha)^{n-1} (\sin\beta)^n \, \mathrm{d}\alpha \mathrm{d}\beta \\ &= \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} x^n \frac{B(\frac{n}{2},\frac{1}{2}) B(\frac{n+1}{2},\frac{1}{2})}{4} \\ &= \frac{\pi}{2} \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^2} x^n \\ &= -\frac{\pi}{2} \operatorname{Li}_2(-x). \end{align*} $$

(Although the series expansion is derived only for $|x| < 1$, the validity of the equality between $I(x)$ and the last line remains true for all $x \in \mathbb{C}\setminus (-\infty, -1]$ by the principle of analytic continuation.) This shows that

$$ I = \operatorname{Re}[ I(i)] = \frac{\pi}{2} \left( \frac{1}{2^2} - \frac{1}{4^2} + \frac{1}{6^2} - \cdots \right) = \frac{\pi^3}{96}. $$

Answer 3

Not a complete answer, but using @Quanto ideas in his solution at this link here

We can reduce the given integal as $$\eqalign{ & I = \int\limits_0^1 {\frac{{{\text{arcsinh}}\left( x \right)\arccos \left( x \right)}}{{\sqrt {1 + {x^2}} }}dx} = \int\limits_0^1 {x\arccos \left( x \right)\left( {\int\limits_0^{\frac{\pi }{2}} {\frac{{\sin \left( t \right)}}{{1 + {x^2}{{\sin }^2}\left( t \right)}}} dt} \right)dx} \cr & = \int\limits_0^{\frac{\pi }{2}} {\sin \left( t \right)\left( {\int\limits_0^1 {\frac{{x\arccos \left( x \right)}}{{1 + {x^2}{{\sin }^2}\left( t \right)}}dx} } \right)dt} = \frac{\pi }{2}\int\limits_0^{\frac{\pi }{2}} {\frac{1}{{\sin \left( t \right)}}\ln \left( {\frac{{\sqrt {1 + {{\sin }^2}\left( t \right)} + 1}}{2}} \right)dt} \cr} $$

Look promising but I can't process anymore.

Answer 4

Solution similar to Sangchul Lee’s

Being reminded to use Quanto’s idea, we can integrate both sides,w.r.t. $x$, $$ \frac{\operatorname{arcsinh} x}{\sqrt{1+x^2}}=\int_0^{\frac{\pi}{2}} \frac{x \sin \alpha}{1+x^2 \sin ^2 \alpha} d \alpha $$ to get $$ \operatorname{arcsinh}{ }^2 x=\int_0^{\frac{\pi}{2}} \frac{\ln \left(1+x^2 \sin ^2 \alpha\right)}{\sin \alpha} d \alpha, $$ and then use $$ \ln \left(1+x^2\right)=\sum_{n=1}^{\infty} \frac{(-1)^n x^{2n}}{n} $$ we have$$ \begin{aligned} I & =\frac{1}{2} \int_0^{\frac{\pi}{2}} \int_0^{\frac{\pi}{2}} \frac{\ln \left(1+\sin ^2 \theta \sin ^2 \alpha\right)}{\sin^2 \alpha} d \alpha d \theta \\ & =\frac{1}{2} \sum_{n=1}^{\infty}\frac{(-1)^n}{n} \int_0^{\frac{\pi}{2}} \int_0^{\frac{\pi}{2}} \sin ^{2 n} \theta \sin ^{2 n-1} \alpha d \alpha d \theta \\ & =\frac{1}{8} \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} B\left(n, \frac{1}{2}\right) B\left(n+\frac{1}{2}, \frac{1}{2}\right) \\ & =\frac{1}{8} \sum_{n=1}^{\infty} \frac{(-1)^n}{n} \frac{\Gamma(n) \Gamma\left(\frac{1}{2}\right)}{\Gamma\left(n+\frac{1}{2}\right)} \cdot \frac{\Gamma\left(n+\frac{1}{2}\right) \Gamma\left(\frac{1}{2}\right)}{\Gamma(n+1)} \\ & =\frac{\pi}{8} \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^2} \\ & =\frac{\pi}{8}\left(\sum_{n=1}^{\infty} \frac{1}{n^2}-2 \sum_{n=1}^{\infty} \frac{1}{(2n)^2}\right)\\&=\frac{\pi^3}{96} \end{aligned} $$ However, we had unavoidably used the series representation of $\ln(1+x^2)$.