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I tried integration by parts

$$\begin{aligned} \int_{-1}^{1}\sin^{-1}\left(x\right)\tan^{-1}\left(x\right)\mathop{\mathrm{d}x}&=2\int_{0}^{1}\sin^{-1}\left(x\right)\tan^{-1}\left(x\right)\mathop{\mathrm{d}x}\\ &=\left.2x\sin^{-1}\left(x\right)\tan^{-1}\left(x\right)\right|_{0}^{1}-2\int_{0}^{1}x\left(\frac{\sin^{-1}\left(x\right)}{1+x^{2}}+\frac{\tan^{-1}\left(x\right)}{\sqrt{1-x^{2}}}\right)\mathop{\mathrm{d}x}\\ &=\frac{\pi^{2}}{4}-2\left(\int_{0}^{1}\frac{x\sin^{-1}\left(x\right)}{1+x^{2}}\mathop{\mathrm{d}x}+\int_{0}^{1}\frac{x\tan^{-1}\left(x\right)}{\sqrt{1-x^{2}}}\mathop{\mathrm{d}x}\right)\\ &=\frac{\pi^{2}}{4}-2\left(I+J\right) \end{aligned}$$

Evaluting $J$ is relatively easy

$$\begin{aligned} J&=-\int_{0}^{1}\tan^{-1}\left(x\right)\mathop{\mathrm{d}{\sqrt{1-x^{2}}}}\\ &=-\left(\left.\tan^{-1}\left(x\right)\sqrt{1-x^{2}}\right|_{0}^{1}-\int_{0}^{1}\frac{\sqrt{1-x^{2}}}{1+x^{2}}\mathop{\mathrm{d}x}\right)\\ &=\int_{0}^{\frac{\pi}{2}}\frac{\cos^2\left(u\right)}{1+\sin^{2}\left(u\right)}\mathop{\mathrm{d}u}\qquad\left(x=\sin\left(u\right)\right)\\ &=\int_{0}^{\frac{\pi}{2}}\left(\frac{2}{1+\sin^2\left(u\right)}-1\right)\mathop{\mathrm{d}u}\\ &=2\int_{0}^{\frac{\pi}{2}}\frac{1}{\cos^2\left(u\right)+2\sin^2\left(u\right)}\mathop{\mathrm{d}u}-\int_{0}^{\frac{\pi}{2}}1\mathop{\mathrm{d}u}\\ &=\sqrt{2}\int_{0}^{\frac{\pi}{2}}\frac{\mathop{\mathrm{d}\left(\sqrt{2}\tan\left(u\right)\right)}}{1+\left(\sqrt{2}\tan\left(u\right)\right)^{2}}-\frac{\pi}{2}\\ &=\left.\sqrt{2}\tan^{-1}\left(\sqrt{2}\tan\left(u\right)\right)\right|_{0}^{\frac{\pi}{2}}-\frac{\pi}{2}\\ &=\frac{\sqrt{2}-1}{2}\pi \end{aligned}$$

However I'm stuck at evaluating $I$. How to evaluate this integral?

Thanks everyone, the final result is

$$\int_{-1}^{1}\sin^{-1}\left(x\right)\tan^{-1}\left(x\right)\mathop{\mathrm{d}x}=\boxed{\frac{\pi^{2}}{4}-\left(\sqrt{2}-1+\ln\left(4-2\sqrt{2}\right)\right)\pi}$$

El Mismo Sol
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    $$I=\int_0^1\frac{x\sin^{-1}x}{1+x^2}dx\overset{IBP}{=}\frac12\sin^{-1}x,\ln(1+x^2)\bigg|0^1-\frac12\int_0^1\frac{\ln(1+x^2)}{\sqrt{1-x^2}}dx$$ $$\overset{x=\cos t}{=}\frac{\pi\ln2}4-\frac12\int_0^{\frac\pi2}\ln(1+\cos^2t)dt\overset{\tan t=x}{=}\frac{\pi\ln2}4-\frac12\Re\int{-\infty}^\infty\frac{\ln(\sqrt2-ix)-\ln(1-ix)}{1+x^2}dx$$ Closing the contour in the upper half-plane $$I=\frac{\pi\ln2}4-\frac\pi2\Big(\ln(1+\sqrt2)-\ln2\Big)=\frac{3\pi}4-\frac\pi2\ln(1+\sqrt2)=0.2487...$$ – Svyatoslav Jun 18 '24 at 06:49
  • The integral is approximately 0.6686484663. I do not know if there is a closed-form expression. – Lucenaposition Jun 18 '24 at 06:51
  • $$\int_0^1\frac{x\sin^{-1}x}{1+x^2}dx=0.2487...$$ https://www.wolframalpha.com/input?i=%5Cint_%7B0%7D%5E%7B1%7D%5Cfrac%7Bx%5Carcsin%28x%29%7D%7B1%2Bx%5E%7B2%7D%7Ddx – Svyatoslav Jun 18 '24 at 06:53
  • @Svyatoslav. Mathematica provides a lonng result – Claude Leibovici Jun 18 '24 at 08:20
  • @Claude Leibovici, this is strange; it seems that the case isn't very complicated. WA (free option) even gives an analytical expression... – Svyatoslav Jun 18 '24 at 08:50
  • @Svyatoslav. Sory, I made a mistake – Claude Leibovici Jun 18 '24 at 10:08

4 Answers4

6

Let $y=\arcsin x$, then $$ \begin{aligned} I & =\int_0^{\frac{\pi}{2}} \frac{y \sin y}{1+\sin ^2 y} \cos y d y \\ & =\frac{1}{2} \int_0^{\frac{\pi}{2}} \frac{y \sin (2 y)}{1+\frac{1-\cos (2 y)}{2}} d y \\ & =\int_0^{\frac{\pi}{2}} \frac{y \sin (2 y)}{3-\cos (2 y)} d y \\ & =\frac{1}{4} \int_0^\pi \frac{y \sin y}{3-\cos y} d y \end{aligned} $$ Via integration by parts, we have $$ \begin{aligned} I & =\frac{1}{4} \int_0^\pi y d(\ln (3-\cos y)) \\ & =\frac{1}{4}[y \ln (3-\cos y)]_0^\pi-\frac{1}{4} \int_0^\pi \ln (3-\cos y) d y\\ & =\frac{\pi}{2}\ln 2- \frac{1}{4} \pi \ln \left(\frac{3+2 \sqrt{2}}{2}\right)\cdots(*)\\ & =\frac{\pi}{4}[3\ln 2-2 \ln (\sqrt{2}+1)] \end{aligned} $$ (*) obtained by putting $b=-1$ and $c=3$ in my post.

Lai
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$$ \begin{aligned} I & =\frac{1}{2}\left[\sin ^{-1} x \ln \left(1+x^2\right)\right]_0^1-\frac{1}{2} \int_0^1 \frac{\ln \left(1+x^2\right)}{\sqrt{1-x^2}}dx \\ & =\frac{\pi}{4} \ln 2-\frac{1}{2} \int_0^{\frac{\pi}{2}} \ln \left(1+\sin ^2 x\right) d x \quad \textrm{ via }x\mapsto \sin x \\ & =\frac{\pi}{4} \ln 2-\frac{1}{2} \int_0^{\frac{\pi}{2}} \ln \left(1+\frac{1-\cos 2 x}{2}\right) d x \\ & =\frac{\pi}{4} \ln 2-\frac{1}{4} \int_0^\pi \ln (3-\cos x) d x \\ & =\frac{\pi}{4} \ln 2-\frac{1}{4} \pi \ln \left(\frac{3+2 \sqrt{2}}{2}\right) \cdots (*) \\ & =\frac{\pi}{4}[3 \ln 2-2 \ln (1+\sqrt{2})] \end{aligned} $$ $(*)$ obtained by putting $b=-1$ and $c=3$ in my post.

Lai
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For the antiderivative $$I=\int \frac{x \sin ^{-1}(x)}{x^2+1}\,dx=\frac 12 \int \Bigg(\frac{\sin ^{-1}(x)}{x-i}+\frac{\sin^{-1}(x)}{x+i} \Bigg)\,dx$$ Now, we need to use $$\sin ^{-1}(x)=-i \log \left(i x+\sqrt{1-x^2}\right)$$ to get the result in terms of polylogarithms and use $$\int\frac{\sin^{-1}(x)}{x+a}\,dx=-\frac i 2 t^2+t \log \left(1-2 i a e^{it}-e^{2 i t}\right)+ $$ $$i \left(\text{Li}_2\left(-\frac{e^{i t}}{ia-\sqrt{1-a^2} }\right)-\text{Li}_2\left(-\frac{e^{i t}}{ia+\sqrt{1-a^2}}\right)\right) $$ where $t=\sin^{-1}(x)$.

Using the bounds $$J=\int_0^1 \frac{x \sin ^{-1}(x)}{x^2+1}\,dx=\frac{ \pi }{4}\left(3 \log (2)-2 \sinh ^{-1}(1)\right)$$

Claude Leibovici
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Further integration by parts on $I$ leads to

$$\frac{\pi^2}4 - \frac\pi2 \log2 + \int_0^1 \frac{\log\left(1+x^2\right)}{\sqrt{1-x^2}} \, dx - 2 \int_0^1 \frac{x \arctan x}{\sqrt{1-x^2}} \, dx$$

and both integrals can be computed by manipulating the binomial series. Some details are omitted, but the general idea is demonstrated in the "even case" outlined here.

$$\begin{align*} \int_0^1 \frac{\log\left(1+x^2\right)}{\sqrt{1-x^2}} \, dx &= \frac{\sqrt\pi}2 \sum_{n\ge1} \frac{(-1)^{n+1}}n \frac{\Gamma\left(n+\frac12\right)}{\Gamma(n+1)} \\ &= \frac\pi2 \sum_{n\ge1} \frac{(-1)^{n+1}}{2^{2n} n} \binom{2n}n \\ &= \pi \left(\log2 - \log\left(1+\sqrt{1+x}\right)\right) \bigg|_{x=1} = \pi \log\left(2\left(\sqrt2-1\right)\right) \\[2ex] \int_0^1 \frac{x \arctan x}{\sqrt{1-x^2}} \, dx &= \frac{\sqrt\pi}2 \sum_{n\ge0} \frac{(-1)^n}{2n+1} \frac{\Gamma\left(n+\frac32\right)}{\Gamma(n+2)} \\ &= \frac\pi4 \sum_{n\ge0} \frac{(-1)^n}{2^{2n}(n+1)} \binom{2n}n \\ &= \frac\pi2 \cdot \frac{\sqrt{1+x^2}-1}{x^2} \bigg|_{x=1} = \frac\pi2\left(\sqrt2-1\right) \end{align*}$$

user170231
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