Computing $\int_{0}^{1} \frac{\sin ^{-1} (x) \ln (1+x)}{x^{2}} d x$

Applying integration by parts splits the integral into 3 integrals, \displaystyle \begin{aligned}I&=\int_{0}^{1} \frac{\sin ^{-1} x \ln (1+x)}{x^{2}} d x\\&=-\int_{0}^{1} \sin ^{-1} x \ln (1+x) d\left(\frac{1}{x}\right) \\&=-\left[\frac{\sin ^{-1} x \ln (1+x)}{x}\right]_{0}^{1}+\underbrace{\int_{0}^{1} \frac{\ln (1+x)}{x \sqrt{1-x^{2}}}}_{K} +\underbrace{\int_{0}^{1}\frac{\sin ^{-1} x}{x}}_{L} d x-\underbrace{\int_{0}^{1} \frac{\sin ^{-1} x}{1+x}}_{M} d x \end{aligned} \tag*{} Letting $$x= \cos \theta$$ for $$K$$ and $$\sin^{-1}x \mapsto x$$ for $$L$$ and $$M$$, yields $$\displaystyle I=-\frac{\pi}{2} \ln 2 +\underbrace{\int_{0}^{\frac{\pi}{2}} \frac{\ln (1+\cos \theta)}{\cos \theta} d \theta}_{K}+\underbrace{\int_{0}^{\frac{\pi}{2}} \frac{x\cos x }{\sin x} d x}_{L}-\underbrace{\int_{0}^{\frac{\pi}{2}} \frac{x\cos x }{1+\sin x} d x }_{M}\tag*{}$$

For the integral $$K,$$putting $$a=1$$ in my post yields $$\displaystyle \boxed{K=\frac{\pi^{2}}{8}}\tag*{}$$

For the integral $$L,$$ integration by parts yields \displaystyle \begin{aligned}L &=\int_{0}^{\frac{\pi}{2}} x d \ln (\sin x) \\&=[x \ln (\sin x)]_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}} \ln (\sin x) d x \\&=\boxed{\frac{\pi}{2} \ln 2}\end{aligned}\tag*{}

For the integral $$M,$$ integration by parts yields \displaystyle \begin{aligned}M &=\int_{0}^{\frac{\pi}{2}} x d \ln (1+\sin x)\\&=[x \ln (1+\sin x)]_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}} \ln (1+\sin x) d x \\&=\frac{\pi}{2} \ln 2-\underbrace{\int_0^{\frac{\pi}{2} }\ln (1+\sin x) d x}_{N}\end{aligned}\tag*{} For the integral $$N,$$ using my post in the second last step yields \displaystyle \begin{aligned}N \stackrel{x\mapsto\frac{\pi}{2}-x}{=} &\int_{0}^{\frac{\pi}{2}} \ln (1+\cos x) d x \\=&\int_{0}^{\frac{\pi}{2}} \ln \left(2 \cos ^{2} \frac{x}{2}\right) d x \\=&\frac{\pi}{2} \ln 2+2 \int_{0}^{\frac{\pi}{2}} \ln \left(\cos \frac{x}{2}\right) d x \\=&\frac{\pi}{2} \ln 2+4 \int_{0}^{\frac{\pi}{4}} \ln (\cos x) d x \\=&\frac{\pi}{2} \ln 2+4\left(\frac{1}{4}(2 G-\pi \ln 2)\right) \\=&\boxed{-\frac{\pi}{2} \ln 2+2 G}\end{aligned}\tag*{} where $$G$$ is the Catalan’s Constant.

Putting them together yields $$\displaystyle \boxed{I=-\pi \ln 2+\frac{\pi^{2}}{8}+2G} \tag*{}$$

Question: Is there any shorter solution?

• I am not satisfied with my solution which is lengthy! Just want to improve it. Do you have any suggestion?
– Lai
Commented Apr 15, 2022 at 10:25
• An old issue of RMM, see here ssmrmh.ro/2021/11/02/integral-calculus-567 Commented Apr 16, 2022 at 11:48

A self-contained solution \begin{align} I=&\int_{0}^{1} \frac{\sin ^{-1} x \ln (1+x)}{x^{2}} d x =\int_{0}^{1} \sin ^{-1}x \>d\left( \ln x -\frac{1+x}x \ln (1+x)\right)\\ \overset{ibp} =&\> -\pi \ln 2-{\int_{0}^{1} \frac{\ln \frac x{1+x}-\frac1x \ln(1+x)}{\sqrt{1-x^{2}}}}\> \overset{x=\frac{2t}{1+t^2}}{dx}\\ =&\>-\pi\ln2 + \int_0^1 \underset{=K}{\frac{\ln\frac{(1+t)^2}{1+t^2}}{t}}dt -2\int_0^1 \underset{=J}{\frac{\ln\frac{2t}{(1+t)^2}}{1+t^2}}dt \end{align} with \begin{align} K=&\>\int_0^1 \frac{\ln (1+t)^2}{t}dt -\int_0^1 \frac{\ln (1+t^2)}{t}\overset{t^2\to t}{dt} =\frac32 \int_0^1 \frac{\ln (1+t)}{t}dt =\frac{\pi^2}8\\ J=& \int_0^1 \frac{\ln t}{1+t^2}dt +\int_0^1 \frac{\ln \frac2{(1+t)^2}}{1+t^2}\overset{t\to\frac{1-t}{1+t}}{dt}= \int_0^1 \frac{\ln t}{1+t^2}dt=-G\\ \end{align} Thus $$I=-\pi \ln 2+\frac{\pi^{2}}{8}+2G$$
Perhaps not 100% satisfactory, as I've performed each step via Mathematica rather than a step-by-step derivation, but the following could be considered a more simple solution. Note that $$\ln(1+x) = \sum_{n=1}^\infty(-1)^{n+1}\frac{x^n}{n}$$ and $$\int_0^1\sin^{-1}x\,x^n dx = \frac{\sqrt{\pi}}{n+1}\left( \frac{\sqrt\pi}{2} - \frac{\Gamma\big(1+\frac n2\big)}{(n+1)\Gamma\big(\frac{n+1}{2}\big)} \right),$$ where, for $$n=-1$$, the above is understood to be $$\frac{\pi\ln2}{2}$$. Performing the infinite sum yields the answer given in the original question.