Is there an easier way to evaluate the integral $I=\int_{0}^{1}\frac{\ln x\sin^{-1}x\cos^{-1}x}{x}dx$?

Question

As I was surfing the Mathematics side of Instagram (as usual), I came across this integral: $$I=\int_{0}^{1}\frac{\ln x\sin^{-1}x\cos^{-1}x}{x}dx$$ It encouraged me to embark on a very satisfying journey to try and evaluate it. The result turned out very nice, of course. I've found that $$I=\frac{1}{24}\ln^{4}2-\frac{\pi^{2}}{24}\ln^{2}2-\frac{49}{2880}\pi^{4}+\text{Li}_{4}\left(\frac{1}{2}\right)$$ using a series of tedious calculations. I will provide my solution below.

My question is as follows:

• Can you think of alternative ways to evaluate this integral?
• Is it possible to generalize it in any way? (e.g. $F(a,b)=\int_{0}^{1}\frac{\ln x\sin^{-1}ax\cos^{-1}bx}{x}dx$)
• At one point, I defined the function $J_{n}=\int_{0}^{\frac{\pi}{2}}\theta^{n}\ln(\sin\theta)\cot\theta\,d\theta\quad ,n\in\mathbb{N}$.Is there a nice closed form for this function?

Cheers!

Answer 1

\begin{align} &\int_{0}^{1}\frac{\ln x\sin^{-1}x\cos^{-1}x}{x}dx\\ =&\ \frac12\int_{0}^{1}\sin^{-1}x\cos^{-1}x\ d(\ln^2x)\\ \overset{ibp}=&\ \frac12\int_{0}^{1}\frac{\ln^2x\ (\sin^{-1}x-\cos^{-1}x)}{\sqrt{1-x^2}}\overset{x=\sin t}{dx}\\ =&-\frac\pi4 \int_0^{\pi/2}\ln^2(\sin t)\ dt + \int_0^{\pi/2}t\ln^2(\sin t)\ dt \end{align} where the first integral $\int_0^{\pi/2}\ln^2(\sin t)\ dt= \frac{\pi^2}{24}+\frac\pi2\ln^22$ is better known and the second is referenced below $$\int_0^{\pi/2} t\ln^2(\sin t)\ d t = \text{Li}_4(\frac12) -\frac{19\pi^4}{2880}+\frac{\pi^2}{12}\ln^22+\frac1{24}\ln^42 $$

Answer 2

We are going to find the definite integral $${I=\int\limits_{0}^{1}\frac{\ln x\sin^{-1}x\cos^{-1}x}{x}dx}$$

Ah, this integral. The more I looked into it the more it looked like it might have a closed form. Arriving at it just feels so satisfying!

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Firstly, by the substitution $x=\sin \theta$, we obtain

$$\begin{align*}I&=\int\limits_{0}^{\frac{\pi}{2}}\theta\left(\frac{\pi}{2}-\theta\right)\ln(\sin\theta)\,\cot\theta d\theta\\&=\frac{\pi}{2}J_{1}-J_{2}\end{align*}$$

where $\displaystyle{J_{n}=\int\limits_{0}^{\frac{\pi}{2}}\theta^{n}\ln(\sin\theta)\cot\theta d\theta}$

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For $J_{1}$, let’s first integrate by parts to get

$$\begin{align*}J_{1}&=\int\limits_{0}^{\frac{\pi}{2}}\theta\ln\sin\theta\,\,d\left(\ln\sin\theta\right)\\&=-\int\limits_{0}^{\frac{\pi}{2}}\theta\cot\theta\ln\sin\theta d\theta-\int\limits_{0}^{\frac{\pi}{2}}\ln^{2}\sin\theta d\theta\\&=-\frac{1}{2}\int\limits_{0}^{\frac{\pi}{2}}\ln^{2}\sin\theta d\theta\end{align*}$$

Now

$\displaystyle{J_{1}=-\frac{1}{2}\int\limits_{0}^{\infty}\frac{\partial^{2}}{\partial a^{2}}\frac{\theta^{a}}{\sqrt{1-\theta^{2}}}d\theta\quad}$when $a=0$,

so since $\int\limits_{0}^{\infty}\frac{\theta^{a}}{\sqrt{1-\theta^{2}}}d\theta=\frac{\Gamma\left(\frac{a+1}{2}\right)}{\Gamma\left(\frac{a}{2}+1\right)}$, by some tedious calculations, we can obtain $$\displaystyle{J_{1}=-\frac{\pi^{3}}{48}-\frac{\pi}{4}\ln^{2}2}$$

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Repeating the same integration-by-parts trick for $J_{2}$ quickly yields $$\begin{align*}J_{2}&=\int\limits_{0}^{\frac{\pi}{2}}\theta^{2}\ln\sin\theta\,\,d\left(\ln\sin\theta\right)\\&=-J_{2}-2\int\limits_{0}^{\frac{\pi}{2}}\theta\ln^{2}\sin\theta d\theta\\&=-\int\limits_{0}^{\frac{\pi}{2}}\theta\ln^{2}\sin\theta d\theta\end{align*}$$ This last integral looks like a beast, but we can first simplify it a bit using the Fourier series $\displaystyle{\ln^{2}\left(2\sin\theta\right)=\left(\frac{\pi}{2}-\theta\right)^{2}+2\sum\limits_{k=1}^{\infty}\frac{H_{k-1}}{k}\cos 2k\theta}$, valid for $0<\theta<\pi$.

Integrating that gets us $$\begin{align*}\int\limits_{0}^{\frac{\pi}{2}}\theta\ln^{2}\left(2\sin\theta\right)d\theta&=\frac{\pi^{4}}{192}+2\sum\limits_{k=1}^{\infty}\frac{H_{k-1}}{k}\int\limits_{0}^{\frac{\pi}{2}}\theta\cos 2k\theta d\theta\\&=\frac{\pi^{4}}{192}+\frac{1}{2}\sum\limits_{k=1}^{\infty}\frac{H_{k-1}}{k^{3}}\left((-1)^{k}-1\right)\\&=\frac{\pi^{4}}{192}+\frac{1}{2}\sum\limits_{k=1}^{\infty}\frac{(-1)^{k}H_{k-1}}{k^{3}}-\frac{1}{2}\sum\limits_{k=1}^{\infty}\frac{H_{k-1}}{k^{3}}\\&=\frac{\pi^{4}}{192}+\frac{1}{2}\sum\limits_{k=1}^{\infty}\frac{(-1)^{k}H_{k}}{k^{3}}-\frac{1}{2}\sum\limits_{k=1}^{\infty}\frac{(-1)^{k}}{k^{4}}-\frac{1}{2}\sum\limits_{k=1}^{\infty}\frac{H_{k-1}}{k^{3}}+\frac{1}{2}\sum\limits_{k=1}^{\infty}\frac{1}{k^{4}}\\&=\frac{\pi^{4}}{192}+\frac{1}{2}\sum\limits_{k=1}^{\infty}\frac{(-1)^{k}H_{k}}{k^{3}}+\frac{1}{2}\eta(4)-\frac{1}{2}\sum\limits_{k=1}^{\infty}\frac{H_{k-1}}{k^{3}}+\frac{1}{2}\zeta(4)\end{align*}$$

where $\eta$ and $\zeta$ represent the Dirichlet Eta function and the Riemann Zeta function respectively.

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$$\sum\limits_{k=1}^{\infty}\frac{H_{k}}{k^{3}}=\frac{\pi^{4}}{72}$$ is a well-known result that we can find as the equation $(53)$ on the Wolfram MathWorld page for harmonic numbers (No idea why this link wouldn't turn in a hyperlink, so I removed it), and $$\sum\limits_{k=1}^{\infty}\frac{(-1)^{k}H_{k}}{k^{3}}=-\frac{11}{4}\zeta(4)+2\lambda(3)\ln2-\frac{1}{2}\ln^{2}2\zeta(2)+\frac{1}{12}\ln^{4}2+2\text{Li}_{4}\left(\frac{1}{2}\right)$$ is an equally fascinating result that we can obtain via modifying equation $[4.85]$ in Cornel Ioan Vălean’s infamous book (Almost) Impossible Integrals, Sums, and Series.

Note that $\lambda$ here denotes the Dirichlet Lambda function (ugly, I know, but you’ll see in a second that it cancels out very nicely).

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With these handy results, we have $$\begin{align*}&\,\int\limits_{0}^{\frac{\pi}{2}}\theta\ln^{2}\left(2\sin\theta\right)d\theta\\&=\frac{\pi^{4}}{192}+\frac{1}{2}\left(-\frac{11}{4}\zeta(4)+2\lambda(3)\ln2-\frac{1}{2}\ln^{2}2\zeta(2)+\frac{1}{12}\ln^{4}2+2\text{Li}_{4}\left(\frac{1}{2}\right)+\eta(4)-\frac{\pi^{4}}{72}+\zeta(4)\right)\\&=-\frac{19}{2880}\pi^{4}+\frac{1}{24}\ln^{4}2-\frac{\pi^{2}}{24}\ln^{2}2+\lambda(3)\ln2+\text{Li}_{4}\left(\frac{1}{2}\right)\end{align*}$$

But on the other hand, $\displaystyle{\ln^{2}\left(2\sin\theta\right)}$ is readily expandable into $\displaystyle{\ln^{2}2+2\ln2\ln\sin\theta+\ln^{2}\sin\theta}$

This means that what we’re dealing with can indeed be rewritten as $$\begin{align*}\int\limits_{0}^{\frac{\pi}{2}}\theta\ln^{2}\left(2\sin\theta\right)d\theta&=\ln^{2}2\int\limits_{0}^{\frac{\pi}{2}}\theta d\theta+2\ln2 \int\limits_{0}^{\frac{\pi}{2}}\theta\ln\sin\theta d\theta+\int\limits_{0}^{\frac{\pi}{2}}\theta\ln^{2}\sin\theta d\theta\\&=\frac{\pi^{2}}{8}\ln^{2}2+2\ln2\underbrace{\int\limits_{0}^{\frac{\pi}{2}}\theta\ln\sin\theta d\theta}_{K_{1}}-J_{2}\end{align*}$$

The bracketed term is easy to evaluate. Using the so-called King Property of definite integrals, we have $$\displaystyle{K_{1}=\int\limits_{0}^{\frac{\pi}{2}}\theta\ln\sin\theta d\theta=\int\limits_{0}^{\frac{\pi}{2}}\left(\frac{\pi}{2}-\theta\right)\ln\cos\theta d\theta}$$

Adding, we get $$\begin{align*}2K_{1}&=\int\limits_{0}^{\frac{\pi}{2}}\theta\ln\tan\theta d\theta+\frac{\pi}{2}\int\limits_{0}^{\frac{\pi}{2}}\ln\cos\theta d\theta\\2K_{1}&=-2\sum\limits_{k=1}^{\infty}\frac{1}{2k-1}\int\limits_{0}^{\frac{\pi}{2}}\theta\cos\left(4k\theta-2\theta\right)d\theta-\frac{\pi}{2}\left(\frac{\pi}{2}\ln 2\right)\\ 2K_{1}&=\sum\limits_{k=1}^{\infty}\frac{1}{\left(2k-1\right)^{3}}-\frac{\pi^{2}}{4}\ln 2\\2\ln 2K_{1}&=\lambda(3)\ln2-\frac{\pi^{2}}{4}\ln^{2}2\end{align*}$$

And therefore, $$\displaystyle{-\frac{19}{2880}\pi^{4}+\frac{1}{24}\ln^{4}2-\frac{\pi^{2}}{24}\ln^{2}2+\lambda(3)\ln2+\text{Li}_{4}\left(\frac{1}{2}\right)=\frac{\pi^{2}}{8}\ln^{2}2+2\ln2+\lambda(3)\ln2-\frac{\pi^{2}}{4}\ln^{2}2-J_{2}}$$ which is equivalent to saying $$J_{2}=\frac{19}{2880}\pi^{4}-\frac{\pi^{2}}{12}\ln^{2}2-\frac{1}{24}\ln^{4}2+\text{Li}_{4}\left(\frac{1}{2}\right)$$

(See? The lambda’s gone!)

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Finally, we can evaluate the original integral as desired. $$\begin{align*}I&=\frac{\pi}{2}J_{1}-J_{2}\\&\boxed{=\frac{1}{24}\ln^{4}2-\frac{\pi^{2}}{24}\ln^{2}2-\frac{49}{2880}\pi^{4}+\text{Li}_{4}\left(\frac{1}{2}\right)_{\blacksquare}}\end{align*}$$

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Luna luceat tibi magis.

ー　如月あやみ　(Kisaragi Ayami)

Answer 3

Let $y=\sin^{-1} x$, then $$ \begin{aligned} I & =\int_0^{\frac{\pi}{2}} \ln (\sin y) y\left(\frac{\pi}{2}-y\right) \cot y d y \\ & =\int_0^{\frac{\pi}{2}}\left(\frac{\pi}{2}-y\right) y \ln (\sin y) d(\ln (\sin y)) \end{aligned} $$ Using integration by parts, we have \begin{aligned} I= & -\int_0^{\frac{\pi}{2}}\left(\frac{\pi}{2}-2 y\right) \ln ^2(\sin y) d y -\underbrace{ \int_0^{\frac{\pi}{2}}\left(\frac{\pi}{2}-y\right) y \ln (\sin y) \cot y d y}_{I} \end{aligned} Rearranging yields $$ \begin{aligned} I & =-\frac{1}{2} \int_0^{\frac{\pi}{2}}\left(\frac{\pi}{2}-2 y\right) \ln ^2(\sin y) d y \\ & =-\frac{\pi}{4} \underbrace{\int_0^{\frac{\pi}{2}} \ln ^2(\sin y) d y}_{J} + \underbrace{ \int_0^{\frac{\pi}{2}} y \ln ^2(\sin y) d y}_{K} \end{aligned} $$

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For the integral $J$, we note that $$ J=\int_0^{\frac{\pi}{2}} \ln ^2(\sin y) d y=\left.\frac{\partial^2}{\partial a^2} J(a)\right|_{a=0} $$ where $$ J(a)=\int_0^{\frac{\pi}{2}} \sin ^a \theta d \theta= \frac{1}{2} B\left(\frac{a+1}{2}, \frac{1}{2}\right) $$ $$\frac{\partial^2}{\partial a^2} J(a)=\frac{1}{8}\left[\left(\psi\left(\frac{a+1}{2}\right)-\psi\left(\frac{a}{2}+1\right)\right)^2+\psi^{\prime}\left(\frac{a+1}{2}\right)-\psi^{\prime}\left(\frac{a}{2}+1\right)\right] B\left(\frac{a+1}{2}, \frac{1}{2}\right) $$ So we get

$$ \begin{aligned} J & =\frac{1}{8}\left[\left(\psi\left(\frac{1}{2}\right)-\psi(1)\right)^2+\psi^{\prime}\left(\frac{1}{2}\right)-\psi^{\prime}(1)\right]B\left( \frac{1}{2},\frac{1}{2} \right) \\ &= \frac{\pi}{8}\left(4 \ln ^2 2+\frac{\pi^2}{3}\right) \\ & =\frac{\pi}{24}\left(\pi^2+12 \ln ^2 2\right) \end{aligned} $$

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For the integral $K$, please refer to the result of the post $$\int_0^{\frac\pi2} y\ln^2(\sin y)\ d y = \text{Li}_4(\frac12) -\frac{19\pi^4}{2880}+\frac{\pi^2}{12}\ln^22+\frac1{24}\ln^42, $$ we can now conclude that $$ \begin{aligned} I= & -\frac{\pi}{4} \cdot \frac{\pi}{24}\left(\pi^2+12 \ln ^2 2\right) +\operatorname{Li_ 4}\left(\frac{1}{2}\right)-\frac{19 \pi^4}{2880}+\frac{\pi^2}{12} \ln ^2 2+\frac{1}{24} \ln ^4 2\\=& \frac{1}{24}\ln^{4}2-\frac{\pi^{2}}{24}\ln^{2}2-\frac{49 \pi^{4}}{2880}+\text{Li}_{4}\left(\frac{1}{2}\right) \end{aligned} $$