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

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!

\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 $$\int_0^{\pi/2} t\ln^2(\sin t)\ d t = \text{Li}_4(\frac12) -\frac{19\pi^4}{2880}+\frac{\pi^2}{12}\ln^22+\frac1{24}\ln^42$$

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!

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

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 $$\displaystyle{J_{1}=-\frac{\pi^{3}}{48}-\frac{\pi}{4}\ln^{2}2}$$

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.

$$\sum\limits_{k=1}^{\infty}\frac{H_{k}}{k^{3}}=\frac{\pi^{4}}{72}$$ 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).

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!)

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*}
Luna luceat tibi magis.

ー　如月あやみ　(Kisaragi Ayami)

• All too easy.... Commented Mar 4 at 22:33
• @MarkViola Indeed, not difficult but just tedious. Coming up with an alternative solution would be very interesting. Commented Mar 4 at 23:03
• @MarkViola All things aside, don't you think it sounds a little too condescending to comment under a post saying nothing but it's "all too easy"? I question the necessity of such a remark. Commented Mar 4 at 23:04
• The comment was not condescending at all. How could one infer that it was? The comment was faceitious since the development was quite obviously NOT "all too easy." Aside, the line "all to easy" was a quote by the character Darth Vader in the 1980 film "The Empire Strikes Back." So it was intended to be a humorous homage. Is this clear? Commented Mar 4 at 23:18
• Perhaps I've been on Quora for too long. Not to bad-mouth the community over there, but after a few years of lurking there, it's hard not to be overly sensitive to the possibility that a plebian in the Dunning-Kruger pit might be smugly attacking another user. Happy to know that Math StackExchange is entirely different from that, with a great sense of humour on the side as well! Commented Mar 4 at 23:29

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}

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}

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}