Tough definite integral: $\int_0^\frac{\pi}{2}x\ln^2(\sin x)~dx$

Question

Any ideas on evaluating the definite integral $$\int_0^\frac{\pi}{2}x\ln^2(\sin x)\ dx$$

The best numerical approximation I could get is $0.2796245358$.

Is there even a closed form solution?

Answer 1

Solution by real methods:

From here we have

$$\frac23\arcsin^4x=\sum_{n=1}^\infty\frac{H_{n-1}^{(2)}(2x)^{2n}}{n^2{2n\choose n}}=\sum_{n=1}^\infty\frac{H_{n}^{(2)}(2x)^{2n}}{n^2{2n\choose n}}-\sum_{n=1}^\infty\frac{(2x)^{2n}}{n^4{2n\choose n}}$$

Set $x=1$ we get

$$\sum_{n=1}^\infty\frac{4^n}{n^4{2n\choose n}}=\sum_{n=1}^\infty\frac{4^nH_{n}^{(2)}}{n^2{2n\choose n}}-\frac{15}{4}\zeta(4)\tag1$$

In this question we showed $$\sum_{n=1}^\infty\frac{4^nH_n}{n^3{2n\choose n}}=-\sum_{n=1}^\infty\frac{4^nH_n^{(2)}}{n^2{2n\choose n}}+12\ln^2(2)\zeta(2)\tag2$$

Adding $(1)$ and $(2)$ yields

$$\sum_{n=1}^\infty\frac{4^n}{n^4{2n\choose n}}=12\ln^2(2)\zeta(2)-\frac{15}{4}\zeta(4)-\sum_{n=1}^\infty\frac{4^nH_n}{n^3{2n\choose n}}$$

By using the Fourier series of $\tan x\ln(\sin x)$, we showed in this solution:

$$\sum_{n=1}^\infty\frac{4^nH_n}{n^3{2n\choose n}}=-8\text{Li}_4\left(\frac12\right)+\zeta(4)+8\ln^2(2)\zeta(2)-\frac{1}{3}\ln^4(2)$$

substitute this result we get

$$\sum_{n=1}^\infty\frac{4^n}{n^4{2n\choose n}}=8\text{Li}_4\left(\frac12\right)-\frac{19}{4}\zeta(4)+4\ln^2(2)\zeta(2)+\frac{1}{3}\ln^4(2)\tag3$$

Now we use the well-known series expansion of $\arcsin^2 x$:

$$\arcsin^2(x)=\frac12\sum_{n=1}^\infty\frac{4^n x^{2n}}{n^2{2n\choose n}}$$

Multiply both sides by $-\frac{\ln x}{x}$ then $\int_0^1$ and use that $-\int_0^1 x^{2n-1}\ln xdx=\frac{1}{4n^2}$ we get

$$\frac18\sum_{n=1}^\infty\frac{4^n}{n^4{2n\choose n}}=-\int_0^1\frac{\ln x\arcsin^2(x)}{x}dx$$

$$\overset{IBP}{=}\int_0^1\frac{\ln^2x\arcsin(x)}{\sqrt{1-x^2}}dx\overset{x=\sin\theta}{=}\int_0^{\pi/2}x\ln^2(\sin x)dx\tag4$$

From $(3)$ and $(4)$ we obtain

$$\int_0^{\pi/2} x\ln^2(\sin x)dx=\frac{1}{2}\ln^2(2)\zeta(2)-\frac{19}{32}\zeta(4)+\frac{1}{24}\ln^4(2)+\operatorname{Li}_4\left(\frac{1}{2}\right)$$

Answer 2

Here is an elementary approach. Denote $ K= \int_0^\infty \frac{\ln t\ln(1+t^2) \tan^{-1}t}{1+t^2}dt $

\begin{align} I= &\int_0^\frac{\pi}{2}x\ln^2(\sin x)\ dx \overset{t=\tan x}= \frac14 \int_0^\infty \frac{\tan^{-1}t \ln^2\frac{t^2}{1+t^2}}{1+t^2}dt\\ =& \int_0^\infty \frac{\tan^{-1}t\ln^2t}{1+t^2}\overset{t\to 1/t}{dt} +\frac14 \int_0^\infty \frac{\tan^{-1}t\ln^2(1+t^2)}{1+t^2}\overset{t\to 1/t}{dt} -K\\ =& \frac\pi8 \int_0^\infty \frac{\ln^2t}{1+t^2}dt + \frac\pi{16}\int_0^\infty \frac{\ln^2(1+t^2)}{1+t^2}dt -I -K\\ =& \frac\pi8 \cdot\frac{\pi^3}8 +\frac\pi{16}\left(\frac{\pi^3}6+2\pi\ln^22\right) -I-K =\frac{5\pi^4}{192}+\frac{\pi^2}8\ln^22-\frac12K\tag1 \end{align} Note that, with the substitution $x=\frac{(1+t^2)y}{1-y}$ \begin{align} \int_0^1 \frac{2t\ln\frac{1-y}y}{1+t^2y^2}dy = \int_0^\infty \frac{2t\ln\frac{1+t^2}x}{(1+x)^2+t^2}dx=\ln(1+t^2)\tan^{-1}t \end{align} Then, integrate $K$ as follows \begin{align} K=& \int_0^\infty \frac{\ln t}{1+t^2}\int_0^1 \frac{2t\ln\frac{1-y}y}{1+t^2y^2}dy \>\overset{t^2\to t}{dt}\\ = &\frac12 \int_0^1 \ln\frac{1-y}y\int_0^\infty \frac{\ln t}{(1+t)(1+y^2t)}dt\>dy =\int_0^1 \ln\frac{1-y}y \frac{\ln^2y}{1-y^2}dy\\ =&\frac{\pi^4}{16}+\frac12\int_0^1\frac{\ln^2y\ln(1-y)}{1-y}dy + \frac12\int_0^1\frac{\ln^2y\ln(1-y)}{1+y}dy \end{align} The pair of integrals above are known and given below \begin{align} &\int_0^1\frac{\ln^2y\ln(1-y)}{1-y}dy= -\frac{\pi^4}{180}\\ &\int_0^1\frac{\ln^2y\ln(1-y)}{1+y}dy= -4Li_4\left(\frac12\right) +\frac{\pi^4}{90}+\frac{\pi^2}6\ln^22-\frac16\ln^42\\ \end{align} Plug them into $K$ and then into (1) to obtain $$\int_0^\frac{\pi}{2}x\ln^2(\sin x)\ dx = Li_4\left(\frac12\right)-\frac{19\pi^4}{2880}+\frac{\pi^2}{12}\ln^22+\frac1{24}\ln^42 $$

Answer 3

Using the Fourier series of $\ln^2(2\sin x):$

$$\ln^2(2\sin x)=\left(\frac{\pi}{2}-x\right)^2+2\sum_{n=1}^\infty\frac{H_{n-1}}{n}\cos(2nx),\quad 0<x<\pi$$

we get $$ \int_0^{\frac{\pi}{2}}x\ln^2(2\sin x)\, dx$$ $$=\int_0^{\frac{\pi}{2}} x\left(\frac{\pi}{2}-x\right)^2\, dx+2\sum_{n=1}^\infty\frac{H_{n-1}}{n}\int_0^{\frac{\pi}{2}} x\cos(2nx)\, dx$$

$$=\frac{15}{32}\zeta(4)+2\sum_{n=1}^\infty\frac{H_{n-1}}{n}\left(\frac{(-1)^n}{4n^2}-\frac{1}{4n^2}\right)$$

$$=\frac{45}{32}\zeta(4)-\frac12\sum_{n=1}^\infty\frac{H_n}{n^3}+\frac12\sum_{n=1}^\infty\frac{(-1)^n H_n}{n^3}$$

$$=\operatorname{Li_4}\left(\frac12\right)-\frac{19}{32}\zeta(4)+\frac78\ln(2)\zeta(3)-\frac14\ln^2(2)\zeta(2)+\frac{1}{24}\ln^4(2).$$

On the other hand, we can write

$$\int_0^{\frac{\pi}{2}}x\ln^2(2\sin x)\, dx$$

$$=\ln^2(2)\int_0^{\frac{\pi}{2}}x\, dx+2\ln(2)\int_0^{\frac{\pi}{2}}x\ln(\sin x)\, dx+\int_0^{\frac{\pi}{2}}x\ln^2(\sin x)\, dx$$

$$=\frac78\ln(2)\zeta(3)-\frac34\ln^2(2)\zeta(2)+\int_0^{\frac{\pi}{2}}x\ln^2(\sin x)\, dx.$$

Thus,

$$\int_0^{\frac{\pi}{2}} x\ln^2(\sin x)\,dx=\operatorname{Li}_4\left(\frac{1}{2}\right)-\frac{19}{32}\zeta(4)+\frac{1}{2}\ln^2(2)\zeta(2)+\frac{1}{24}\ln^4(2).$$