Find: $\int_0^\pi x^2\ln(\sin(x))dx$

Question

I've been working on a few log-sine integrals. So far I have found $$\int_0^\pi \ln(\sin(x))dx=-\pi\ln(2)$$ $$\int_0^\pi x\ln(\sin(x))dx=-\frac{\pi^2\ln(2)}{2}$$ ...but I am struggling with the integral $$\int_0^\pi x^2\ln(\sin(x))dx$$ and I can't figure it out... however, I do know that the answer will contain $\zeta(3)$. Any hints?

Answer 1

Rewrite the integral as: $$\begin{aligned}\int_0^\pi {{x^2}\ln (\sin x)dx} &= \int_{ - \pi /2}^{\pi /2} {{{(x + \frac{\pi }{2})}^2}\ln (\cos x)dx} \\ &= 2\int_0^{\pi /2} {{x^2}\ln (\cos x)dx} + \frac{{{\pi ^2}}}{2}\underbrace{\int_0^{\pi /2} {\ln (\cos x)dx}}_{-\pi \ln 2 / 2} \end{aligned}$$ To evaluate the first integral, we use the Fourier expansion of $\ln(\cos x)$: $$\ln (\cos x) = - \ln 2 - \sum\limits_{k = 1}^\infty {\frac{{{{( - 1)}^k}\cos 2kx}}{k}} $$ Plug it into the integral (it is legitimate to interchange order of summation and integration, because the resulting series is absolutely convergent), we obtain $$\begin{aligned} \int_0^{\pi /2} {{x^2}\ln (\cos x)dx} &= - \ln 2\int_0^{\pi /2} {{x^2}dx} - \sum\limits_{k = 1}^\infty {\frac{{{{( - 1)}^k}}}{k}\int_0^{\pi /2} {{x^2}\cos 2kxdx} } \\& = - \frac{{{\pi ^3}\ln 2}}{{24}} - \sum\limits_{k = 1}^\infty {\frac{{{{( - 1)}^k}}}{k}\frac{{\pi {{( - 1)}^k}}}{{4{k^2}}}} \\ &= - \frac{{{\pi ^3}\ln 2}}{{24}} - \frac{\pi }{4}\zeta (3) \end{aligned}$$

The final result is, $$\int_0^\pi {{x^2}\ln (\sin x)dx} = -\frac{\pi^3}{3}\ln 2 -\frac{\pi}{2}\zeta(3)$$

Answer 2

Hint: we know $$\int_a^b f(x)dx= \int_a^b f(a+b-x)dx$$ With $a=0,~~~b=\pi$ set

$$I=\int_0^\pi \ln(\sin(x))dx~~~~and ~~~~~ J=\int_0^\pi x\ln(\sin(x))dx$$$$A=\int_0^\pi x^3\ln(\sin(x))dx~~~~~and~~~~~B =\int_0^\pi x^2\ln(\sin(x))dx $$ From above formula we have \begin{split}\int_0^\pi x^3\ln(\sin(x))dx &=&-\int_0^\pi (x-\pi)^3\ln(\sin(\pi-x))dx \\ &=&-\int_0^\pi x^3\ln(\sin(x))dx +3\pi \int_0^\pi x^2\ln(\sin(x))dx \\&-&3\pi^2\int_0^\pi x\ln(\sin(x))dx+\pi^3 \int_0^\pi \ln(\sin(x))dx\end{split}

that is $$2A -3\pi B = -3\pi^2J+\pi^3 I$$ Do this again with

$$\int_0^\pi x^4 \ln(\sin(x))dx =\int_0^\pi (x-\pi)^4\ln(\sin(\pi-x))dx$$

Which is equivalent after doing as above to $$ -4\pi A+6\pi^2B -4\pi^3J+\pi^4 I=0$$

then you will get A and B by solving $$2A -3\pi B = -3\pi^2J+\pi^3 I$$ $$ -4\pi A+6\pi^2B =4\pi^3J-\pi^4 I$$

Answer 3

Here is a beautiful generalization from D. Orr's, Generalized Log-Sine Integrals and Bell polynomials, where he gives the following identity.

$$\int\limits_{0}^{\pi} x^n \log(\sin x)\ \mathrm dx = - \pi^{n+1}\left(\frac{\log 2}{n+1} - \sum_{j=1}^{\lfloor{n/2}\rfloor} \frac{n! (-1)^j \zeta(2j + 1)}{(n+1-2j)! (2\pi)^{2k}}\right)$$

Answer 4

For the general case of $$I_n=\int_0^\pi {x^n}\log (\sin (x))\,dx=\int_{-\frac \pi 2}^{+\frac \pi 2}\left(x+\frac{\pi }{2}\right)^n\,\log (\cos (x))\,dx$$ and $$\log (\cos (x))=\sum_{p=1}^\infty (-1)^p\, \frac{2^{2 p-1} \left(4^p-1\right) B_{2 p}}{p\, (2 p)!}\, x^{2p}$$ $$J_{n,p}=\int_{-\frac \pi 2}^{+\frac \pi 2}\left(x+\frac{\pi }{2}\right)^n\,x^{2p}\,dx$$ $$J_{n,p}=\left(\frac{\pi}{2 }\right)^{n+2 p+1}\Bigg(\frac{\Gamma (n+1) \Gamma (2 p+1)}{\Gamma (n+2 p+2)} +\frac{\, _2F_1(-n,2 p+1;2 (p+1);-1)}{2 p+1} \Bigg)$$ where appears the Gaussian hypergeometric function.

This gives $$I_n=-\frac{\pi^{n+1}}{n+1} \log(2)-A_n$$ with $$\left( \begin{array}{cc} n & A_n \\ 1 & 0 \\ 2 & \frac{\pi }{2} \zeta (3) \\ 3 & \frac{3 \pi ^2 }{4}\zeta (3) \\ 4 & \pi ^3 \zeta (3)-\frac{3 \pi }{2} \zeta (5) \\ 5 & \frac{5 \pi ^4 }{4}\zeta (3)-\frac{15 \pi ^2 }{4}\zeta (5) \\ 6 & \frac{3 \pi ^5 }{2}\zeta (3)-\frac{15 \pi ^3}{2} \zeta (5)+\frac{45 \pi }{4}\zeta (7) \\ 7 & \frac{7 \pi ^6 }{4}\zeta (3)-\frac{105 \pi ^4 }{8}\zeta (5)+\frac{315 \pi ^2 }{8}\zeta (7) \\ \end{array} \right)$$

Answer 5

Note that $\int_0^{\pi/2} \ln(2\sin x)dx= \int_0^{\pi/2} \ln(2\cos x)dx=0$ \begin{align} J &=\int_0^{\pi/2} x\ln(2\sin x)dx\>\>\>\>\>\>\>({x\to \frac\pi2-x})\\ &= -\int_0^{\pi/2}x \ln(2\cos x)dx= \frac12 \int_0^{\pi/2} x\ln(\tan x)dx\\ &=\frac12 \int_0^{\pi/2} \int_0^1\frac{\tan x\ln(\tan x)}{1+t^2 \tan^2x}dt\ dx\>\>\>\>\>\>\>(\tan x =\frac1{t\tan y})\\ &= \frac14 \int_0^1 \int_0^{\pi/2} \frac{\tan y\ln t}{1+t^2 \tan^2y}dy\ dt = \frac14 \int_0^1\frac{\ln^2t}{1-t^2}dt =\frac{7}{16}\zeta(3)\\ \\ I &=\int_0^{\pi/2} x^2\ln(2\sin x)dx\>\>\>\>\>\>\>({x\to \frac\pi2-x})\\ &= \pi J+ \int_0^{\pi/2}x^2 \ln(2\cos x)dx = \pi J-I + \int_0^{\pi/2}x^2 \ln(\overset{2x\to x}{2\sin 2x})dx\\ &= \pi J-\frac78 I + \frac18 \int_{\pi/2}^\pi x^2 \ln({2\sin x})dx\>\>\>\>\>\>\> (x\to \pi -x)\\ &=\frac{3\pi}{4}J -\frac34 I =\frac{3\pi}7 J=\frac{3\pi}{16}\zeta(3) \end{align} As a result \begin{align} &\int_0^{\pi} x^2\ln(\sin x)dx \\= &-\frac{\pi^3}3\ln2+ \int_0^{\pi} x^2\ln(2\sin x)dx\>\>\>\>\>\>\>\bigg(\> \int_{\pi/2}^\pi\cdot \overset{x\to\pi-x}{dx}\bigg)\\ = & -\frac{\pi^3}3\ln2 + 2I-2\pi J= -\frac{\pi^3}3\ln2-\frac\pi2\zeta(3) \end{align}