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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?

Guy Fsone
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Franklin Pezzuti Dyer
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2 Answers2

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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 $$2A -3\pi B = -3\pi^2J+\pi^3 I$$ $$ -4\pi A+6\pi^2B =4\pi^3J-\pi^4 I$$

Guy Fsone
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    My question was about $x^2$, not $x^3$, but thanks anyways! This is still helpful! :D – Franklin Pezzuti Dyer Oct 07 '17 at 18:19
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    Typo, use $-3\pi \int_0^\pi x^2\ln(\sin(x))dx$ – gammatester Oct 07 '17 at 18:23
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    Unfortunately, the two equations you obtained are linearly dependent, as you can easily check. Also, since answer is $$B=-\frac{1}{6} \pi \left(3 \zeta (3)+\pi ^2 \log (4)\right)$$ your answer, if correct, would result in an elementary evaluation of $\zeta(3)$, which is extremely unlikely. – pisco Jan 22 '18 at 05:42
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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, $$\int_0^\pi {{x^2}\ln (\sin x)dx} = -\frac{\pi^3}{3}\ln 2 -\frac{\pi}{2}\zeta(3)$$

pisco
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