How evaluate this integral? $$I=\int_0^{\pi/2}\frac{x^2\log^2{(\sin{x})}}{\sin^2x}\,dx$$ Note: $$\int_0^{\pi/2}\frac{x^2\log{(\sin x)}}{\sin^2x}dx=\pi\ln{2}-\frac{\pi}{2}\ln^22-\frac{\pi^3}{12}.$$

  • 4
    $\begingroup$ It would be useful to see how you derived the last identity. Fourier series, differentiation under the integral sign or what else? $\endgroup$ Feb 10, 2015 at 22:10
  • $\begingroup$ $$u=log{\,x}, du=cot{\,x}; v=\int\frac{x^2}{sin^2x}=\operatorname{CI_2}{(2x)}+2x\log({2sin{\,x})}-x^2cot{\,x}$$ $\endgroup$
    – user178256
    Feb 10, 2015 at 23:19
  • $\begingroup$ sorry,$$u=log{sinx}$$ $\endgroup$
    – user178256
    Feb 10, 2015 at 23:21
  • $\begingroup$ Do you know how to find the integral $\int_0^{\pi/2}\frac{x^2\log{(\sin x)}}{\sin^2x}dx=\pi\ln{2}-\frac{\pi}{2}\ln^22-\frac{\pi^3}{12}.$ $\endgroup$
    – science
    Feb 11, 2015 at 18:39
  • $\begingroup$ Below I have just computed the Fourier series of $\log\sin x,\log^2\sin x$ and $\frac{x^2}{\sin^2 x}$. $\endgroup$ Feb 11, 2015 at 19:06

3 Answers 3




Let's start out with the auxiliary result \begin{equation*} \int_0^{\pi/2}\frac{x^2\log{(\sin(x))}}{\sin^2(x)}dx=\pi\ln{(2)}-\frac{\pi}{2}\ln^2(2)-\frac{\pi^3}{12}. \end{equation*}

By the integration by parts all reduces to $$\int_0^{\pi/2} \cot(x) (x^2 \cot(x)+2 x\log(\sin(x))) \ dx=\int_0^{\pi/2} x^2 \cot^2(x) \ dx+2\int_0^{\pi/2} x\cot(x) \log(\sin(x))) \ dx.$$ For the first integral in the right-hand side we apply the integration by parts that yields \begin{equation*} \begin{aligned} \int_0^{\pi/2} x^2 \cot^2(x) \ dx &=-\frac{\pi^3}{8}+\int_0^{\pi/2} 2 x (x+\cot (x)) \ dx \\ &=-\frac{\pi^3}{24}+2\int_0^{\pi/2}x \cot (x) \ dx \\ &=-\frac{\pi^3}{24}-2\int_0^{\pi/2} \log(\sin(x)) \ dx \\ &=-\frac{\pi^3}{24}-\int_0^{\pi} \log(\sin(x)) \ dx \\ \end{aligned} \end{equation*} where in the penultimate equality we used again the integration by parts, and then the symmetry.

Then, \begin{equation*} \begin{aligned} \int_0^{\pi} \log(\sin(x)) \ dx &=\int_0^{\pi} \log(2\sin(x/2)\cos(x/2)) \ dx \\ &=\pi \log(2)+ \int_0^{\pi} \log(\sin(x/2)) \ dx+ \int_0^{\pi} \log(\cos(x/2)) \ dx. \end{aligned} \end{equation*} Letting $x/2=y$ in both integrals in the right-hand side, we obtain that \begin{equation*} \begin{aligned} \int_0^{\pi} \log(\sin(x)) \ dx &=\pi \log(2) + 2\int_0^{\pi/2} \log(\sin(x)) \ dx+2\int_0^{\pi/2} \log(\cos(x)) \ dx \\ &=\pi \log(2) + 4\int_0^{\pi/2} \log(\sin(x)) \ dx \\ &=\pi \log(2) + 2\int_0^{\pi} \log(\sin(x)) \ dx \\ \end{aligned} \end{equation*} whence we get that
$$\int_0^{\pi} \log(\sin(x)) \ dx =-\pi\log(2).$$ Then, $$\int_0^{\pi/2} x^2 \cot^2(x) \ dx=\pi\log(2)-\frac{\pi^3}{24}.$$

On the other hand, the integration by parts yields that \begin{equation*} \begin{aligned} 2\int_0^{\pi/2}x\cot(x) \log(\sin(x)) \ dx &=-2\int_0^{\pi/2} (\log ^2(\sin (x))+ x \cot (x) \log (\sin (x))) \ dx \\ &=-2\int_0^{\pi/2} \log ^2(\sin (x)) \ dx -2\int_0^{\pi/2} x \cot (x) \log (\sin (x)) \ dx \end{aligned} \end{equation*}

whence we have that $$\int_0^{\pi/2}x\cot(x) \log(\sin(x)) \ dx=-\frac{1}{2}\int_0^{\pi/2} \log ^2(\sin (x)) \ dx.$$ According to the trigonometric form of the beta function, we have that $$\int_0^{\pi/2} \sin^a(x)\cos^b(x) \ dx=\frac{1}{2}B \left(\frac{1}{2}(a+1),\frac{1}{2}(b+1)\right).$$ Differentiating $2$ times with respect to $a$ and then letting $a\to 0$ and $b\to 0$, we obtain that \begin{equation*} \begin{aligned} \int_0^{\pi/2} \log ^2(\sin (x)) \ dx &=\frac{1}{2} \lim_{b \to 0} \lim_{a \to 0} \frac{\partial^2}{\partial a^2}\left(B \left(\frac{1}{2}(a+1),\frac{1}{2}(b+1)\right)\right) \\ &=\frac{1}{24} \left(\pi ^3+12 \pi \log ^2(2)\right). \end{aligned} \end{equation*}

Thus, $$\int_0^{\pi/2}x\cot(x) \log(\sin(x)) \ dx=-\frac{1}{48} \left(\pi ^3+12 \pi \log ^2(2)\right).$$

and finally our auxiliary result is proved. $$\int_0^{\pi/2}\frac{x^2\log{(\sin(x))}}{\sin^2(x)}dx=\pi\ln{(2)}-\frac{\pi}{2}\ln^2(2)-\frac{\pi^3}{12}.$$

Now, we prove the main result,

\begin{equation*} \int_0^{\pi/2}\frac{x^2\log^2{(\sin(x))}}{\sin^2(x)}dx=\left(\frac{\pi ^3 }{6} +2 \pi \right) \log (2)+\frac{1}{3} \pi \log ^3(2)+\frac{1}{8}\pi \zeta (3)-\frac{\pi ^3}{6}-\pi \log ^2(2). \end{equation*}

Applying the integration by parts, we get $$2\int_0^{\pi/2} x^2 \cot ^2(x) \log (\sin (x)) \ dx+2 \int_0^{\pi/2} x \cot (x) \log ^2(\sin (x)) \ dx$$ For the integral in the left side we make use of the integration by parts that yields $$2\int_0^{\pi/2} x^2 \cot ^2(x) \log (\sin (x)) \ dx$$ $$=2 \int_0^{\pi/2} x^3 \cot (x) \ dx + 2 \int_0^{\pi/2} x^2 \cot ^2(x) \ dx + 4\int_0^{\pi/2} x^2 \log (\sin (x)) \ dx+4 \int_0^{\pi/2} x \cot (x) \log (\sin (x)) \ dx$$ $$=\frac{2}{3} \int_0^{\pi/2} x^3 \cot (x) \ dx + 2 \int_0^{\pi/2} x^2 \cot ^2(x) \ dx +4 \int_0^{\pi/2} x \cot (x) \log (\sin (x)) \ dx$$ and since the last $2$ integrals are already compute (see the auxiliary result), we obtain $$\frac{2}{3} \int_0^{\pi/2} x^3 \cot (x) \ dx + 2 \int_0^{\pi/2} x^2 \cot ^2(x) \ dx +4 \int_0^{\pi/2} x \cot (x) \log (\sin (x)) \ dx$$ $$=\frac{2}{3} \int_0^{\pi/2} x^3 \cot (x) \ dx+2\pi\log(2)-\pi\log^2(2)-\frac{\pi^3}{6}$$ and integrating by parts again, we get $$-2\int_0^{\pi/2} x^2 \log(\sin(x)) \ dx+2\pi\log(2)-\pi\log^2(2)-\frac{\pi^3}{6}.$$

Using that $\displaystyle \log(\sin(x))=-\log(2)-\sum_{n=1}^{\infty} \frac{\cos(2 n x)}{n}$, we obtain
$$-2\int_0^{\pi/2} x^2\left(-\log(2)-\sum_{n=1}^{\infty} \frac{\cos(2 n x)}{n}\right)\ dx+2\pi\log(2)-\pi\log^2(2)-\frac{\pi^3}{6}$$ $$=2\log(2)\int_0^{\pi/2} x^2\ dx+2\int_0^{\pi/2} x^2 \sum_{n=1}^{\infty} \frac{\cos(2 n x)}{n} \ dx+2\pi\log(2)-\pi\log^2(2)-\frac{\pi^3}{6}$$ $$=2\int_0^{\pi/2} x^2 \sum_{n=1}^{\infty} \frac{\cos(2 n x)}{n} \ dx+2\pi\log(2)-\pi\log^2(2)-\frac{\pi^3}{6}+\frac{1}{12} \pi ^3 \log (2)$$ $$=2 \sum_{n=1}^{\infty} \int_0^{\pi/2} x^2 \frac{\cos(2 n x)}{n} \ dx+2\pi\log(2)-\pi\log^2(2)-\frac{\pi^3}{6}+\frac{1}{12} \pi ^3 \log (2)$$ $$=-\frac{3}{8} \pi \zeta(3)+2\pi\log(2)-\pi\log^2(2)-\frac{\pi^3}{6}+\frac{1}{12} \pi ^3 \log (2).$$

Therefore, we have that $$2\int_0^{\pi/2} x^2 \cot ^2(x) \log (\sin (x)) \ dx=-\frac{3}{8} \pi \zeta(3)+2\pi\log(2)-\pi\log^2(2)-\frac{\pi^3}{6}+\frac{1}{12} \pi ^3 \log (2).$$

For the remaining integral, we use the integration by parts again that yields $$2 \int_0^{\pi/2} x \cot (x) \log ^2(\sin (x)) \ dx=-2\int_0^{\pi/2} \log ^3(\sin (x)) \ dx-4 \int_0^{\pi/2} x \cot (x) \log ^2(\sin (x)) \ dx$$ and thus $$2 \int_0^{\pi/2} x \cot (x) \log ^2(\sin (x)) \ dx=-\frac{2}{3}\int_0^{\pi/2} \log ^3(\sin (x)) \ dx.$$

According to the trigonometric form of the beta function, we know that $$\int_0^{\pi/2} \sin^a(x)\cos^b(x) \ dx=\frac{1}{2}B \left(\frac{1}{2}(a+1),\frac{1}{2}(b+1)\right).$$ Differentiating $3$ times with respect to $a$ and then letting $a\to 0$ and $b\to 0$, we obtain that \begin{equation*} \begin{aligned} \int_0^{\pi/2} \log ^3(\sin (x)) \ dx &=\frac{1}{2} \lim_{b \to 0} \lim_{a \to 0} \frac{\partial^3}{\partial a^3}\left(B \left(\frac{1}{2}(a+1),\frac{1}{2}(b+1)\right)\right) \\ &=-\frac{3 \pi }{4}\zeta (3)-\frac{1}{2} \pi \log ^3(2)-\frac{1}{8} \pi ^3 \log (2). \end{aligned} \end{equation*}

Thus, $$2 \int_0^{\pi/2} x \cot (x) \log ^2(\sin (x)) \ dx=\frac{1}{2}\pi \zeta (3)+\frac{1}{3} \pi \log ^3(2)+\frac{1}{12} \pi ^3 \log (2).$$

Hence, $$\int_0^{\pi/2}\frac{x^2\log^2{(\sin(x))}}{\sin^2(x)}dx=\left(\frac{\pi ^3 }{6} +2 \pi \right) \log (2)+\frac{1}{3} \pi \log ^3(2)+\frac{1}{8}\pi \zeta (3)-\frac{\pi ^3}{6}-\pi \log ^2(2).$$


  • $\begingroup$ This question is about to be deleted (see discussion here). $\endgroup$
    – user1729
    Aug 31, 2018 at 15:18

Since $$\log(2\sin x)=-\sum_{k\geq 1}\frac{\cos(2kx)}{k}\tag{1}$$ it follows that: $$\log^2(2\sin x)=\sum_{j,k\geq 1}\frac{\cos(2kx)\cos(2jx)}{kj}=\frac{1}{2}\sum_{j,k\geq 1}\frac{\cos(2(k+j)x)+\cos(2(k-j)x)}{kj}$$ where: $$\sum_{j,k\geq 1}\frac{\cos(2(k+j)x)}{kj}=\sum_{n\geq 2}\cos(2nx)\sum_{h=1}^{n-1}\frac{1}{h(n-h)}=\sum_{n\geq 2}\frac{2 H_{n-1}}{n}\cos(2nx),$$ $$\sum_{k\geq j\geq 1}\frac{\cos(2(k-j)x)}{kj}=\sum_{n\geq 0}\cos(2nx)\sum_{j=1}^{+\infty}\frac{1}{j(n+j)}=\zeta(2)+\sum_{n\geq 1}\frac{H_n}{n}\cos(2nx),$$ so: $$\log^2(2\sin x) = \frac{\pi^2}{3}+\sum_{n\geq 1}\frac{H_{n-1}+H_n}{n}\,\cos(2nx)\tag{2}$$ and we just need to find: $$ J_n = \frac{4}{\pi}\int_{0}^{\pi/2}\frac{x^2}{\sin^2 x}\cos(2nx)\,dx \tag{3}$$ to turn the computation of the integral into a computation of a series. Now:

$$ J_n = -\frac{16}{\pi}\cdot\Re\int_{0}^{\pi/2}\frac{x^2 e^{2nix}}{(e^{ix}-e^{-ix})^2}\,dx\tag{4}$$ and since: $$\frac{x^2 e^{2(n-1)ix}}{(1-e^{-2ix})^2}=x^2 e^{2(n-1)ix}\left(1+2e^{-2ix}+3e^{-4ix}+4e^{-6ix}+\ldots\right),$$ $$\Re\int_{0}^{\pi/2}x^2 e^{2mix}\,dx = (-1)^m\frac{\pi}{m^2},$$ it follows that: $$\begin{eqnarray*} J_n &=&4\log 2-n\frac{\pi^2}{3}-4\sum_{k=1}^{n-1}(n-k)\frac{(-1)^k}{k^2}\\&=&4n\sum_{k\geq n}\frac{(-1)^k}{k^2}-4\sum_{k\geq n}\frac{(-1)^k}{k}\tag{5}\end{eqnarray*} $$ The plan, now, is to exploit partial summation through the identities: $$ \sum_{k=1}^{n}\frac{H_{k-1}}{k}=\frac{1}{2}\left(H_n^2-H_n^{(2)}\right), \qquad \sum_{k=1}^{n}\frac{H_{k}}{k}=\frac{1}{2}\left(H_n^2+H_n^{(2)}\right), $$ $$ \sum_{k=1}^{n}H_k = n H_n - \sum_{k=1}^{n-1}\frac{k}{k+1} = n H_n -n + H_n.$$ Continues.


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