# How to compute $\int_0^\pi \ln^2(\sin x)dx$ using complex analysis?

I would like to know if the integral $$\int_0^\pi \ln^2(\sin x)dx$$ could be attacked with contour integration. I have seen the integral evaluated using the Fourier series for $$\ln(\sin x)$$ and Cauchy product for infinite sums, but this method gets extremely hard for higher degrees than 2, which is why I want to see it evaluated using complex analysis.

• Is this $\ln(\ln(\sin(x))$ or is this $\ln(\sin(x))*\ln(\sin(x)))$? Jan 30, 2022 at 20:16
• It’s squared. It’s quite widely accepted notation from what I knew. Jan 30, 2022 at 20:17
• @intellect4 - both are widely accepted notation. Hence the question. For instance, while $\sin^2x$ usually means $(\sin(x))^2$, $d^2y$ usually means $d(d(y))$. Jan 31, 2022 at 0:03
• For "higher degrees" use $\int_0^\pi\ln^n\sin x\,dx=f^{(n)}(0)$ where $f(\lambda)=\int_0^\pi\sin^\lambda x\,dx$ is a "beta" integral; this gives expressions in terms of the polygamma functions (or, finally, values of the $\zeta$-function). Jan 31, 2022 at 4:35
• Thanks, I’ll look into it Feb 1, 2022 at 0:24

There is a method to evaluate your integral using contour integration, though I'm unsure if it generalises neatly to higher powers. First, let $$\begin{equation} \begin{split} I&=\int_0^\pi\log^2(\sin(\theta))d\theta\\ &=\frac12 \int_0^{2\pi}\log^2(\sin(\theta /2))d\theta\\ &=\frac12 \int_0^{2\pi}\left[\log(1-e^{i\theta})-\log(2) -\frac{i}{2}(\theta-\pi)\right]^2d\theta\\ \end{split} \end{equation}$$ (Note that we're taking the principal branch of the logarithm) Expanding the square, and taking real parts (which we may do since $$I$$ is real valued), we see that $$I=\frac{1}{2}(J_1+J_2+J_3)$$, where $$\begin{equation} \begin{split} J_1&=\mathfrak{R}\int_0^{2\pi}\left[\log^2(2)-\frac14(\theta-\pi)^2\right]d\theta\\ J_2&=\mathfrak{R}\int_0^{2\pi}[-i\log(1-e^{i\theta})(\theta-\pi)]d\theta\\ J_3&=\mathfrak{R}\int_0^{2\pi}[\log^2(1-e^{i\theta})-2\log(2)\log(1-e^{i\theta})]d\theta\\ \end{split} \end{equation}$$ It is easy to evaluate $$J_1=2\pi\log^2(2)-\frac{\pi^3}{6}$$. To evaluate $$J_2$$, note that $$\mathfrak{R}[-i\log(1-e^{i\theta})]=\frac{1}{2}(\theta-\pi)$$ for $$\theta\in(0,2\pi)$$, so $$\begin{equation} J_2=\int_0^{2\pi}\frac{1}{2}(\theta-\pi)^2d\theta=\frac{\pi^3}{3} \end{equation}$$ Finally, to evaluate $$J_3$$, note that $$\begin{equation} \begin{split} J_3&=\mathfrak{R}\lim_{r\rightarrow 1^{-}}\int_0^{2\pi}[\log^2(1-re^{i\theta})-2\log(2)\log(1-re^{i\theta})]d\theta\\ &=\mathfrak{R}\lim_{r\rightarrow 1^{-}}\oint_{|z|=r}-\frac{i}{z}[\log^2(1-z)-2\log(2)\log(1-z)]dz\\ \end{split} \end{equation}$$ Since the above integrand is analytic in the open unit disc (apart from a removable singularity at $$z=0$$), by Cauchy's integral theorem, the above integrals vanish for each $$r\in(0,1)$$, and thus $$J_3=0$$. Combining our work, we have that $$\begin{equation} I=\pi\log^2(2)+\frac{\pi^3}{12} \end{equation}$$