Solving the integral $\int_{0}^{\pi}{\log(\sin x)}\,\mathrm dx$

Question

Solve this integral$$\int_{0}^{\pi}{\log(\sin x)}\,\mathrm dx$$

I have tried this by breaking the limit from $0$ to $\pi/2$ and $\pi/2$ to $\pi$, but I am unable to solve the 2nd part.

Answer 1

Splitting the integral as you mentioned works:

$$ \begin{aligned} \int\limits_{0}^{\pi}\ln (\sin(x))dx &= \int\limits_{0}^{\pi/2}\ln(\sin(x))dx+\int\limits_{\pi/2}^{\pi}\ln(\sin(x))dx\\ &= \int\limits_{0}^{\pi/2}\ln(\sin(x))dx+\int\limits_{0}^{\pi/2}\ln(\cos(x))dx\\ &= \int\limits_{0}^{\pi/2}\ln\left(\frac{1}{2}\sin(2x)\right)dx\\ &=\int\limits_{0}^{\pi/2}\ln\left(\sin(2x)\right)dx- \int\limits_{0}^{\pi/2}\ln(2)dx\\ &= \frac{1}{2}\int\limits_{0}^{\pi}\ln(\sin(x))dx-\frac{\pi}{2}\ln 2 \end{aligned} $$

and therefore:

$$\int\limits_{0}^{\pi}\ln (\sin(x))dx =-\pi\ln 2$$

Answer 2

Hint:

Use Euler's formula:

$$\begin{align}\int\ln(\sin(x))\,\mathrm dx&\equiv\int\ln\left(\dfrac{\mathrm i}2\left(\exp(-\mathrm ix) - \exp(\mathrm ix)\right)\right)\,\mathrm dx \\ &= \int\ln(\exp(-\mathrm ix) - \exp(\mathrm ix)) + \ln\left(\dfrac{\mathrm i}2\right)\,\mathrm du\end{align}$$

Let $u = \mathrm ix\implies\mathrm du = \mathrm i\mathrm dx$. Therefore,

$$\begin{align}\int\ln(\exp(-\mathrm ix) - \exp(\mathrm ix)) + \ln\left(\dfrac{\mathrm i}2\right)\,\mathrm du&\equiv-\mathrm i\int\ln(\exp(-u) + \exp(u)) + \ln\left(\dfrac{\mathrm i}2\right)\,\mathrm du \\ &=-\mathrm i\int\ln\left(\exp(-u) + \exp(u)\right)\,\mathrm du -\mathrm i\int\ln\left(\dfrac{\mathrm i}2\right)\,\mathrm du\end{align}$$

Solve the integral $$\int\ln\left(\exp(-u) + \exp(u)\right)\,\mathrm du$$ by parts by setting $$\begin{align}f = \ln(\exp(-u) + \exp(u))&\implies f' = \dfrac{-\exp(-u) - \exp(u)}{\exp(-u) - \exp(u)}, \\ g' = 1 &\implies g = u.\end{align}$$

Answer 3

Hint: use integration by parts: $$x\ln\left(\sin\left(x\right)\right)\Big|_0^\pi-\int_{0}^{\pi}x\cot\left(x\right)$$