How to solve the integral $\int_0^\pi\ln(\sin(\theta))\sin(\theta)d\theta$?

Question

Could you help me with a comprehensive one? I have tried to solve it but I can't reach the result. The integral I am trying to evaluate is $$\int_0^\pi\ln(\sin(\theta))\sin(\theta)d\theta$$and according to Wolfram alpha the result should be $\ln(4)-2$.

I already have tried to evaluate it using integration by parts but, when doing so, I find a term of $\cos(\theta)\ln(\sin(\theta)) $ which is evaluated from $0$ to $\pi$ and this is where I stay because those values make the result ​​indeterminate. I have tried to make variable changes but I have not been able to find a suitable one that allows me to avoid indeterminate results. I would like you to provide me with a suggestion to avoid the indeterminacies.

Answer 1

You can do this: \begin{align*}&\ \ \ \ \int_0^\pi\ln(\sin(\theta))\sin(\theta)d\theta \\&= \frac{1}{2}\int_0^\pi\ln(1-\cos^2(\theta))\sin(\theta) d\theta= \\ &=\frac{1}{2}\left(\int_0^\pi\ln(1-\cos(\theta))d(1-\cos(\theta))-\int_0^\pi\ln(1+\cos(\theta))d(1+\cos(\theta))\right) \\ &=\frac{1}{2}\left((1-\cos(\theta))(\ln(1-\cos(\theta))-1)|_{0}^{\pi}-(1+\cos(\theta))(\ln(1+\cos(\theta))-1)|_{0}^{\pi}\right) \\ &=\frac{1}{2}(2\ln 2-2+2\ln 2 -2)\\&= \ln(4)-2 \end{align*}

Answer 2

This integral is related to the beta function. Substitute $\theta=\tau+\dfrac\pi2$ and take advantage of symmetry to write

$$\begin{align*} & \int_0^\pi \ln(\sin \theta) \sin \theta \, d\theta \\ &= \int_{-\tfrac\pi2}^{\tfrac\pi2} \ln\left(\sin \left(\tau+\frac\pi2\right)\right) \sin \left(\tau+\frac\pi2\right) \, d\tau \\ &= \int_{-\tfrac\pi2}^{\tfrac\pi2} \ln(\cos \tau) \cos \tau \, d\tau \\ &= 2 \int_0^{\tfrac\pi2} \ln(\cos \tau) \cos \tau \, d\tau \\ &= 2 \lim_{a\to1} \frac{\partial}{\partial a} \int_0^{\tfrac\pi2} \cos^a\tau \, d\tau \\ &= \lim_{a\to1} \frac{\partial}{\partial a} \operatorname{B}\left(\frac12,\frac a2+\frac12\right) \\ &= \frac12 \lim_{a\to1} \operatorname{B}\left(\frac12,\frac a2+\frac12\right) \left[\psi\left(\frac a2+\frac12\right)- \psi\left(\frac a2+1\right)\right] \\ &= -\gamma - (-\gamma+2-\ln4) \end{align*}$$

and the result follows. $\gamma$ is the Euler-Mascheroni constant and $\psi$ denotes digamma.

Answer 3

$$ \begin{aligned} \int_0^\pi \ln (\sin \theta) \sin \theta d \theta &=2 \int_0^{\frac{\pi}{2}} \ln (\sin \theta) \sin \theta d \theta \\ = & -2 \int_0^{\frac{\pi}{2}} \ln (\sin \theta) d(\cos \theta-1) \quad \textrm{ (IBP)} \\ = & \underbrace{ -2[(\cos \theta-1) \ln (\sin \theta)]_0^{\frac{\pi}{2}}}_{=0} +2 \int_0^{\frac{\pi}{2}} \frac{\cos \theta-1}{\sin \theta} \cdot \cos \theta d \theta\\=& 2\int_0^{\frac{\pi}{2}} \frac{\cos \theta-1}{\sin \theta} d(\sin \theta)\\=& 2[\cos \theta-1] _0^{\frac{\pi}{2}}-2\int _0^{\frac{\pi}{2}}\frac{\cos \theta-1}{\sin \theta} d \theta\\=&-2 +2 \int_0^{\frac{\pi}{2}} \tan \frac{\theta}{2} d \theta\\=&-2 -4\left[\ln \left(\cos \frac{\theta}{2}\right)\right]_0^{\frac{\pi}{2}}\\=&2(\ln2-1) \end{aligned} $$