Integral of $\log(\sin(x))$ using contour integrals

Question

I know the integral is possible with a simple fourier series expansion of $-\log(\sin(x))$

But I am interested in complex analysis, so I want to try this.

$$I = \int_{0}^{\pi} \log(\sin(x)) dx$$

The substitution $x = \arcsin(t)$ first comes to mind.

But that substitution isnt valid as,

The upper and lower bounds would both be $0$ because $\sin(\pi) = \sin(0) = 0$

What is a workaround using inverse trig or some other way?

Inverse sine is good, because that gives us a denominator from which we will be able to find poles.

$$x = arctan(t)$$

Is also good, but it would hard to do.

Idea?

Answer 1

We have

\begin{align}\int_0^{\pi} \log(\sin x)\, dx &= \int_0^{\pi} \log\left|\frac{e^{ix}-e^{-ix}}{2i}\right|\, dx\\ &= \int_0^{\pi} \log\left|\frac{e^{2ix}-1}{2}\right|\, dx\\ &= \int_0^{\pi} \log|e^{2ix}-1|\, dx - \int_0^{\pi} \log 2\, dx\\ &= \frac{1}{2}\int_0^{2\pi} \log|1 - e^{ix}|\, dx - \pi\log 2\\ &= \lim_{r\to 1^{-}} \frac{1}{2}\int_0^{2\pi} \log|1 - re^{ix}|\, dx - \pi\log 2\\ &= \lim_{r\to 1^{-}}\bigl(\pi\log|1 - z||_{z = 0}\bigr) - \pi \log 2 \quad (\text{by Gauss's mean value theorem})\\ &= -\pi\log 2. \end{align}

Answer 2

Another solution is as follows.

Let

$$ I = \int_0^{\pi} \log(\sin x)\, dx. $$

First note by symmetry,

$$ I = 2\int_0^{\pi/2} \log(\sin x)\, dx. $$

Now letting $x = 2u$, $dx = 2du$, and $\sin{2u} = 2\sin{u}\cos{u}$,

\begin{align} 2\int_0^{\pi/2} \log(\sin x)\, dx &= 4\int_0^{\pi/4} \log(\sin 2u)\, du\\ &= 4\int_0^{\pi/4} \log(2\sin{u}\cos{u})\, du\\ &= \pi \log{2} + 4\int_0^{\pi/4} \log(\sin u)\, du + 4\int_0^{\pi/4} \log(\cos u)\, du. \end{align}

In the last integral, letting $z = \pi/2 - u$, gives $dz = -du$ and $\cos u = \sin(z)$ and hence

\begin{align} I &= \pi \log{2} + 4\int_0^{\pi/4} \log(\sin u)\, du - 4\int_{\pi/2}^{\pi/4} \log(\sin z)\, dz\\ &= \pi \log{2} + 4\int_0^{\pi/4} \log(\sin u)\, du + 4\int_{\pi/4}^{\pi/2} \log(\sin z)\, dz\\ &= \pi \log{2} + 2I. \end{align}

Solving for $I$ shows

$$ I = -\pi \log{2}. $$

Answer 3

Consider $f(z)=\ln(1-z)$. Then $f$ is analytic in $|z|<r<1$. Thus by Cauchy's Formula, we have $$\int_{|z|=r}\frac{\ln(1-z)}{z}dz= 2\pi i 0=0.$$

Letting $z=re^{i\theta}$, we have $$ \int_0^{2\pi}r\ln\left(1-re^{i\theta}\right) \,d\theta=0. $$

Letting $r\to0^-$, we have $$ \int_0^{2\pi}\ln\left(1-e^{i\theta}\right) \,d\theta=0 $$

so that $$ \Re\int_0^{2\pi}\ln\left(1-e^{i\theta}\right) \,d\theta = \int_0^{2\pi}\Re\ln\left(1-e^{i\theta}\right) \,d\theta = \int_0^{2\pi}\ln\left|1-e^{i\theta}\right| \,d\theta=0. $$

But \begin{eqnarray} |1-e^{i\theta}|&=&|1-\cos\theta-i\sin\theta|=4\sin^2\frac{\theta}{2} \end{eqnarray}

and hence $$ \int_0^{2\pi}\ln\left(4\sin^2\frac{\theta}{2}\right) \,d\theta=0.$$

Changing variable $\theta\to\frac{\theta}{2}$, we have $$\int_0^\pi\ln(\sin\theta) \,d\theta=-\pi\ln2. $$