I want to show that

$$\int_0^{\pi} \log(2 - 2 \cos x) = 0$$

However, I cannot do this. I tried splitting the integral into $\int_0^{\pi/3} \log(2 - 2 \cos x)\,dx + \int_{\pi/3}^{\pi} \log(2 - 2 \cos x) \,dx$ and showing that the two parts were negatives of one another. Wolframalpha does not give very a simple antiderivative. I was wondering if there was a nice way to do this.

Other attempts: using $\int_0^a f(x) \,dx = \int_0^{a} f(a-x) \,dx$, trying to change the $\cos$ to $\sin$ by some substitution like $u = \pi/2 - x$ and trying to get things to cancel.

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    $\begingroup$ $$2sin^{2}(\frac{x}{2})= [1−cos(x)]$$ $\endgroup$ – user29418 Jul 23 at 1:20
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    $\begingroup$ It's improper at $0$. $\endgroup$ – Randall Jul 23 at 1:26
  • $\begingroup$ $$\ln\left(2 - 2\cos(x)\right) = \ln\left(2\left(1 - \cos(x)\right)\right) = \ln(2) + \ln\left(1 - \cos(x)\right) $$ Thus, $$I = \ln(2)\pi + \int_0^\pi \ln\left(1 - \cos(x)\right)\:dx $$ For the remaining integral use the Weirerstrauss Substitution $t = \tan\left( x/2\right)$ $\endgroup$ – user679268 Jul 23 at 1:36

First we prove

$$\displaystyle\int_0^{\pi/2} \log \left(\sin x\right)\,dx =-\frac12\pi\log2$$

Proof $\ $ From the substitution $x \to \frac{\pi}{2}-x$ , we get $$ \displaystyle\int_0^{\pi/2} \log( \sin x )\,dx = \displaystyle\int_0^{\pi/2} \log ( \cos x )\,dx $$

Thus \begin{align} 2\displaystyle\int_0^{\pi/2} \log( \sin x )\,dx &= \displaystyle\int_0^{\pi/2} \log (\sin x \cos x )\,dx \\ &= \displaystyle\int_0^{\pi/2} \log (\sin 2x )\,dx-\frac{1}{2}\pi\log 2 \\ &=\frac12\displaystyle\int_0^{\pi} \log (\sin x )\,dx-\frac{1}{2}\pi\log 2 \\ &= \displaystyle\int_0^{\pi/2} \log( \sin x )\,dx -\frac{1}{2}\pi\log 2 \end{align} Then we arrive at the conclusion.

To calculate the original integral, we have \begin{align} \int_0^\pi \log(2-2\cos x)\,dx &= \int_0^\pi \log(4\sin^2 \frac{x}{2})\,dx \\ &= 2\pi \log 2 + 2 \int_0^\pi \log (\sin \frac{x}{2}) \,dx \\ &= 2\pi \log 2 + 4\int_0^{\pi/2} \log \left(\sin x\right)\,dx \\ &= 0 \end{align}


$$\log(2-2\cos x)=\log(2(1-\cos x))=\log2+\log(2\sin^2 x/2)=2\log2+2\log\sin x/2$$

If we show that $\int_0^\pi\log\sin(x/2)~\mathrm dx=-\pi\log2$, we are done.

With $t\mapsto x/2$, the above claim is,

$$\int_0^{\pi/2}\log\sin t~\mathrm dt=-\frac \pi2\log2$$

This is a pretty well known integral. Can you take it from here?


$$\int_0^\pi \log \left(4\sin^2\left(\frac{x}{2}\right)\right) dx$$ is equivalent. Use logarithm rules to make it - $$\int_0^\pi \left[ \log 4 + 2 \log\left(\sin\left(\frac{x}{2}\right)\right) \right] dx$$

The rightmost part of the integral has already been solved, which is linked here.

  • $\begingroup$ sorry for bad math jax, if anyone could edit (i’m on mobile) $\endgroup$ – Zach Jul 23 at 1:42
  • $\begingroup$ No worries - already done :-) $\endgroup$ – user679268 Jul 23 at 2:27

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