4
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ $$2sin^{2}(\frac{x}{2})= [1−cos(x)]$$ $\endgroup$ – user29418 Jul 23 at 1:20
  • 1
    $\begingroup$ 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
3
$\begingroup$

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}

$\endgroup$
2
$\begingroup$

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

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?

$\endgroup$
1
$\begingroup$

$$\int_0^\pi \log \left(4\sin^2\left(\frac{x}{2}\right)\right) dx$$ 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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.