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- Jun 20, 2014

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Prove that $$\int_{-\pi}^{\pi}\ln\lvert 1 - e^{it}\rvert\, dt = 0$$

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- Thread starter
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- #1

- Jun 20, 2014

- 1,892

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Prove that $$\int_{-\pi}^{\pi}\ln\lvert 1 - e^{it}\rvert\, dt = 0$$

-----

Remember to read the POTW submission guidelines to find out how to submit your answers!

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- Jun 20, 2014

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Let \(\displaystyle J = \int_0^{\pi/2}\ln(\sin x)\,dx\).This is an improper integral because of the singularity at $x=0$. But since $\sin x \approx x$ when $x$ is close to $0$, the integral converges by comparison with \(\displaystyle \int\ln x\,dx = x\ln x - x + C\), which converges at $x = 0.$

To evaluate $J$, let $y = \frac12x$. Then $\sin x = 2\sin y\cos y $, and $$J = \int_0^{\pi/4}\ln(2\sin y \cos y)\,2dy = 2\int_0^{\pi/4} \bigl(\ln2 + \ln\sin y + \ln\cos y\bigr)\,dy = \tfrac\pi2\ln2 + 2\int_0^{\pi/4}\ln\sin y\,dy + 2\int_0^{\pi/4}\ln\cos y\,dy.$$ In the last of those integrals, make the substitution $y\to \frac\pi2-y.$ It then becomes \(\displaystyle 2\int_{\pi/4}^{\pi/2}\ln\sin y\,dy\) (because $\cos\bigl(\frac\pi2 - y\bigr) = \sin y$). Therefore $$J = \tfrac\pi2\ln2 + 2\int_0^{\pi/4}\ln\sin y\,dy + 2\int_{\pi/4}^{\pi/2}\ln\sin y\,dy = \tfrac\pi2\ln2 + 2J.$$ So $J = -\tfrac\pi2\ln2.$ It follows that \(\displaystyle \int_0^{\pi/2}\ln(2\sin x)\,dx = \int_0^{\pi/2}\ln2\,dx + \int_0^{\pi/2}\ln(\sin x)\,dx = \tfrac\pi2\ln2 + J = 0.\) From the first paragraph of the proof, this shows that \(\displaystyle \int_{-\pi}^\pi \ln|1-e^{it}|\,dt = 0.\)

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