# Problem of the Week #248 - Apr 25, 2017

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#### Euge

##### MHB Global Moderator
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This week's problem was solved correctly by Opalg . You can read his solution below.

If $u = \frac12t$ then $1-e^{it} = 1 - \cos t - i\sin t = 2\sin^2u - 2i\sin u\cos u = 2\sin u(-ie^{iu})$, and so $|1-e^{it}| = 2|\sin u|.$ Therefore $$\int_{-\pi}^\pi \ln|1-e^{it}|\,dt = \int_{-\pi/2}^{\pi/2}\ln|2\sin u|\,2du = 4\int_0^{\pi/2}\ln(2\sin u)\,du$$ (since the integral from $-\frac12\pi$ to $0$ is the same as the integral from $0$ to $\frac12\pi$).

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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