오늘의 계산 51 - An Integral

이전에 비슷한 적분을 실해석적으로 푼 적이 있는데, 이번에는 깔끔하게 복소로 계산해보았습니다. 제 복소적분 실력도 못 써먹을 수준은 아니군요.

Problem. Prove that^[1] $$ \int_0^{\pi/2} \frac{\log(\cos x)}{x^2+\log^2(\cos x)} \, dx = \frac{\pi}{2}\left(1-\frac{1}{\log 2}\right) $$

Proof. Note that for $|x| < \frac{\pi}{2}$, we have

$$ \frac{\log\cos x}{\log^2 \cos x + x^2} = \Re \frac{1}{\log\left( \frac{1+e^{2ix}}{2} \right)}. $$

Thus if $I$ denotes the given integral, we have

\begin{align*} I &= \frac{1}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \Re \frac{1}{\log\left( \frac{1+e^{2ix}}{2} \right)} \, dx = \frac{1}{4} \int_{-\pi}^{\pi} \Re \frac{1}{\log\left( \frac{1+e^{ix}}{2} \right)} \, dx \\ &= \frac{1}{4} \mathrm{PV}\int_{-\pi}^{\pi} \Re \frac{1}{\log\left( \frac{1+e^{ix}}{2} \right)} \, dx = \frac{1}{4} \Re \mathrm{PV}\int_{-\pi}^{\pi} \frac{1}{\log\left( \frac{1+e^{ix}}{2} \right)} \, dx. \end{align*}

Now let $C_{\epsilon}$ be the counter-clockwised contour consisting of the circle of radius 1 centered at the origin, with two semicircular indents $\gamma_{1,\epsilon}$ around $1$ and $\gamma_{2,\epsilon}$ around $-1$ as follows:

By writing

\begin{align*} I = \frac{1}{4} \Re \lim_{\delta\to0^{+}}\left( \int_{-\pi+\delta}^{-\delta} + \int_{\delta}^{\pi-\delta} \right) \frac{1}{\log\left( \frac{1+e^{ix}}{2} \right)} \, dx \end{align*}

and plugging the substitution $z = e^{ix}$, we observe that

\begin{align*} I = \frac{1}{4} \Im \lim_{\epsilon\to 0^{+}}\left(\oint_{C_{\epsilon}} - \int_{\gamma_{1,\epsilon}} - \int_{\gamma_{2,\epsilon}} \right) \frac{dz}{z \log\left(\frac{1+z}{2}\right)} \end{align*}

Let

\begin{align*} f(z) = \frac{1}{z \log\left(\frac{1+z}{2}\right)}. \end{align*}

It is plain from the logarithmic singularity that

\begin{align*} \lim_{\epsilon \to 0^{+}} \int_{\gamma_{2,\epsilon}} f(z) \, dz = 0. \end{align*}

Also it follows that

\begin{align*} \lim_{\epsilon\to 0^{+}} \oint_{C_{\epsilon}} f(z) \, dz &= 2\pi i \operatorname{Res}_{z=0} f(z) = -\frac{2\pi i}{\log 2}, \\ \lim_{\epsilon\to 0^{+}} \int_{\gamma_{1,\epsilon}} f(z) \, dz &= -\pi i \operatorname{Res}_{z=1} f(z) = -2\pi i. \end{align*}

Therefore we have

\begin{align*} I &= \frac{1}{4} \Im \left( 2\pi i - \frac{2\pi i}{\log 2} \right) = \frac{\pi}{2} \left( 1 - \frac{1}{\log 2} \right). \end{align*}

References

1. Definite Integral $\int_0^{\pi/2} \frac{\log(\cos x)}{x^2+\log^2(\cos x)}dx = \frac{\pi}{2}\left(1-\frac{1}{\log 2}\right)$ in "Math StackExchange"