오늘의 계산 31 - An Integral (With Logarithmic Behavior)

Problem. Evaluate the following integral. \begin{equation}\label{eqn:wts} \int_{0}^{1}\frac{\log \cos ( \frac{\pi x}{2} )}{x (x+1)} \, dx \end{equation}

We first introduce a lemma which will play the key role to our calculation.

Lemma. Define the function $F(s)$ by \begin{equation*} F(s) = \int_{0}^{1} \frac{\log | \cos (sx) | }{x} \, dx - \int_{1}^{2} \frac{\log | \sin (sx) | }{x} \, dx. \end{equation*} Then $F(s)$ satisfies the functional equation \begin{equation*} F(s) = F\left( \tfrac{s}{2} \right) - \log^2 2, \end{equation*} where $s > 0$ and $\log^2 x = (\log x)^2$.

Proof. It is a consequence of direct calculation, combined with double angle formula for sine. \begin{align*} F(s) &= \int_{0}^{1} \frac{\log | \cos (sx) | }{x} \, dx - \int_{1}^{2} \frac{\log | \sin (sx) | }{x} \, dx \\ & = \int_{0}^{2} \frac{\log | \cos (sx/2) | }{x} \, dx - \int_{1}^{2} \frac{\log | \sin (sx) | }{x} \, dx \qquad (x \mapsto x/2) \\ & = \int_{0}^{1} \frac{\log | \cos (sx/2) | }{x} \, dx - \int_{1}^{2} \frac{\log | \sin (sx) | - \log | \cos (sx/2) |}{x} \, dx \\ & = \int_{0}^{1} \frac{\log | \cos (sx/2) | }{x} \, dx - \int_{1}^{2} \frac{\log 2 + \log | \sin (sx/2) |}{x} \, dx \\ &= F\left( \tfrac{s}{2} \right) - \log^2 2. \end{align*}

Now we are ready to solve the problem.

Solution of the Problem. Let $I$ denote the integral \eqref{eqn:wts}. The partial fraction decomposition shows that \begin{align*}I & = \int_{0}^{1} \frac{\log \cos \frac{\pi x}{2}}{x} \, dx - \int_{0}^{1} \frac{\log \cos \frac{\pi x}{2}}{x+1} \, dx \\ &= \int_{0}^{1} \frac{\log \cos \frac{\pi x}{2}}{x} \, dx - \int_{1}^{2} \frac{\log \sin \frac{\pi t}{2}}{t} \, dt \qquad (x+1 = t) \\ &= F\left( \tfrac{\pi}{2} \right). \end{align*} This implies that \begin{align*}I &= F( \pi 2^{-n} ) - (n-1) \log^2 2 \\ & = \int_{0}^{1} \frac{\log \cos (\pi 2^{-n} x) }{x} \, dx - \int_{1}^{2} \frac{\log \sin (\pi 2^{-n} x) }{x} \, dx - (n-1) \log^2 2 \\ & = \int_{0}^{\pi 2^{-n}} \frac{\log \cos t }{t} \, dt - \int_{1}^{2} \left[ \log \left( \frac{\sin (\pi 2^{-n} x)}{\pi 2^{-n} x} \right) - n \log 2 + \log \pi + \log x \right] \, \frac{dx}{x} \\ & \quad - (n-1) \log^2 2 \\ & \to \frac{1}{2} \log^2 2 - \log \pi \log 2 \end{align*} as $n \to \infty$. This completes the calculation.

p.s. 혹자는 제가 계산을 잘 한다고 하는데, 그건 제가 성공한 계산만 올리기 때문에 그렇게 보이는 것일 뿐입니다 =ㅁ=;; 아는 것 쥐뿔에 모르는 것 태산이죠...