수학 얘기/계산
오늘의 계산 31 - An Integral (With Logarithmic Behavior)
sos440
2010. 3. 10. 19:41
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...