Calculation of Integral with log term and some lessons

By | June 17, 2015

Problem. Evaluate

    \[ \int_{0}^{\infty}\frac{\left(\ln x\right)^{2}}{1+x^{2}}dx. \]

이 값을 구하는 과정에서 다음과 같은 결과가 필요하다.

Proposition. We have

    \[ \int_{0}^{\infty}\frac{\ln x}{1+x^{2}}dx=0. \]

Proof. Define f\left(z\right)=\frac{\log z}{1+z^{2}} where \left|z\right|>0 and -\frac{\pi}{2}<\arg z<\frac{3\pi}{2}. Consider the following contour:

Untitled-2

Here 0<\rho<1<R. Then inside the contour, f has simple pole at i. So

    \[ \mathrm{Res} (f;i)=\frac{\log i}{2i}=\frac{i\left(\frac{\pi}{2}\right)}{2i}=\frac{\pi}{4}. \]

So by Residue theorem, we get

    \[ \int_{\rho}^{R}f\left(x\right)dx+\int_{C_{R}}f\left(z\right)dz+\int_{-R}^{-\rho}\frac{\ln\left(-x\right)+i\pi}{1+x^{2}}dx-\int_{C_{\rho}}f\left(z\right)dz=2\pi i\left(\frac{\pi}{4}\right). \]

First, note that

    \[ \int_{-R}^{-\rho}\frac{\ln\left(-x\right)+i\pi}{1+x^{2}}dx=\int_{\rho}^{R}\frac{\ln x+i\pi}{1+x^{2}}dx=\int_{\rho}^{R}f\left(x\right)dx+i\pi\int_{\rho}^{R}\frac{1}{1+x^{2}}dx. \]

Also, we have

    \[ \left|\int_{C_{R}}f\left(z\right)dz\right|\le\pi R\cdot\frac{\sqrt{\left(\ln R\right)^{2}+\pi^{2}}}{R^{2}-1}. \]

Note that R\cdot\frac{\sqrt{\left(\ln R\right)^{2}+\pi^{2}}}{R^{2}-1}=O\left(\frac{\ln R}{R}\right) as R\rightarrow\infty and R\ln R\rightarrow0 as R\rightarrow0^{+}. Hence, as \rho\rightarrow0 and R\rightarrow\infty, we have

    \[ 2\int_{0}^{\infty}f\left(x\right)dx+i\pi\int_{0}^{\infty}\frac{1}{1+x^{2}}dx=\pi i\left(\frac{\pi}{2}\right). \]

Since i\pi\int_{0}^{\infty}\frac{1}{1+x^{2}}dx=\pi i\left(\frac{\pi}{2}\right), we get \int_{0}^{\infty}f\left(x\right)dx=0.


이제 본격적으로 문제의 적분을 계산해보도록 하자.

Solution to Problem. Define f\left(z\right)=\frac{\left(\log z\right)^{2}}{z^{2}+1} where \left|z\right|>0 and -\frac{\pi}{2}<\arg z<\frac{3\pi}{2}. Consider the following contour:

Untitled-2

Here 0<\rho<1<R. Then inside the contour, f has simple pole at i. So

    \[ \mathrm{Res} (f,i)=\frac{\left(i\frac{\pi}{2}\right)^{2}}{2i}=\frac{\pi^{2}i}{8}. \]

Hence, by Residue theorem, we get

(1)   \begin{equation*} \int_{\rho}^{R}f\left(x\right)dx+\int_{C_{R}}f\left(z\right)dz+\int_{-R}^{-\rho}\frac{\left(\ln\left(-x\right)+i\pi\right)^{2}}{1+x^{2}}dx-\int_{C_{\rho}}f\left(z\right)dz=2\pi i\left(\frac{\pi^{2}i}{8}\right)=-\frac{\pi^{3}}{4}. \end{equation*}

Note that

(2)   \begin{align*} \int_{-R}^{-\rho}\frac{\left(\ln\left(-x\right)+i\pi\right)^{2}}{1+x^{2}}dx & =\int_{\rho}^{R}\frac{\left(\ln x\right)^{2}+2\pi i\left(\ln x\right)-\pi^{2}}{1+x^{2}}dx\\ & =\int_{\rho}^{R}f\left(x\right)dx+2\pi i\int_{\rho}^{R}\frac{\ln x}{1+x^{2}}dx-\pi^{2}\int_{\rho}^{R}\frac{1}{1+x^{2}}dx.\nonumber \end{align*}

Also, we have

    \[ \left|\int_{C_{R}}f\left(z\right)dz\right|\le\frac{\pi R\left(\left(\ln R\right)^{2}+\pi^{2}\right)}{R^{2}-1}. \]

So the essential part of our calculation is calculating

    \[ \lim_{R\rightarrow\infty}\frac{R\left(\ln R\right)^{2}}{R^{2}-1}=\lim_{R\rightarrow\infty}\frac{\frac{\left(\ln R\right)^{2}}{R}}{1-\frac{1}{R^{2}}}=0 \]

since \lim_{R\rightarrow\infty}\frac{\left(\ln R\right)^{2}}{R}=\lim_{R\rightarrow\infty}\frac{2\left(\ln R\right)/R}{1}=0.
Similarly,

    \[ \lim_{\varepsilon\rightarrow0}\frac{\varepsilon\left(\ln\varepsilon\right)^{2}}{\varepsilon^{2}-1}=0. \]

So as \rho\rightarrow0 and R\rightarrow\infty, from (1) and (2) and by Proposition, we get

    \[ 2\int_{0}^{\infty}f\left(x\right)dx-\pi^{2}\int_{0}^{\infty}\frac{1}{1+x^{2}}dx=-\frac{\pi^{3}}{4}. \]

Since \int_{0}^{\infty}\frac{1}{1+x^{2}}dx=\frac{\pi}{2}, we get

    \[ \int_{0}^{\infty}f\left(x\right)dx=\frac{\pi^{3}}{8}. \]

이 문제가 오늘 시험문제에 나왔었는데, 나는 간단한 극한 계산 마무리를 그때 당시에 잘못 했었다. Estimate를 정교하지 못하게 했나라는 생각이 들었었는데, 쓸데없는 생각이 나를 막았던 것 같다.

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