5
$\begingroup$

I am trying to evaluate

$$\int_0^\infty \frac{\log(1+x^2)}{x^2} dx $$

by using contour integration. It is possible to compute this integral using real techniques; integration by parts yields the result:

$$\int_0^\infty \frac{\log(1+x^2)}{x^2} dx = \pi$$

I have been attempting to use the integrand $f(z)= \frac{\log(1+z^2)}{z^2}$ (with any suitable branch cut in the lower half plane), and have been trying to integrate around an indented semi-circle in the upper half plane, but I feel this is perhaps the wrong approach. Firstly, when summing the integrals along the real axis, the imaginary part seems to not converge. Furthermore, the indentation integral seems difficult to evaluate. Any help would be appreciated.

$\endgroup$
  • 1
    $\begingroup$ Almost sure it is a duplicate. $\endgroup$ – Jack D'Aurizio Jan 8 '15 at 14:42
  • 1
    $\begingroup$ I spent a good half hour searching, with no success. I'm happy to see an original though, if you can find a link. $\endgroup$ – CunningTF Jan 8 '15 at 14:53
  • 1
    $\begingroup$ $$\frac{1}{2}\int^\infty_{-\infty}\frac{\ln(1+x^2)}{x^2}dx=\pi i\operatorname*{Res}_{z=0}\frac{\ln(1-iz)}{z^2}=\pi$$ $\endgroup$ – M.N.C.E. Jan 8 '15 at 14:54
  • $\begingroup$ What contour are you using? I understand how to expand $\log(1+z^2)$, but not how to proceed. $\endgroup$ – CunningTF Jan 8 '15 at 15:09
  • 1
    $\begingroup$ I am just using the classic semicircular contour. $\frac{\ln(1-iz)}{z^2}$ is meromorphic in the upper-half plane. $\endgroup$ – M.N.C.E. Jan 8 '15 at 15:13
4
$\begingroup$

See this answer for a way to deal with such logs using contour integration. Note that, about each branch point, we replace the offending log with a factor of $-i 2 \pi$ to account for the phase jump introduced by traversing around the branch point in a clockwise direction.

In this case, you can use a semicircular contour of radius $R$ in the upper half plane, with a detour downward from the semicircle at $\theta=\pi/2$ that avoids the branch point at $z=i$, as pictured below:

enter image description here

Thus, in the limit as $R \to \infty$, the contour integral is equal to

$$\int_{-\infty}^{\infty} dx \frac{\log{(1+x^2)}}{x^2} - i 2 \pi \int_{e^{i \pi/2}}^{\infty e^{i \pi/2}} \frac{dy}{y^2} $$

By Cauchy's Theorem, the contour integral is zero because there are no poles inside this contour. Thus, we have

$$\int_{-\infty}^{\infty} dx \frac{\log{(1+x^2)}}{x^2} = \frac{i 2 \pi}{i} = 2 \pi $$

$\endgroup$
  • 1
    $\begingroup$ Thank you Ron, very insightful answer! $\endgroup$ – CunningTF Jan 8 '15 at 15:29
4
$\begingroup$

Notice that by integrating by parts we have: $$ I = \int_{0}^{+\infty}\frac{\log(1+x^2)}{x^2}\,dx = -\left.\frac{1}{x}\log(1+x^2)\right|_{0}^{+\infty}+\int_{0}^{+\infty}\frac{2\,dx}{1+x^2}=\color{red}{\int_{-\infty}^{+\infty}\frac{dz}{1+z^2}} $$ and the last integral is very easy to evaluate with complex or real-analytic techniques.

$\endgroup$
  • $\begingroup$ Yeah I'm aware of this method. Is there a method which doesn't rely on integration by parts first? I found the integral in a complex analysis book, and the phrasing of the question makes it seem like there should be a purely complex method. $\endgroup$ – CunningTF Jan 8 '15 at 14:55
  • 1
    $\begingroup$ @CunningTF: M.N.C.E. just gave you one in the comments. However, I think that integration by parts make things a lot easier, anyway. $\endgroup$ – Jack D'Aurizio Jan 8 '15 at 15:06
1
$\begingroup$

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ Note that \begin{align}&\color{#66f}{\large% \int_{0}^{\infty}{\ln\pars{1 + x^{2}} \over x^{2}}\,\dd x} =\half\int_{0}^{\infty}{\ln\pars{1 + x} \over x^{3/2}}\,\dd x =\half\int_{1}^{\infty}\pars{x - 1}^{-3/2}\,\,\ln\pars{x}\,\dd x \qquad\pars{1} \end{align} We evaluate the integral, in the RHS, by performing the contour integral $$ \oint_{\gamma}\pars{z - 1}^{-3/2}\,\,\ln\pars{z}\,\dd z $$ along the path $\ds{\gamma}$ depicted below:

enter image description here

which shows two branch cuts. Namely: \begin{align} \begin{array}{rclrcll} \pars{z - 1}^{-3/2} & = &\verts{z - 1}^{-3/2} \exp\pars{-\,{3 \over 2}\,\,{\rm Arg}\pars{z - 1}\ic}\,, & z & \not= & 1 \,, & \phantom{-}0 < \,{\rm Arg}\pars{z - 1} < 2\pi \\[5mm] \ln\pars{z}&=&\ln\pars{\verts{z}} + \ic\,{\rm Arg}\pars{z}\,,& z &\not = & 0\,, & -\pi < \,{\rm Arg}\pars{z} < \pi \end{array} \end{align} Obviously, the integral along $\ds{\gamma}$ vanishes out because there are not any poles inside the contour. For simplicity, we omit the contribution of the arcs $\ds{C_{R}}$, of radius $\ds{R}$, and the 'small' semicircles, of radius $\ds{\epsilon}$, around $\ds{z=0}$ and $\ds{z=1}$: They don't yield any contribution in the limits $\ds{R\ \to\ \infty\,,\ \epsilon\ \to\ 0^{+}}$. Then, \begin{align} 0&=\oint_{\gamma}\pars{z - 1}^{-3/2}\,\,\ln\pars{z}\,\dd z =\overbrace{\int_{1}^{\infty}\pars{x - 1}^{-3/2}\ln\pars{x}\,\dd x} ^{\ds{\mbox{along}\ \dsc{C_{++}}}} \\[5mm]&+\overbrace{% \int_{-\infty}^{0}\pars{1 - x}^{-3/2}\expo{-3\pi\ic/2} \bracks{\ln\pars{x} + \ic\pi}\,\dd x} ^{\ds{\mbox{along}\ \dsc{C_{-+}}}}\ +\ \overbrace{% \int^{-\infty}_{0}\pars{1 - x}^{-3/2}\expo{-3\pi\ic/2} \bracks{\ln\pars{x} - \ic\pi}\,\dd x} ^{\ds{\mbox{along}\ \dsc{C_{--}}}} \\[5mm]&+\overbrace{\int^{1}_{\infty}\pars{x - 1}^{-3/2} \expo{-3\pi\ic}\ln\pars{x}\,\dd x} ^{\ds{\mbox{along}\ \dsc{C_{+-}}}}\ =\ 2\int_{1}^{\infty}\pars{x - 1}^{-3/2}\ln\pars{x}\,\dd x \\[5mm]&\phantom{=}+2\pi\ic\expo{-3\pi\ic/2}\int_{-\infty}^{0} \pars{1 - x}^{-3/2}\,\dd x =2\int_{1}^{\infty}\pars{x - 1}^{-3/2}\ \ln\pars{x}\,\dd x -\left.{4\pi \over \root{1 - x}}\,\right\vert_{\, x\ \to\ -\infty}^{\, x\ =\ 0} \\[5mm]&=2\int_{1}^{\infty}\pars{x - 1}^{-3/2}\ \ln\pars{x}\,\dd x - 4\pi \quad\imp\quad \boxed{\ds{\quad\int_{1}^{\infty}\pars{x - 1}^{-3/2}\ \ln\pars{x}\,\dd x =2\pi\quad}} \end{align}

Replacing this result in expression $\pars{1}$, we'll find:

\begin{align}&\color{#66f}{\large% \int_{0}^{\infty}{\ln\pars{1 + x^{2}} \over x^{2}}\,\dd x} =\color{#66f}{\Large\pi} \end{align}

$\endgroup$
  • $\begingroup$ Another really good answer. Thank you. $\endgroup$ – CunningTF Jan 9 '15 at 11:31
  • $\begingroup$ @CunningTF I'm glad you found it useful. Thanks. $\endgroup$ – Felix Marin Jan 9 '15 at 21:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.