9
$\begingroup$

How do I compute

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

What I am doing is take

$$f(z)=\frac{(\log z)^2}{1+z^2}$$

and calculating

$\text{Res}(f,z=i) = \dfrac{d}{dz} \dfrac{(\log z)^2}{1+z^2}$ which came out to be $\dfrac{\pi}{2}-\dfrac{i\pi^2}{8}+\dfrac{i\pi}{2}$

Im not too sure how to move on from here. the given answer is $\dfrac{\pi^3}{8}$

Any help will be appreciated. thank you in advanced.

$\endgroup$
  • 2
    $\begingroup$ Thank you all for the input. They were all very helpful and I was able to find the solution (: $\endgroup$ – alice Nov 20 '14 at 22:43
8
$\begingroup$

For this integral, you want to define you branch cut along the negative x axis and use a contour with a semi circle around the origin. enter image description here

Also, note we are taking the real Cauchy Principal value. Then $$ \int_0^{\infty}\frac{(\ln(z))^2}{1+z^2}dz = \int_{\Gamma}f(z)dz+\int_{\gamma}f(z)dz + \int_{-\infty}^0f(z)dz + \int_0^{\infty}f(z)dz $$ Let $R$ be the radius of big semi circle, $\Gamma$ and $\delta$ the radius of the small semi circle $\gamma$. When $R\to\infty$, $\int_{\Gamma}\to 0$ and $\int_{\gamma}\to 0$ when $\delta\to 0$ by the estimation lemma. Then $$ \int_{-\infty}^0f(z)dz + \int_0^{\infty}f(z)dz = 2\pi i \sum\text{Res}_{\text{UHP}} $$ where UHP is the upper half plane. There is only one pole in the upper half plane and that is $z = i$ $$ \text{Re PV}\biggl[\int_{-\infty}^0\frac{(\ln|z| + i\pi)^2}{z^2+1}dz + \int_0^{\infty}\frac{(\ln|z| + i\cdot 0)^2}{z^2+1}dz\biggr] = 2\pi i\sum\text{Res}_{\text{UHP}} $$ Recall that $\ln(z) = \ln|z| + i\arg(z)$. Can you take it from here?


Mouse over for solution.

\begin{align}\text{Re PV}\biggl[\int_{-\infty}^0\frac{\ln^2|z| + 2\pi i\ln|z| - \pi^2}{z^2 + 1}dz +\int_0^{\infty}\frac{\ln^2|z|}{z^2 + 1}dz\biggr]&= 2\pi i\lim_{z\to i}(z - i)\frac{(\ln|z| + i\pi/2)^2}{z^2+1}\\2\int_0^{\infty}\frac{\ln^2(x)}{x^2 + 1}dx - \pi^2\int_0^{\infty}\frac{dx}{x^2+1} &= -\frac{\pi^3}{4}\\2\int_0^{\infty}\frac{\ln^2(x)}{x^2 + 1}dx - \frac{\pi^3}{2} &= -\frac{\pi^3}{4}\\\int_0^{\infty}\frac{\ln^2(x)}{x^2 + 1}dx &= \frac{\pi^3}{4} - \frac{\pi^3}{8}\\&= \frac{\pi^3}{8}\end{align}

$\endgroup$
  • $\begingroup$ @ Dustin Just a technical question: how did you create the "mouse over" part of your contribution? $\endgroup$ – Dr. Wolfgang Hintze Apr 29 at 14:24
  • $\begingroup$ @Dr.WolfgangHintze click edit my post and you can see the code. $\endgroup$ – dustin Apr 30 at 6:31
6
$\begingroup$

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{}$ \begin{align} \overbrace{\color{#66f}{\large\int_{0}^{\infty}{\ln^{2}\pars{x} \over 1 + x^{2}}\,\dd x}}^{\ds{\color{#c00000}{x^{2}\ \mapsto\ x}}}&\ =\ \int_{0}^{\infty}{\ln^{2}\pars{x^{1/2}} \over 1 + x}\,\half\,x^{-1/2}\,\dd x ={1 \over 8}\int_{0}^{\infty}{x^{-1/2}\ln^{2}\pars{x} \over 1 + x}\,\dd x \\[5mm]&={1 \over 8}\,\lim_{\mu\ \to\ -1/2}\totald[2]{}{\mu} \color{#c00000}{\int_{0}^{\infty}{x^{\mu} \over 1 + x}\,\dd x}\tag{1} \end{align}

\begin{align}&\color{#c00000}{% \int_{0}^{\infty}{x^{\mu} \over 1 + x}\,\dd x} =2\pi\ic\expo{\pi\mu\ic} -\int_{\infty}^{0}{x^{\mu}\expo{2\pi\mu\ic} \over 1 + x}\,\dd x =2\pi\ic\expo{\pi\mu\ic} +\expo{2\pi\mu\ic}\color{#c00000}{\int_{0}^{\infty}{x^{\mu} \over 1 + x}\,\dd x} \\[5mm]&\imp\color{#c00000}{\int_{0}^{\infty}{x^{\mu} \over 1 + x}\,\dd x} ={2\pi\ic\expo{\pi\mu\ic} \over 1 - \expo{2\pi\mu\ic}} =\pi\,{2\ic \over \expo{-\pi\mu\ic} - \expo{\pi\mu\ic}}=-\pi\csc\pars{\pi\mu} \end{align}

Replacing in $\pars{1}$: \begin{align} &\color{#66f}{\large\int_{0}^{\infty}{\ln^{2}\pars{x} \over 1 + x^{2}}\,\dd x} ={1 \over 8}\,\lim_{\mu\ \to\ -1/2}\totald[2]{\bracks{-\pi\csc\pars{\pi\mu}}}{\mu} \\[5mm]&={1 \over 8}\,\lim_{\mu\ \to\ -1/2}\braces{-\pi\bracks{% \pi^{2}\cot^{2}\pars{\pi\mu}\csc\pars{\pi\mu} + \pi^{2}\csc^{3}\pars{\pi\mu}}} \\[5mm]&={1 \over 8}\braces{-\pi\bracks{\pi^{2}\csc^{3}\pars{-\,{\pi \over 2}}}} =\color{#66f}{\large{\pi^{3} \over 8}} \end{align}

$\endgroup$
4
$\begingroup$

Using $\beta(3)=\frac{\pi^3}{32}$ from this answer, we can apply $x\mapsto e^{-x}$ to get $$ \begin{align} \int_0^\infty\frac{\log(x)^2}{1+x^2}\,\mathrm{d}x &=\int_{-\infty}^\infty\frac{x^2}{1+e^{-2x}}\,e^{-x}\,\mathrm{d}x\\ &=2\int_0^\infty\frac{x^2}{1+e^{-2x}}\,e^{-x}\,\mathrm{d}x\\ &=2\sum_{k=0}^\infty\int_0^\infty x^2(-1)^ke^{-(2k+1)x}\,\mathrm{d}x\\ &=2\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)^3}\Gamma(3)\\[6pt] &=2\beta(3)\Gamma(3)\\[9pt] &=\frac{\pi^3}8 \end{align} $$

$\endgroup$
3
$\begingroup$

I would consider the following contour integral

$$\oint_C dz \frac{\log^3{z}}{1+z^2} $$

where $C$ is a keyhole contour of outer radius $R$ and inner radius $\epsilon$, about the positive real axis. Without explicitly showing that the integrals over the outer and inner circles vanish as $R\to\infty$ and $\epsilon \to 0$, the integral is equal to

$$\int_0^{\infty} dx \frac{\log^3{x}-(\log{x}+i 2 \pi)^3}{1+x^2} = -i 6 \pi \int_0^{\infty} dx \frac{\log^2{x}}{1+x^2} + 12 \pi^2 \underbrace{\int_0^{\infty} dx \frac{\log{x}}{1+x^2}}_{0} + \\i 8 \pi^3 \underbrace{\int_0^{\infty} dx \frac{1}{1+x^2}}_{\pi/2} $$

The contour integral is also equal to the sum of the residues of the integrand at the poles $z=\pm i$:

$$-i 6 \pi \int_0^{\infty} dx \frac{\log^2{x}}{1+x^2} + i 4 \pi^4 = i 2 \pi \left (\frac{-i \pi^3/8}{2 i} + \frac{-i 27\pi^3/8}{-2 i} \right ) = i \frac{13 \pi^4}{4}$$

Note that the keyhole contour demands that $\arg{z} \in [0,2\pi]$. Thus,

$$\int_0^{\infty} dx \frac{\log^2{x}}{1+x^2} = \frac{\pi^3}{8} $$

$\endgroup$
3
$\begingroup$

We have $$\int_1^{\infty} \dfrac{\ln^2(x)}{1+x^2}dx = \int_1^0 \dfrac{\ln^2(1/x)}{1+1/x^2} \dfrac{-dx}{x^2} = \int_0^1 \dfrac{\ln^2(x)}{1+x^2}dx$$ Hence, our integral is $$I = 2\int_0^1 \dfrac{\ln^2(x)}{1+x^2}dx$$ From complex analysis, since $\dfrac1{1+z^2}$ is analytic in the open unit disk, we have $$\dfrac1{1+z^2} = \sum_{k=0}^{\infty}(-1)^kz^{2k}$$ Hence, $$I = 2\sum_{k=0}^{\infty}(-1)^k \int_0^1 x^{2k}\ln^2(x)dx = 2\sum_{k=0}^{\infty}(-1)^k \dfrac2{(2k+1)^3} = 2 \cdot 2 \cdot \dfrac{\pi^3}{32} = \dfrac{\pi^3}8$$ where the last equality follows from the fact that $$\text{Li}_n(e^{2\pi ix}) + (-1)^n \text{Li}_n(e^{-2\pi ix}) = -\dfrac{(2 \pi i)^n}{n!} B_n(x)$$ taking $n=3$, $x=1/4$.

$\endgroup$

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.