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Anyone remember the method for this? I think this should been done on the site $$\int_0^{\infty}\frac{\ln x}{x^2+a^2}\mathrm{d}x$$

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    $\begingroup$ Make the substitution $u=a^2/x$ and look for the integral to reappear on the right side $\endgroup$ – kiwi Jan 30 '13 at 0:45
  • $\begingroup$ Nicely done, @kiwi! Write it as an answer? $\endgroup$ – Julien Jan 30 '13 at 1:09
  • $\begingroup$ @kiwi : Thx kiwi! Mind if you write it done? :) $\endgroup$ – Ryan Jan 30 '13 at 1:35
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    $\begingroup$ A related problem. $\endgroup$ – Mhenni Benghorbal Jan 30 '13 at 7:32
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I think @kiwi ment $$ \fbox {$I$} = \int_0^\infty \frac {\ln x}{x^2+a^2} dx = \left | u = \frac {a^2}x \Longrightarrow\left\{\begin{array}{c} \ln x = 2 \ln a - \ln u \\ dx = -\frac {a^2du}{u^2} \end{array}\right\} \right | = -\int_\infty^0 \frac{2\ln a - \ln u}{\frac {a^4}{u^2}+a^2}\frac {a^2 du}{u^2} = \\ 2\int_0^\infty \frac{\ln a}{u^2+a^2}du-\int_0^\infty \frac{\ln u}{u^2+a^2}du = \fbox{$2\int_0^\infty \frac{\ln a}{u^2+a^2}du - I$} $$ From last part it's clear that $$ I = \int_0^\infty \frac{\ln a}{u^2+a^2}du $$ This integral can be easily found $$ I = \ln a\int_0^\infty \frac {du}{u^2+a^2} = \frac {\ln a}a \ \left.\mbox{atan}\ \frac ua \right|_0^\infty = \frac {\pi \ln a}{2a} $$

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Related problems: (I), (II), (III), $(4)$. Let us consider the integeral

$$ \int_{0}^{\infty}\frac{x^{s-1}}{x^2+a^2}\text{d}x, $$

which is nothing but the Mellin transform of the function $ \frac{1}{x^2+a^2}$ and it is given by

$$ F(s)=\int_{0}^{\infty}\frac{x^{s-1}}{x^2+a^2}\text{d}x = \frac{1}{2}\frac{\pi a^{s-2}}{\sin(\pi s/2)} $$

$$ \implies F'(s)=\int_{0}^{\infty}\frac{x^{s-1}\ln(x)}{x^2+a^2}\text{d}x = \frac{d}{ds}\frac{1}{2}\frac{\pi a^{s-2}}{\sin(\pi s/2)}. $$

Taking the limit as $s \to 1$ the desired result follows

$$ \int_{0}^{\infty}\frac{\ln(x)}{x^2+a^2}\text{d}x = \frac{\pi \ln(a)}{2a}. $$

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  • $\begingroup$ hehe, I also thought of posting this solution (+1). $\endgroup$ – user 1357113 Jan 30 '13 at 7:59
  • $\begingroup$ It is a general approach. $\endgroup$ – Mhenni Benghorbal Jan 30 '13 at 8:04
  • $\begingroup$ @ Mhenni Benghorbal: Mellin transform is very powerful when dealing with many ugly integrals. Unfortunately, less used. $\endgroup$ – user 1357113 Jan 30 '13 at 8:06
  • $\begingroup$ $\large a < 0\ ?$ with $\large\ln\left(a\right)$. $\endgroup$ – Felix Marin May 12 '14 at 19:56
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Or simply let $x=ay$

$$\int_0^{\infty}\frac{\ln x}{x^2+a^2}\mathrm{dx}=\frac{\ln a}{a}\int_0^{\infty}\frac{1}{y^2+1}\mathrm{dy}+\frac{1}{a}\int_0^{\infty}\frac{\ln y}{y^2+1}\mathrm{dy}=\frac{\pi\ln a}{2a}$$ because letting $y=1/z$ in $\int_0^{1}\frac{\ln y}{y^2+1}\mathrm{dy}=-C$ (Catalan's constant) , we get $\int_1^{\infty}\frac{\ln z}{z^2+1}\mathrm{dz}=C$. Now, add up the $2$ integrals and get that $\int_0^{\infty}\frac{\ln y}{y^2+1}\mathrm{dy}=0$.

Chris.

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    $\begingroup$ Nice answer (+1). $\endgroup$ – Mhenni Benghorbal Jan 30 '13 at 8:03
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    $\begingroup$ I just saw this question, and this was almost the answer I thought of. Instead of evaluating $\int_0^\infty\frac{\log(y)\,\mathrm{d}y}{1+y^2}$ I was going to note that $y\mapsto\frac1y$ sends the integral to its negative, so it is $0$. (+1) $\endgroup$ – robjohn Oct 25 '14 at 16:21
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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$

$\ds{\large\tt\mbox{It's almost done without any integral evaluation !!!.}}$

\begin{align} &\color{#00f}{\large\int_{0}^{\infty}{\ln\pars{x} \over x^{2} + a^{2}}\,\dd x} ={1\over \verts{a}}\int_{0}^{\infty}{\ln\pars{\verts{a}x} \over x^{2} + 1}\,\dd x \\[3mm]&={\ln\pars{\verts{a}}\over \verts{a}}\ \overbrace{\int_{0}^{\infty}{\dd x \over x^{2} + 1}}^{\ds{=\ {\pi \over 2}}}\ +\ {1\over \verts{a}}\int_{0}^{\infty}{\ln\pars{x} \over x^{2} + 1}\,\dd x \\[3mm]&={\pi\ln\pars{\verts{a}} \over 2\verts{a}}\quad +\quad{1 \over \verts{a}}\ \underbrace{\quad\bracks{\int_{0}^{1}{\ln\pars{x} \over x^{2} + 1}\,\dd x +\ \overbrace{% \int_{1}^{0}{\ln\pars{1/x} \over 1/x^{2} + 1}\,\pars{-\,{\dd x \over x^{2}}}} ^{\ds{=-\int_{0}^{1}{\ln\pars{x} \over x^{2} + 1}\,\dd x}}}\quad} _{\ds{\color{#c00000}{\LARGE =\ 0}}} \\[3mm]&=\color{#00f}{\large{\pi\ln\pars{\verts{a}} \over 2\verts{a}}} \end{align}

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Let the considered integral be $I$. Use the substitution $x=a\tan\theta \,d\theta$ to obtain: $$I=\frac{1}{a}\int_0^{\pi/2} \ln(a\tan\theta)\,d\theta=\frac{1}{a}\int_0^{\pi/2} \ln a\,d\theta+\frac{1}{a}\int_0^{\pi/2}\ln(\tan\theta)\,d\theta$$ It is easy to see that $$\int_0^{\pi/2}\ln(\tan\theta)\,d\theta=0$$ (The above can be shown by using the substitution $\theta=\pi/2-t$).

Hence, $$I=\frac{\pi}{2a}\ln a$$

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  • $\begingroup$ It is worth adding how one 'sees' that the $\ln \tan$ integral is zero: by substituting $\theta \mapsto \pi/2-\theta$ and getting minus the original integral. $\endgroup$ – user111187 May 12 '14 at 20:08
  • $\begingroup$ Done! Thanks. :) $\endgroup$ – Pranav Arora May 12 '14 at 20:12

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