$\int_0^{\infty}\frac{\ln x}{x^2+a^2}\mathrm{d}x$ Evaluate Integral

Question

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

Answer 1

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

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

Answer 2

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 $$ I = \ln a\int_0^\infty \frac {du}{u^2+a^2} = \frac {\ln a}a \ \left.\mbox{atan}\ \frac ua \right|_0^\infty = \frac {\pi \ln a}{2a} $$

Answer 3

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.

Answer 4

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

\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}

Answer 5

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