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How to compute the integral of $\log x/(x+17)^2$ , x from 0 to $\infty$, using contour integration? $$\int_{0}^\infty \frac{\log x}{(x+17)^2}dx$$

Thanks!

aghost
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user145722
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1 Answers1

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Contour Integration

Let $\gamma=[0,Re^{i\epsilon}]\cup Re^{i[\epsilon,2\pi-\epsilon]}\cup[Re^{-i\epsilon},0]$. The integral over the arc vanishes and the contour circles one pole at $x=-17$.

$\hspace{3.5cm}$enter image description here

$$ \begin{align} \int_\gamma\frac{\log(z)^2}{(z+17)^2}\,\mathrm{d}z&=\int_0^\infty\frac{\log(x)^2-(\log(x)+2\pi i)^2}{(x+17)^2}\,\mathrm{d}x\\ &=\int_0^\infty\frac{-4\pi i\log(x)+4\pi^2}{(x+17)^2}\,\mathrm{d}x\\ \end{align} $$ The residue of $\frac{\log(z)^2}{(z+17)^2}$ at $z=-17$ is $\frac{2\log(17)+2\pi i}{-17}$, so the integral above must be $$ \frac{4\pi^2-4\pi i\log(17)}{17} $$ Taking the imaginary parts we get $$ \int_0^\infty\frac{\log(x)}{(x+17)^2}\,\mathrm{d}x=\frac{\log(17)}{17} $$

Integration by Parts $$ \begin{align} \int_0^\infty\frac{\log(x)}{(x+17)^2}\,\mathrm{d}x &=\lim_{a\to0}\int_a^\infty\log(x)\,\mathrm{d}\frac{-1}{x+17}\\ &=\lim_{a\to0}\left(\int_a^\infty\frac1{x(x+17)}\,\mathrm{d}x -\left[\frac{\log(x)}{x+17}\right]_a^\infty\right)\\ &=\lim_{a\to0}\left(\frac1{17}\log(a+17)\color{#C00000}{-\frac1{17}\log(a)+\frac{\log(a)}{a+17}}\right)\\ &=\frac{\log(17)}{17}\color{#C00000}{-\lim_{a\to0}\frac{a\log(a)}{17(a+17)}}\\ &=\frac{\log(17)}{17} \end{align} $$

robjohn
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  • Thank you!! Wonderful, although this is a standard routine. I want to kill myself after I read your solution. Should I be that dumb $%(&)_(^%$*&%(&)_ – user145722 Apr 25 '14 at 17:41
  • I can’t understand why is the integral over the arc vanishes? – BeesaFangirl DOTO Feb 27 '20 at 08:48
  • The integrand has size $O!\left(\frac{\log(R)}{R^2}\right)$ and the path has length $2\pi R$. Thus, the integral has size $O!\left(\frac{\log(R)}{R}\right)$, which vanishes as $R\to\infty$. – robjohn Feb 27 '20 at 12:35