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How do I solve $\int_{0}^{\infty}\frac{\cos\left( ax\right) }{x^{2}+b^{2}}dx$? I found the solution on Gradshteyn Table of Integral (3.723) which is $\frac{\pi e^{ab}}{2b}$ but I would like to know the step-by-step solution. Couldn't find the first reference. Thanks

Pierre
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  • Let $f(z)=\mathrm{e}^{iaz}$ and use a semicircular contour in the upper half of the complex plane. – poweierstrass Sep 06 '16 at 23:23
  • Almost the same as http://math.stackexchange.com/questions/140580/computing-int-infty-infty-frac-cos-xx2-a2dx-using-residue, you will need a couple of minor adjustments. – David Sep 06 '16 at 23:33

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Assuming $a,b>0$ we have $$\int_{0}^{+\infty}\frac{\cos(ax)}{x^2+b^2}\,dx = \frac{1}{2}\,\text{Re}\int_{-\infty}^{+\infty}\frac{e^{iax}}{x^2+b^2}\,dx \tag{1}$$ and since $$ \text{Res}\left(\frac{e^{iax}}{x^2+b^2},x=ib\right)=-\frac{i}{2b} e^{-ab}\tag{2} $$ by the residue theorem and the ML lemma it follows that $$\int_{0}^{+\infty}\frac{\cos(ax)}{x^2+b^2}\,dx = \frac{1}{2}\,\text{Re}\left(\frac{2\pi}{2b}e^{-ab}\right) = \color{red}{\frac{\pi}{2b\, e^{ab}}}.\tag{3}$$

Jack D'Aurizio
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