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I'm trying to solve this using Laplace transforms by using the relation

$\int_0^\infty \frac{f(x)}{x} dx = \int_0^\infty F(p) dp$

where $F(p) = L[f(x)]$

But recasting it transforms it into a very messy integral.

Any help would be highly appreciated.

Jean Marie
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user34304
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    Have a look at: http://math.stackexchange.com/questions/272622/integral-evaluation-int-infty-infty-frac-cos-ax-pi-1x2dx?noredirect=1&lq=1, http://math.stackexchange.com/questions/9402/calculating-the-integral-int-0-infty-frac-cos-x1x2-mathrmdx-wit – StubbornAtom Nov 14 '16 at 11:16

1 Answers1

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Hint. Let's consider the Laplace transform of $\displaystyle I(a):=\int_{0}^{\infty}\frac{\cos(ax)}{x^2+1}\:dx$. We have $$ \begin{aligned} \mathcal{L}\left(I(a)\right)(s)&=\mathcal{L}\left(\int_{0}^{\infty}\frac{\cos(ax)}{x^2+1}\:dx\right)(s) \\& = \int_{0}^{\infty}\int_{0}^{\infty}\frac{\cos(ax)}{x^2+1}e^{-as}\:da\:dx \\&= \int_{0}^{\infty}\frac{s}{(x^2+1)(s^2+x^2)}\;{dx} \\&= \frac{\pi}{2(s+1)} \end{aligned}\tag1 $$giving $$ I(a)=\int_{0}^{\infty}\frac{\cos(ax)}{x^2+1}\:dx=\mathcal{L}^{-1}\left(\frac{\pi}{2(s+1)}\right) =\frac{\pi}{2}e^{-a},\qquad a>0, \tag2$$

Jan Eerland
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