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Compute $\displaystyle\int_{0}^{\infty} \frac{\cos(x)}{x^2 + a^2} \mathrm{dx}$, for $a\in \mathbb{R}$ using the Laplace Transform.


I'm not sure on how to start with this problem. I tried to first compute the Laplace Transform but apparently this is not possible (WolframAlpha gives me a solution with respect to the sine integral $\mathrm{Si(x)}$ and the cosine integral $\mathrm{Ci(x)}$).

Thank you for your help.

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  • $\begingroup$ Well, what is the laplace transform that you know of? $\endgroup$ – Ali Caglayan Mar 22 '15 at 19:54
  • $\begingroup$ It seems possible to compute the inverse Laplace transform (with respect to $a$), then calculate the integral and than apply the Laplace transform. If you really are forced to use the Laplace transform, that is. $\endgroup$ – mickep Mar 22 '15 at 20:20
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$$ \int_0^\infty \frac{\cos xt}{1+t^2}\,\mathrm{d}t = \frac{\pi}{2}e^{-|x|}$$

To transform your integral into this form use $x \mapsto t \cdot a$. Now we can begin by taking the laplace transform of the integral $$ \mathcal{L}(I) = \int_{0}^{\infty} \left( \int_{0}^{\infty} \frac{\cos(xt)}{1+t^2} \, \mathrm{d}t\right)e^{-sx}\,\mathrm{d}x $$ The next step is to interchange the limits (fubinis theorem). Since $|I|$, converges in the Riemann sense, so does $I$. $$ \begin{align*} \mathcal{L}(I) = \int_{0}^{\infty} \left(\int_{0}^{\infty} \frac{\cos(xt)}{1+t^2} e^{-sx}\,\mathrm{d}x \right)\mathrm{d}t & = \int_{0}^{\infty} \frac{1}{1+t^2}\left(\int_{0}^{\infty} \cos(xt) e^{-sx}\,\mathrm{d}x \right)\mathrm{d}t \\ & = \int_{0}^{\infty} \frac{1}{1+t^2} \frac{s}{s^2+t^2} \mathrm{d}t \end{align*} $$ Where we used the laplace tranform of $\cos(\omega x)$ in the last transition. The last expression can easilly be solved by partial fractions $$ \frac{1}{1+t^2} \frac{s}{s^2+t^2} = \frac{s}{1-s^2} \left( \frac{1}{s^2+t^2} - \frac{1}{1+t^2} \right) $$ To obtain the final answer you have to take the inverse-laplace transform of your integral.


\begin{align*} |I| \leq \int_{0}^{\infty} \int_{0}^{\infty} \frac{e^{-sx}}{1+t^2} \mathrm{d}t\,\mathrm{d}x = \int_{0}^{\infty} \left[ \frac{\pi}{2}e^{-sx} \right]_0^\infty \mathrm{d}t\,\mathrm{d}x = \biggl[ -\frac{\pi}{2s} e^{-sx} \biggr]_0^\infty = \frac{\pi}{2s} \end{align*}

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There are two ways of trying to do this:

  1. Use the inverse Laplace transform. Using the relation $2\cos{x} = e^{ix}+e^{-ix}$, we can rewrite the integral as $$ \frac{1}{2}\int_{-\infty}^{\infty} \frac{e^{ix}}{x^2+a^2} \, dx. $$ Changing variables to $s=ix$, we find $$ -\frac{1}{2i}\int_{-i\infty}^{i\infty} e^{s} \frac{1}{s^2-a^2} \, ds = \frac{1}{2i (2a)}\int_{-i\infty}^{i\infty} \frac{e^{s}}{s+a} ds - \frac{1}{2i (2a)}\int_{-i\infty}^{i\infty} \frac{e^{s}}{s-a} ds, $$ where the contours are the imaginary axis. Now, this is almost the Bromwich integral for the functions $(s\pm a)^{-1}$, but evaluated with the imaginary axis as a contour. Therefore the second one is zero, shifting the contour away to the left and bounding it by $e^{s}$. The first one, on the other hand, really is the inverse Laplace transform of $$ \frac{\pi}{2a(s-a)}, $$ and the function that has this Laplace transform is $$ \frac{\pi}{2a}e^{-ax}, $$ and so this with $x=1$ is the value of the original integral, using the inversion theorem.

  2. Alternatively, we can compute the Laplace transform of $$ \frac{1}{x^2+a^2} $$ and evaluate it at $\pm i$, since $e^{ix}+e^{-ix} = 2\cos{x}$. This requires that we compute $$ L(s) = \int_0^{\infty} \frac{e^{-sx}}{x^2+a^2} \, dx. $$ Probably the easiest way to do this is to differentiate under the integral sign a couple of times: we find $$ L''(s) = \int_0^{\infty} \frac{x^2}{x^2+a^2} e^{-sx} \, dx = -a^2 L(s) + \int_0^{\infty} e^{-sx} \, dx = -a^2 L(s) + \frac{1}{s}. $$ Now we solve the differential equation $$ L''(s) + a^2L(s) = \frac{1}{s}, $$ which, as you noted, requires the cosine and sine integrals. Therefore the question probably wants the first way.

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