Unfortunately, I forgot how to solve the following improper integral.
\begin{equation} \int_{-\infty}^\infty \frac{\cos{x}}{x^2+4}\;dx \end{equation}
Could you give some hints?
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1Contour integration of $e^{iz}/(z^2+4)$ over the usual semicircle. – Angina Seng Oct 12 '17 at 19:07
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Your integral is a special case of this one (set $a=2$): Computing $\int_{-\infty}^{\infty} \frac{\cos x}{x^{2} + a^{2}}dx$ using residue calculus – projectilemotion Oct 12 '17 at 19:10
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the result should be $$\frac{\pi}{2e^2}$$ – Dr. Sonnhard Graubner Oct 12 '17 at 19:15
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@Sonnhard thanks for providing this unique new information – tired Oct 12 '17 at 19:59
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HINT:
Let $I(a)=\int_{-\infty}^\infty \frac{\cos(ax)}{x^2+4}\,dx$ and show (See This Answer, and This One) that $I''(a)=4I(a)$ along with $I(0)=\pi/2$ and $I'(0)=-\pi$.
Find $I(a)$ as solution of the ODE and initial conditions then set $a=1$.
Mark Viola
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@xpaul Thank you for catching the typo.... $I''(a)=4I(a)\ne I(a)$ ... I've edited. – Mark Viola Oct 12 '17 at 20:00
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@captain23 Please let me know how I can improve my answer. I really want to give you the best answer I can. – Mark Viola Oct 19 '17 at 02:26
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@xpaul No, $I''(a)$ does not diverge! $I(a)$ is clearly twice differentiable inasmuch as $I(a)=\frac\pi2 e^{-2a}$. To find $I''(a)$, one cannot simply interchange the integral with the second derivative. Read the answers in the links. – Mark Viola Oct 19 '17 at 15:21
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@xpaul Read the answers in the links! Note $$I''(a)\ne \int_{-\infty}^{\infty}\frac{x^2\cos(ax)}{x^2+4},dx$$Why do you insist on nonsense? – Mark Viola Oct 19 '17 at 20:32
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@xpaul Have you read the answers in the links yet? If so, do they resolve your issue? – Mark Viola Oct 20 '17 at 13:18
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@xpaul So,.was this answer and the ones in the link not useful to you then? – Mark Viola Oct 20 '17 at 18:19
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@xpaul Pleased to hear. And feel free to up vote answers as you see fit. ;-) – Mark Viola Oct 21 '17 at 13:52