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$ \displaystyle \int_0^\infty \dfrac{\cos xdx}{1+x^2} $

How to calculate the above using differentiation under the integral sign??

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  • $\begingroup$ what do u want? contour integration or differentiation under the integral sigen? Or both? the tags you set are not really fitting to the question $\endgroup$ – tired Sep 20 '15 at 10:21
  • $\begingroup$ by the way what have you tried ? what are your thoughts...? $\endgroup$ – tired Sep 20 '15 at 10:24
  • $\begingroup$ I have tried the standard methods like ibp, putting it into summation of series..I know the answer but differentiating under the integral sign is what I need! $\endgroup$ – Kunal Gupta Sep 20 '15 at 10:25
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    $\begingroup$ Doesn't "The Feynman way" imply differentiation under the integral sign? @Kunal: Maybe you should change the title to "...: Differentiation under the integral sign. I think not everyone knows that this is often called "The Feynman way" :). $\endgroup$ – MrYouMath Sep 20 '15 at 10:26
  • $\begingroup$ @tired: doesn't differentiating your integral twice wrt $a$ produce a non convergent integral? $\endgroup$ – Ron Gordon Sep 20 '15 at 10:35
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Introduce the function $$L(s) = \int_{0}^{\infty}\frac{\cos(x)}{1+x^{2}}\,e^{-sx}\, dx$$ for $s\geq 0$

You can indeed show that it is well-defined and legitimate to differentiate under the integral sign (at least for $s>0$). By doing so you will get $$L''(s)= \int_{0}^{\infty}\frac{x^{2}\cos(x)}{1+x^{2}}\,e^{-sx}dx= \int_{0}^{\infty}\cos(x)\,e^{-sx}dx-L(s)$$

Hence you might want to solve the IVP $$\left\{L''(s)+L(s) = \frac{s}{1+s^{2}} \, \,, \,\lim_{s \rightarrow \infty}L(s)=0 \, \,, \, \, \lim_{s\rightarrow \infty}L'(s)=0\right\}$$

Honestly I think this route is far more complicated than using residues, since we are forced to work with a global problem for the purpose of obtaining information at a local point $s=0$ i.e seeking a solution to an ODE for $s\geq 0$ only to find $L(0)$

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    $\begingroup$ Awesome :). Thank you for sharing! $\endgroup$ – MrYouMath Sep 20 '15 at 10:38
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    $\begingroup$ Just curious, how did you figure that out? Are there some standard modifications to the integral? Did you just use it to convert the integral to a Laplace transform? $\endgroup$ – MrYouMath Sep 20 '15 at 10:45
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    $\begingroup$ Cool!! figured it out ;) @TheOscillator $\endgroup$ – Kunal Gupta Sep 20 '15 at 10:47
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Another way is to note that, if we put $$I\left(a\right)=\int_{0}^{\infty}\frac{\sin\left(ax\right)}{x\left(1+x^{2}\right)}dx,\,a>0$$ we have $$I'\left(a\right)=\int_{0}^{\infty}\frac{\cos\left(ax\right)}{1+x^{2}}dx $$ and now follow this answer to find $$I\left(a\right)=\frac{\pi}{2}\left(1-e^{-a}\right)$$ and so $$\lim_{a\rightarrow1}I'\left(a\right)=\frac{\pi }{2e}.$$

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