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I want to integrate $$\int_{-\infty}^{\infty} \frac{e^{itx}}{{1+x^2}} dx.$$ I don't see how substitution or integration by parts could help here. Does anybody know how to do this?

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  • $\begingroup$ I happen to remember that it is $\pi e^{-|t|}$. $\endgroup$ – Stephen Montgomery-Smith May 13 '14 at 19:24
  • $\begingroup$ See my answers here and here. $\endgroup$ – Tunk-Fey May 14 '14 at 16:29
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Consider the complex function $f(z,t)=\frac{e^{itz}}{1+z^2}$ and consider the complex upper semicircle, from $+\infty$ to $-\infty$, the only pole in that region is a simple pole, $z=i$, so we can summarise our integral $\int_{-\infty}^\infty\frac{e^{itx}}{1+x^2}=2\pi iRes(f(z,t),z=i))$

Here $Res(f(z,t),z=i))=\lim_{z\to i}(z-i)f(z,t)=\lim_{z\to i}\frac{e^{itz}}{(z+i)}=\frac{e^{-t}}{2i}$

Thus:

$\int_{-\infty}^\infty\frac{e^{itx}}{1+x^2}=2\pi i\frac{e^{-t}}{2i}=\pi e^{-t}$

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  • $\begingroup$ wow... why the downvote? $\endgroup$ – Ellya May 13 '14 at 19:50
  • $\begingroup$ if there is a problem please say so... $\endgroup$ – Ellya May 13 '14 at 20:22
  • $\begingroup$ it was not me...actually, I just upvoted yours... $\endgroup$ – user66906 May 13 '14 at 20:28
  • $\begingroup$ @Lipschitz :) it was a comment at the community, really if you downvote you should put some kind of reasoning, or suggestion for improvement $\endgroup$ – Ellya May 13 '14 at 20:33
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    $\begingroup$ @Lipschitz, so we consider the semi circle, defined by $|z|=a$ for $a\gt 1$ then we have $|z|^2=|z^2|=|z^2+1-1|\le|z^2+1|+1$ therefore, $|z^2+1|\ge|z|^2-1=|a|^2-1=|a^2-1|$, so $\frac{1}{|z^2+1|}\le\frac{1}{|a^2-1|}$ $\endgroup$ – Ellya May 13 '14 at 21:03
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Do you know how to do a contour integral? This specific problem is actually solved on Wikipedia's page on this subject.

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