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Can somebody remind me how to prove the equality $$\frac{2}{\pi}\int_0^{\infty}\frac{\cos tx}{1+x^2}dx=e^{-|t|}?$$ I found this in a book, where it is considered as obvious. As always, it is obvious once you know the trick, so could somebody remind me of the specific trick?

Thanks in advance !

yannis
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    Hint: use parity, Jordan's lemma and residue theorem. – Start wearing purple Apr 14 '14 at 11:42
  • Note that you can write $\int_0^{\infty}$ as $\frac12\int_{-\infty}{\infty}$. This can then be computed by considering the contour integral around a semicircle of radius $R$ in the upper half plane whose base is the interval $[-R,R]$. Use the residue theorem and note that the only pole within the contour is at $z=i$. – MPW Apr 14 '14 at 11:43

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Highlights: define

$$C_R:=[-R,R]\cup\gamma_R\;,\;\;\gamma_R:=\{z\in\Bbb C\;;\;z=Re^{i\theta}\,\,,\,0\leq \theta\leq \pi\;,\;\;R\in\Bbb R^+\}\;,\;\;f(z):=\frac{e^{itz}}{z^2+1}$$

For $\;R>1\;$ , we have only one simple pole of $\;f\;$ inside the domain determined by the above curve, and

$$\text{Res}_{z=1}(f)=\lim_{z\to i}(z-i)f(z)=\lim_{z\to i}\frac{e^{itz}}{z+i}=\frac{e^{-t}}{2i}$$

and by the Residue Theorem

$$\int\limits_{C_R}f(z)dz=2\pi i\frac{e^{-t}}{2i}=\frac\pi{e^t}$$

Also, by Jordan's Lemma

$$\int\limits_{\gamma_R}f(z)dz\xrightarrow[R\to\infty]{}0$$

So passing to the limit:

$$\frac\pi{e^t}=\lim_{R\to\infty}\int\limits_{C_R}f(z)=\int\limits_{-\infty}^\infty\frac{e^{ixt}}{x^2+1}dx$$

Now take real parts, use the fact that the real function is even and etc.

DonAntonio
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