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I am doing a Fourier transform of the function $f(t)=\int_{0}^{\infty}\frac{e^{-x}}{1+(t-x)^2} dx$ and I can see that I can use convolution. The convolution are of $e^{-x}$ and $\frac{1}{1+x^2}$. Then when I look in the Fourier transform table I get $F[\frac{1}{1+x^2}]=\sqrt{\frac{\pi}{2}}e^{-|w|}$ but I can't find $F[e^{-u}]$. Where have I gone wrong?

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  • $\begingroup$ It is not $e^{-u} $, but $e^{-u} 1_{(0,\infty)} $. I think one compute the Fourier transform of this function by hand using the definition. $\endgroup$ – PhoemueX Oct 29 '16 at 17:16
  • $\begingroup$ I am now confusing myself. Is the second convolution right, $\frac{1}{1+x^2}$ giving the transformation $\sqrt{\frac{\pi}{2}}e^{-|w|}$ $\endgroup$ – S.n Nov 1 '16 at 12:18
  • $\begingroup$ With the answer $F(1e^{-u})\cdot F(1/(1+u^2)$ gives the answer $\frac{1}{2}\frac{e^{-|u|}}{(1-iw)}$ $\endgroup$ – S.n Nov 1 '16 at 12:33
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You made a small mistake. This is the convolution between $\frac{1}{1+x^2}$ and $e^{-x}\chi_{[0,+\infty)}(x)$ and you can find the fourier transform of the second one with the caracteristic function.

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Using the Heaviside theta, you can find

$$\mathcal{F}(\theta(u)e^{-u})(k) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty} \theta(u) e^{-u}e^{-iku}\ \text{d}u = \int_{0}^{+\infty} e^{-u}e^{-iku}\ \text{d}u $$

The integral is trivial; the integrand becomes

$$e^{-u(1 + ik)}$$

Which integrated gives

$$\frac{1}{1+ik}$$

(Don't forget the $\frac{1}{\sqrt{2\pi}}$ term.)

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  • $\begingroup$ So, putting the two together will get $\sqrt{\frac{\pi}{2}}e^{-|w|}\cdot \frac{1}{\sqrt{2\pi}}\frac{1}{1+iw}$ $\endgroup$ – S.n Oct 30 '16 at 19:26

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