Fourier transform of $f(t)=\int_{0}^{\infty}\frac{e^{-x}}{1+(t-x)^2} dx$

Question

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?

Answer 1

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.

Answer 2

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.)