TITLE: Fourier transform of $f(x)=\frac{1}{x^2+6x+13}$
QUESTION [5 upvotes]: How to find the Fourier transform of the following function:
$$f(x)=\frac{1}{x^2+6x+13}$$

REPLY [2 votes]: If you have Rudin's "Real and Complex Analysis", look at p. 183 in Chapter 9. As he points out, by directly calculating, the inverse Fourier transform of $e^{-\lambda |\xi|}$ for $\lambda > 0$ is seen to be
$$\sqrt{2 \over \pi} {\lambda \over \lambda^2 + x^2}$$
This means the Fourier transform of ${\displaystyle {1 \over 4 + x^2}}$ is given by
$$\sqrt{\pi} {e^{-2|\xi|} \over 2\sqrt{2}}$$
Since ${\displaystyle {1 \over x^2 + 6x + 13} = {1 \over (x + 3)^2 + 4}}$, the Fourier transform of ${\displaystyle {1 \over x^2 + 6x + 13}}$ is then
$$\sqrt{\pi} {e^{-2|\xi|+3\xi} \over 2\sqrt{2}}$$