<s>[INST] Fourier transform of $f(x)=\frac{1}{x^2+6x+13}$ How to find the Fourier transform of the following function:
$$f(x)=\frac{1}{x^2+6x+13}$$
 [/INST] Breaking down to partial fractions yields
$$
\begin{align}
\frac1{x^2+6x+13}
&=\frac1{(x+3)^2+4}\\
&=\frac1{(x+3-2i)(x+3+2i)}\\
&=\frac1{4i}\left(\frac1{x+3-2i}-\frac1{x+3+2i}\right)
\end{align}
$$
Therefore, for $\xi\ge0$,
$$
\begin{align}
\int_{-\infty}^\infty\frac1{x^2+6x+13}e^{-i2\pi x\xi}\,\mathrm{d}x
&=\frac1{4i}\int_{-\infty}^\infty\left(\frac1{x+3-2i}-\frac1{x+3+2i}\right)e^{-i2\pi x\xi}\,\mathrm{d}x\\
&=\frac1{4i}\int_{\gamma}\left(\frac1{z+3-2i}-\frac1{z+3+2i}\right)e^{-i2\pi z\xi}\,\mathrm{d}z\\
&=\frac{2\pi i}{4i}e^{-i2\pi(-3-2i)\xi}\\
&=\frac\pi2e^{(-4+6i)\pi\xi}\\
&=\frac\pi2e^{(-4|\xi|+6i\xi)\pi}\quad\text{for all }\xi
\end{align}
$$
where $\gamma$ is the curve passing left to right along the real axis then circles back clockwise around the lower half-plane.
Thus, the Fourer Transform is
$$
\frac\pi2e^{-4\pi|\xi|}(\cos(6\pi\xi)+i\sin(6\pi\xi))
$$
</s>