So I want to solve the following integral:
$$ I(w) = \int_{-\infty}^{\infty} \frac{e^{-j \omega t}}{t^2 + a} dt $$
where $a \in \mathbb{R}_{++}$. This is just the Fourier transform of $\frac{1}{t^2 + a}$, which, if we consult practically any transform table, or just utilizie a simple half circle contour, we know is:
$$ I(w) = \frac{e^{-\sqrt{a} |w|}}{\sqrt{a}} $$
However, I want to do it a different way using the integral representation for $\frac{1}{t^2 + a}$. Namely:
$$ I(w) = \int_{-\infty}^{\infty} \int_{0}^{\infty} e^{-u(t^2 + a)} e^{-j \omega t} du dt $$
Switching the integrals gives us:
$$ I(w) = \int_{0}^{\infty} e^{-ua - \frac{w}{4u}} \int_{-\infty}^{\infty} e^{-u(t + j \frac{w}{2u})} dt du $$
The inner integral is just a Gaussian integral, so we end up with:
$$ I(w) = \sqrt{\pi} \int_{0}^{\infty} u^{-\frac{1}{2}} e^{-ua - \frac{w^2}{4u}} du $$
This is where I get stuck. Not only am I having trouble solving the integral, but Mathematica gives me a weird answer when I evaluate this inner integral to give finally give me:
$$ I_{Mathematica}(w) = \pi \frac{e^{- \sqrt{a} |w|}}{\sqrt{a}} $$
Questions:
- How can I solve $I(w) = \sqrt{\pi} \int_{0}^{\infty} u^{-\frac{1}{2}} e^{-ua - \frac{w^2}{4u}} du $?
- Where does the extra $\pi$ come from in Mathematica's answer?
Thanks!
