Find the Fourier transform of $\frac1{1+t^2}$ Find the Fourier transform of $$f(t)=\frac1{1+t^2}$$ using contour integration that
$$F\{f(t)\}=\int^\infty_{-\infty}\frac1{1+t^2}e^{2\pi ft}dt$$
How can I do this?
 A: First, I don't think you should have a function of $t$ given by $f(t)$ and then to use $f$ as your transformation variable. That is just going to cause problems.
Anyway, let
$$ g(t) = e^{-a\lvert t \rvert} $$ 
(you will see why we pick this function soon).
Then
$$\begin{align}
\hat g(f) & = \int_{-\infty}^{\infty} e^{-a \lvert t \rvert} e^{-2\pi ift} dt \\
& = \int_{-\infty}^{0} e^{at} e^{-2\pi ift} dt  +  \int_{0}^{\infty} e^{-at} e^{-2\pi ift} dt \\
& = \frac{e^{t(a-2\pi if)}}{a-2\pi if} \Bigg \rvert_{-\infty}^{0} + \frac{e^{-t(a+2\pi if)}}{-(a+2\pi if)} \Bigg \rvert_{0}^{\infty} \\
& = \frac{2a}{a^{2}+4\pi^{2}f^{2}} 
\end{align}$$
Using the inversion theorem
$$ \implies e^{-a\lvert t \rvert} = \int_{-\infty}^{\infty} \frac{2a}{a^{2}+4\pi^{2}f^{2}} e^{2\pi ift} df $$
Noting that $a$ is a constant so can be taken out the front of the integral, and then we will divide both sides by $2a$
$$ \implies \frac{e^{-a\lvert t \rvert}}{2a} = \int_{-\infty}^{\infty} \frac{1}{a^{2}+4\pi^{2}f^{2}} e^{2\pi ift} df $$
Set $a=2\pi$
$$ \implies \frac{e^{-2\pi \lvert t \rvert}}{4\pi} = \frac{1}{4\pi^{2}}\int_{-\infty}^{\infty} \frac{1}{1+f^{2}} e^{2\pi ift} df $$
Multiplying both sides by $4\pi^{2}$
$$ \implies \pi e^{-2\pi \lvert t \rvert} = \int_{-\infty}^{\infty} \frac{1}{1+f^{2}} e^{2\pi ift} df $$
Let $t \mapsto -t$
$$\begin{align}
\implies \pi e^{-2\pi \lvert -t \rvert} & = \pi e^{-2\pi \lvert t \rvert} \\
& = \int_{-\infty}^{\infty} \frac{1}{1+f^{2}} e^{-2\pi ift} df
\end{align} $$
Therefore,
$$ \mathscr{F}\Bigg(\frac{1}{1+t^{2}}\Bigg) = \pi e^{-2\pi \lvert f \rvert} $$
