TITLE: Calculate the Fourier transform of $b(x) =\frac{1}{x^2 +a^2}$
QUESTION [20 upvotes]: I need help to calculate the  Fourier transform of this funcion
$$b(x) =\frac{1}{x^2 +a^2}\,,\qquad a > 0$$
Thanks.

REPLY [28 votes]: One way to attack this is via the residue theorem. Consider
$$\oint_C dz \frac{e^{i k z}}{z^2+a^2}$$
where $C$ is a semicircular contour in the upper half plane of radius $R$.  Note that, to use the residue theorem, we expect the integral over the circular arc to vanish as $R \to \infty$; however, this only happens when $k > 0$.  (I leave it to the reader to show this.)  The residue at the pole $z=i a$ is $e^{-k a}/(i 2 a)$, so, by the residue theorem,
$$\int_{-\infty}^{\infty} dx \frac{e^{i k x}}{x^2+a^2} = i 2 \pi \frac{e^{-a k}}{i 2 a} = \frac{\pi}{a} e^{-a k}$$
when $k > 0$.  When $k < 0$, however, we must use the semicircular contour in the lower half plane, rather than the upper half plane.  Thus, we now consider the pole at $z=-i a$, and the integral takes the value $(\pi/a) e^{a k}$ when $k < 0$.  Putting this together, we have
$$\int_{-\infty}^{\infty} dx \frac{e^{i k x}}{x^2+a^2} = \frac{\pi}{a} e^{- a |k|}$$