I am trying to prove the inverse Fourier transform relation of a Lorentzian
$$F(\omega) = \dfrac{2b}{(\omega-a)^2+b^2} = \dfrac{1}{b+i(\omega-a)}+\dfrac{1}{b-i(\omega-a)}$$
using the relation
$$f(t) = \int_{-\infty}^\infty \dfrac{d\omega}{2\pi}e^{-i\omega t} F(\omega)$$
with the method of residues. (From Wikipedia, the solution should be $f(t)=e^{-b|t|-iat}$)
The first term in $F(\omega)$ has a pole at $\omega=a+ib$ so the semicircle contour goes anticlockwise on the upper half of the complex plane. The second term has a pole at $\omega=a-ib$ so the semicircle contour goes clockwise on the lower half of the complex plane. Then calculating the residues:
$2\pi i \sum$ (residues in upper half of plane) – $2\pi i \sum$ (residues in lower half of plane)
we get
$$2\pi i \left( \dfrac{e^{-i(a+ib)t}}{b-i(a+ib-a)} \right) – 2\pi i \left( \dfrac{e^{-i(a-ib)t}}{b+i(a-ib-a)} \right)$$
and the solution is
$$f(t) = 2b \dfrac{e^{-iat}}{2b} (e^{bt}-e^{-bt})$$
What is the justification for letting $e^{bt}-e^{-bt} \to e^{-b|t|}$? Is it related to how I draw the contours on the complex plane?
Best Answer
Let $f(z)$ be your integrand.
This integral has to be evaluated case-by-case:
For $t>0$:
$e^{-iwt}$ converges to zero at infinity on the lower half plane.
Thus, we take an infinitely large semi-circle contour on the lower half plane, centered at origin.
By residue theorem, $$\underbrace{\int_{\text{arc}}f(z)dz}_{\to0}-\int^\infty_{-\infty}f(z)dz=2\pi i\text{Res}_{z=a-bi}f(z)\overbrace {=}^{\text{as you calculated}}e^{-bt-iat}$$
Therefore, $$\color{blue}{\int^\infty_{-\infty}f(z)dz=e^{-bt-iat}}$$
For $t<0$:
Now take an infinitely large semicircle on the upper half plane as contour, centered at zero.
$$\underbrace{\int_{\text{arc}}f(z)dz}_{\to0}+\int^{\infty}_{-\infty}f(z)dz=2\pi i\text{Res}_{z=a+bi}f(z)=e^{bt-iat}$$
Therefore, $$\color{blue}{\int^\infty_{-\infty}f(z)dz=e^{bt-iat}}$$
Gluing two cases together:
$$\color{red}{\frac1\pi\int^\infty_{-\infty}\frac{be^{-ixt}}{(x-a)^2+b^2}dx=e^{-b|t|-iat}}$$