It is well known that the Fourier transform $\mathcal{F(f(t))} = \int\limits_{-\infty}^\infty f(t) e^{-j2\pi f t} dt$ of
$$e^{-at}, t \geq 0$$ ($a$ is a constant) is given by
$$\dfrac{1}{a+j2\pi f}$$
What is the technique one would use to perform the inverse transform?
I have tried: $\begin{align} \mathcal{F^{-1}(\dfrac{1}{a+j2\pi f})} & = \int\limits_{-\infty}^\infty \dfrac{1}{a+j2\pi f} e^{j2\pi f t} df\\ & = \int\limits_{-\infty}^\infty \dfrac{a+j2\pi f}{a^2+(2\pi f)^2} e^{j2\pi f t} df \end{align}$
However, even though this breaks into real and imaginary parts, it has significantly increased the difficulty of computing the inverse fourier transform.
If you're familiar with complex analysis, one way is to perform contour integration. Consider the complex function:
$$ F(z) = \frac{e^{2\pi itz}}{a + 2\pi iz} $$
which has a simple pole at $z = \frac{a}{2\pi}i$. We define a closed contour $C$ consisting of a line from $z=-R$ to $z=R$ on the real line and an upper semicircle enclosing the pole. It looks something like this.
Then we have $$ \int_C F(z)\ dz = \int_{-R}^{R} F(x)\ dx + \int_{C_R} F(z)\ dz $$
where $C_R$ denotes the circular arc. Using Jordan's lemma, we can prove
$$ \lim_{R\to \infty}\int_{C_R} F(z)\ dz = 0 $$
Then using the residue theorem, we can compute the closed loop integral
$$ \begin{align} \int_C F(z)\ dz &= 2\pi i\operatorname{Res}\left(F(z),\frac{ai}{2\pi}\right) \\ &= 2\pi i\lim_{z\to \frac{ai}{2\pi}} \left(z- \frac{ai}{2\pi}\right) F(z) \\ &= \lim_{z\to \frac{ai}{2\pi}} e^{2\pi itz} \\ &= e^{-at} \end{align} $$
Thus
$$ \lim_{R\to\infty} \int_{-R}^{R} F(z) \ dz = \int_C F(z) \ dz = e^{-at} $$