I'm working through Big Rudin's (Real and Complex Analysis) Fourier Transform chapter, and the following complex integral is part of a discussion on the Inverse Transform that Rudin mentions briefly without resolving, and I'm very curious how this is done.
$\int_{-\infty}^{\infty}e^{-|\lambda t|}e^{itx}dt$
What I remember from my Complex Analysis days, is to take the limit of the integral evaluated around a contour from -R to R, and then around $C_R$, the semicircle connecting -R and R. Since this is closed I can use the Cauchy Residue Theorem, and by Jordan's Lemma hopefully the integral around $C_R$ will vanish, so I'm left with
$\int_{-\infty}^{\infty}e^{-|\lambda t|}e^{itx}dt=2\pi*\{\text{sum of residues}\}$
However, this integral has no residues, so simply taking the limit of this integral from -R to R should work, but my rusty integration techniques aren't helping and Mathematica is not giving me any results.
Any help in solving this is much appreciated!
Let $F(x)$ be the function defined by the Fourier Integral
$$F(x)=\int_{-\infty}^\infty e^{-|\lambda t|}e^{itx}\,dt$$
Since the integral converges for $\lambda \ne 0$, we can write the integral as
$$F(x)=\lim_{R\to \infty}\int_{-R}^R e^{-|\lambda t|}e^{itx}\,dt$$
One might be tempted to evaluate this integral be closing the real line contour with a semi-circle, centered at the origin and with radius $R$, in the upper-half or lower-half plane and then using the Residue Theorem. However, this approach is fatally flawed since $e^{-|\lambda||z|}$ is not an Analytic Function. Therefore, we proceed along a different way forward.
First, we use Euler's Formula and exploit even and odd components of the integrand to reveal
$$F(x)=2\int_0^\infty e^{-|\lambda|t}\cos (xt)\,dt$$
Then, Integrating by Parts twice yields
$$\begin{align} F(x)&=\lim_{R\to \infty}\left.\left(\frac{e^{-|\lambda|t}\left(-|\lambda|\cos (xt)+x\sin(x t)\right)}{x^2+|\lambda|^2}\right)\right|_{t=0}^{t=R}\\\\ &=\bbox[5px,border:2px solid #C0A000]{\frac{|\lambda|}{x^2+|\lambda|^2}} \end{align}$$