Consider the function $$ f(y,\alpha):=\int_0^\infty dx\ \frac{e^{- i y x}}{x-i\alpha} $$ for $y>0$ and $\alpha \in \mathbb{R}$. This is a Fourier transform over half the real line.
I have tried to compute the above with a contour integration by integrating over the quarter-circle that runs along $[0,\infty] \cup \{ R e^{i \theta } | \theta \in [0,\pi/2] \} \cup [i\infty,0]$ but I am unable to make progress for the case when $\alpha > 0$. Is there a simpler way to compute this integral?
Evaluating the integral with Mathematica gives the following rather ugly result consisting of two MeijerG terms.
$$\int_0^\infty\frac{e^{-i x y}}{x-i \alpha}\,dx=\frac{1}{2 \sqrt{\pi}}\left(G_{3,1}^{1,3}\left(\frac{2 i}{y \alpha},\frac{1}{2}| \begin{array}{c} \frac{1}{2},1,1 \\ 1 \\ \end{array} \right)-i G_{3,1}^{1,3}\left(\frac{2 i}{y \alpha},\frac{1}{2}| \begin{array}{c} \frac{1}{2},\frac{1}{2},1 \\ \frac{1}{2} \\ \end{array} \right)\right)\tag{1}$$
I tried separating out the real and imaginary parts of the integrand as illustrated in (2) below and integrating them separately as illustrated in (3) below and Mathematica then gave the somewhat simpler result illustrated in (4) below where the second integral in the right-hand side of (3) evaluated to zero and Shi and Chi in (4) below represent the SinhIntegral and CoshIntegral functions respectively.
$$\frac{e^{-i x y}}{x-i \alpha}=\frac{\alpha \sin(x y)+x \cos(x y)}{\alpha^2+x^2}+i \frac{\alpha \cos(x y)-x \sin(x y)}{\alpha^2+x^2}\tag{2}$$
$$\int_0^\infty\frac{e^{-i x y}}{x-i \alpha}\,dx=\int_0^\infty\frac{\alpha \sin (x y)+x \cos (x y)}{\alpha^2+x^2}\,dx+i \int_0^\infty \frac{\alpha \cos(x y)-x \sin(x y)}{\alpha^2+x^2}\,dx\tag{3}$$
$$\int_0^\infty\frac{e^{-i x y}}{x-i \alpha}\,dx=\frac{1}{2} e^{\alpha y} \left(\log(\alpha)+\log\left(\frac{1}{\alpha}\right)+2 \text{Shi}(y \alpha)-2 \text{Chi}(y \alpha)\right)\tag{4}$$
Mathematica verifies the right-hand sides of (1) and (4) above are equivalent assuming $y>0\land\alpha\in\mathbb{R}\land\alpha\neq 0$ which are the assumptions I specified on all integrals above.
The following figure illustrates the right-hand side of formula (4) above evaluated at $\alpha=1$.
Figure (1): Illustration of formula (4) evaluated at $\alpha=1$