How to compute the integral $\frac{1}{2\pi i} \int^\infty_{-\infty} \frac{e^{ixy}}{y - i} \, dy$ I am reading an essay that says that it is true that
\begin{equation}
\frac{1}{2\pi i} \int^\infty_{-\infty} \frac{e^{ixy}}{y - i} \, dy = 
\begin{cases}  e^{-x} & \text{for x }>0 \\
               0 & \text{for x} < 0
\end{cases}
\end{equation}
I would like to know how to compute this integral, can anyone help please ?
Many thanks!
 A: You need knowledge of complex analysis. First, you should locate singularities for the function you are integrating. Since ${e^{ixy}}$ is entire, only singularity is at  $y = i$. Second, you need Residue theorem http://en.wikipedia.org/wiki/Residue_theorem. The third step is closing the contour of integration, as is described here http://en.wikipedia.org/wiki/Residue_theorem#Example. The fourth step is showing that integral by arc is going to 0 as we limit the radius of contour to infinity.
For your problem, it would go like this:
$$\displaystyle \frac{1}
{{2\pi i}}\int_{ - a}^a {\frac{{{e^{ixy}}}}
{{y - i}}dy}  + \frac{1}
{{2\pi i}}\int_{{\text{arc}}} {\frac{{{e^{ixy}}}}
{{y - i}}dy}  = \frac{1}
{{2\pi i}}\int_C {\frac{{{e^{ixy}}}}
{{y - i}}dy}  = {\text{res}}\left( {\frac{{{e^{ixy}}}}
{{y - i}},i} \right) = {e^{ - x}} \text{ for }a > 1$$
Now, we will need Jordan's lemma http://en.wikipedia.org/wiki/Jordan%27s_lemma to obtain $$\displaystyle \left| {\int_{{\text{arc}}} {\frac{{{e^{ixy}}}}
{{y - i}}dy} } \right| \leqslant \frac{\pi }
{x}\frac{1}
{{a - 1}}$$ (point on imaginary line is closest to $i$ so the maximum of $\displaystyle \frac{1}
{{\left| {y - i} \right|}}$ is obtained in that point) from which it follows that 
$$\displaystyle \int_{{\text{arc}}} {\frac{{{e^{ixy}}}}
{{y - i}}dy}  \leqslant \left| {\int_{{\text{arc}}} {\frac{{{e^{ixy}}}}
{{y - i}}dy} } \right| \leqslant \frac{\pi }
{x}\frac{1}
{{a - 1}}$$ and
$$\displaystyle \mathop {\lim }\limits_{a \to  + \infty } \frac{\pi }
{x}\frac{1}
{{a - 1}} = 0 \Rightarrow \mathop {\lim }\limits_{a \to  + \infty } \left| {\int_{{\text{arc}}} {\frac{{{e^{ixy}}}}
{{y - i}}dy} } \right| = 0 \Rightarrow \mathop {\lim }\limits_{a \to  + \infty } \int_{{\text{arc}}} {\frac{{{e^{ixy}}}}
{{y - i}}dy}  = 0$$
So, $$\displaystyle {e^{ - x}} = \mathop {\lim }\limits_{a \to  + \infty } \left( {\int_{ - a}^a {\frac{{{e^{ixy}}}}
{{y - i}}dy}  + \int_{{\text{arc}}} {\frac{{{e^{ixy}}}}
{{y - i}}dy} } \right) = \int_{ - \infty }^\infty  {\frac{{{e^{ixy}}}}
{{y - i}}dy} $$
For $x < 0$ everything is the same, except that part of contour on a real line has opposite orientation, so we close the contour with an arc on a lower half-plane. Since our function doesn't have any singularities in the lower half plane, sum of residues in that area is 0, and so is the value of the integral in that case.
