What does $\int_{0}^{\infty} e^{y(iu-\alpha)}dy = ?$ Please note $i$ is a complex variable, $\alpha$ and $u $ are constants.
I know this integral evaluates to:
$$\left.\frac{e^{y(iu-\alpha)}}{iu-\alpha}\right|_0^\infty$$
I'm unsure how this evaluates due to the terms inside the parenthesis of the exponetial for which $\infty$ will have to be distributed to. So you end up with a $-\infty$ and $+\infty$ term inside that exponential, which is throwing me off. How do you evaluate and make sense out of this?
I am assuming that $\alpha$ is real.
Write $e^{y(iu-\alpha)} =e^{iuy}e^{-y\alpha} $.
If $\alpha>0$, $e^{-y\alpha} \to 0 $, so the limit at $\infty$ is $0$.
If $\alpha=0$, $e^{iuy} =\cos(uy)+i\sin(uy) $ oscillates as $y \to \infty$, so the limit does not exist.
If $\alpha < 0$, it blows up as $y \to \infty$.