Contour integration $\frac{e^{iz}}{2\sqrt{z}}$

Question

When $z=u+iv$,

I would like to compute the integral of $\frac{e^{iz}}{2\sqrt{z}}$ along above curve.

1. The imaginary axis $$\frac{1}{2}\int_{R}^{0} \frac{e^{-v}}{\sqrt{iv}}d(iv)$$ $R$ goes to $\infty$.

Here because of $\sqrt{iv}$. I confuse to use the change of variable $v=y^{2}$. Anyone can helps me about this?

2. the integral over the circular arc goes to zero as $R\rightarrow \infty$. How can I show it?

Answer 1

1. The integral converges, so you can already take the limit $R\rightarrow \infty$. As you mention, the change of variables $y=\sqrt{v}$ yields a gaussian integral you can simply evaluate.

2. Generally for the function $\frac{e^{iz}}{z^s}$ with $s>0$ you can proceed as follows: Parametrize $z=Re^{it}$ with $t\in[0,\pi/2]$. Then $$\left| \int_C \frac{e^{iz}}{z^s} \, {\rm d}z \right| \leq R^{1-s} \int_0^{\pi/2} e^{-R\sin(t)} \, {\rm d}t \stackrel{\sin(t)\geq 2t/\pi}{\leq} R^{1-s} \int_0^{\pi/2} e^{-2Rt/\pi} \, {\rm d}t \, .$$