Integrating $\frac {e^{-iz}}{z}$ over a semicircle counter-clockwise around $0$ of infinite radius $r$

Question

How do I find $\lim_{r\to +\infty}$ ${∫_\gamma}_r$ $\frac {e^{-iz}}{z}dz$ where $\gamma_r$ is the counter-clockwise semicircle centred at 0 with radius $r$. I think it's equal to $2πi$, but how do I find this integral without using the Sine integral? Any help is appreciated.

Answer 1

I came up with an answer. Let's define ${\gamma}$ the circle with radius r and centered at the origin, and ${\gamma}_s$ is such that ${\gamma}$ = ${\gamma}_r$ + ${\gamma}_s$.

So, $\lim_{r\to +\infty}$ ${∫_\gamma}_r$ $\frac {e^{-iz}}{z}dz$ = $\lim_{r\to +\infty}$ ${∫_\gamma}$ $\frac {e^{-iz}}{z}dz$ - $\lim_{r\to +\infty}$ ${∫_\gamma}_s$ $\frac {e^{-iz}}{z}dz$

By Cauchy or Residue Theorem, $\lim_{r\to +\infty}$ ${∫_\gamma}$ $\frac {e^{-iz}}{z}dz$ = $\lim_{r\to +\infty}$ 2πi * Res($\frac {e^{-iz}}{z}$,0) = 2πi * 1 = 2πi

And, $\lim_{r\to +\infty}$ ${∫_\gamma}_s$ $\frac {e^{-iz}}{z}dz$ = $\lim_{r\to +\infty}$ $\int_π^{2π}$ $\frac {e^{-ire^{it}}}{re^{it}}ire^{it}dz$ = i * $\lim_{r\to +\infty}$ $\int_π^{2π}$ $e^{-ire^{it}}dz$ = i * $\lim_{r\to +\infty}$ $\int_π^{2π}$ $e^{-ir(cos(t)+isin(t))}dz$ = i * $\lim_{r\to +\infty}$ $\int_π^{2π}$ $e^{rsin(t)}e^{-ircos(t)}dz$. But |$e^{-ircos(t)}$|< 1 and $\lim_{r\to +\infty}$ $e^{rsin(t)} = 0$ for π≤t≤2π. So, i * $\lim_{r\to +\infty}$ $\int_π^{2π} e^{rsin(t)}e^{-ircos(t)}dz$ = 0

Finally, $\lim_{r\to +\infty}$ ${∫_\gamma}_r$ $\frac {e^{-iz}}{z}dz$ = 2πi - 0 = 2πi