Bounding of $\frac{e^{iz}-1}{z}$ over a semi-circle

Question

I am trying to solve the integral $\int_0^{\infty} \frac{\sin{x}}{x}$ only by using complex analysis. Note, there are several posts already on this site about such an integral (and done via contour-integration).

To solve this question via contour integration, we have to consider the integral $\frac{e^{iz}-1}{z}$ over the indented semi-circle centered at 0. I am having trouble integrating over the outer semi-circle which is usually called $\gamma_R^+$ (positive orientation and radius is $R$).

I will outline my attempt and show that it is producing an incorrect answer but I am having trouble seeing where I am wrong.

We want to solve $\int_{\gamma_R^+} \frac{e^{iz}-1}{z}$ as $R\to\infty$. I know that $|e^{iz}|\leq 1$ for all complex numbers in the upper half plane. Therefore, $|e^{iz}-1|=|1-e^{iz}|\leq 2$ so that $\frac{e^{iz}-1}{z} \leq \frac{2}{|z|}$. Now, let $z(t) = Re^{it}$ so that $\frac{2}{|z|} = \frac{2}{Re^{it}}$ which then goes to 0 as $R\to\infty$. Looking at other correct solutions, it seems to be the case that $\int_{\gamma_R^+} f(z)dz = -\pi i$ (as $R\to\infty$) and not 0. Therefore, I have went wrong somewhere but cannot seem to figure out where. Any help is appreciated.

Also a further question, if you have time:

1) I am still not sure why we choose our function to be $\frac{e^{iz}-1}{z}$ and not something else. The only reason I knew to choose this function was because it was a hint in the book that this question was asked.

Some posts where the question on how to solve this integral has been asked:

Integrating $(e^{iz}-1)/z$ about a semi-circle to evaluate Dirichlet integral.

I am trying to show $\int^\infty_0\frac{\sin(x)}{x}dx=\frac{\pi}{2}$

Answer 1

I realized there are two main ways in which my method goes wrong. I will highlight one of them here and other @Lord Shark the Unknown has already pointed out (for which I'm grateful).

If I let $z(t) = Re^{it}$, then note that $dt = Rie^{it}$. By the change of variables formula, if I plug in $z(t)$ into $\int \frac{2}{|z|}dz$, then this integral becomes $\int \frac{2}{Re^{it}}Rie^{it}dt = \int 2i dt$ which would clearly not be equal to 0 as $R\to\infty$.

Thank you to everyone who helped!

Answer 2

$$ \int_{\gamma_R^+}\frac{e^{iz}-1}{z}\,dz =i\int_0^\pi(\exp(iR\cos t-R\sin t)-1)\,dt =-\pi i+\int_0^\pi e^{-R\sin t}\exp(iR\cos t)\,dt $$ This shows where the $-\pi i$ comes from. The remaining integral is bounded in absolute value by $$I_R=\int_0^\pi e^{-R\sin t}\,dt.$$ Now $I_R\to0$ as $R\to\infty$. This is immediate from the Dominated Convergence Theorem. There is a tradition in complex analysis textbooks to assume the reader is unfamiliar with the DCT and to use Jordan's inequality to get an explicit upper bound on $I_R$ instead. One can avoid both of these by using a rectangular contour instead of a semicircular contour.

The reason for using $(e^{iz}-1)/z$ instead of $e^{iz}/z$ is to avoid having a pole at the origin (actually on this given contour). Indeed most texts use this function but have to indent the contour at the origin to avoid the pole. In this approach, that is where the $-\pi i$ comes from.