Context: I am trying to show that $\int_0^\infty x^{-1}\sin x dx = \frac{\pi}{2}$ using complex analysis, by first integrating $\oint_{\Gamma} z^{-1}e^{iz}$, where $\Gamma$ is a closed contour consisting of two portions of real axis $-R$ to $\epsilon$ and $\epsilon$ to $R$ and two semicircles $\Gamma_1(0,R)$ and $\Gamma_2(0,\epsilon)$.
I am almost done. I just need to show that $$\lim_{R\to\infty}\int_{\Gamma_1} \frac{e^{iz}}{z} = 0.$$
I don't need a rigorous proof. I am new to complex analysis and know almost no theorems. I just want to understand things "intuitively", if it's possible. So far, I know only Cauchy formula/theorem and residue theorem.
I found following solution, which I don't understand very well:
Integrate by parts $\int_{\Gamma_1} \frac{e^{iz}}{z} = \frac{e^{iz}}{iz}|_{-R}^{R} + \int_{\Gamma_1}\frac{e^{iz}}{z^2}dz$. The first turns out to be $\frac{2i}{R}\sin R \to 0$. Then absolute value of the integral is bounded by $\pi R \max_{\Gamma_1}{|\frac{e^{iz}}{z^2}|}\leq\frac{\pi}{R} \to 0$.
Why does it suffice to evalute at $R$ and $-R$? Why is the integral bounded by this?
Thank you!
Answer
One uses here the inequalilty
$$\left|\int_\Gamma f(z)\,dz\right|\leq|\Gamma|\max\limits_{z\in\Gamma}|f(z)|$$
($|\Gamma|$ being the length of path $\Gamma$), which is almost obvious from the definition of integral of complex variable.
In your case $|\Gamma_1|=\pi R$ (half of a circle of radius $R$) and $|\frac{e^{iz}}{z^2}|=\frac{1}{|z|^2}=\frac{1}{R^2}$ for all $z\in\Gamma_1$, so $\max\limits_{z\in\Gamma_1}|\frac{e^{iz}}{z^2}|=\frac{1}{R^2}$.
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