calculus - Confirm that $int_{0}^{infty}t^{-1}sin t dt=pi/2$

Confirm that $\int_{0}^{\infty}t^{-1}\sin t dt=\pi/2$.

The guide book I am using gives the following help:

Consider $\int_{\gamma}z^{-1}e^{iz}dz$, where for $0

Exercise IV$.4.20.$ For $r$ with $0

Using the hint, I know that $\int_{\gamma}z^{-1}e^{iz}dz=0$ for the Cauchy theorem, with which $\int_{[s,r]}z^{-1}e^{iz}dz+\int_{\gamma_r}z^{-1}e^{iz}dz+\int_{[-r,-s]}z^{-1}e^{iz}dz-\int_{\gamma_s}z^{-1}e^{iz}dz=0$, but I do not know what else to do here, could someone help me please? Thank you very much.

Answer

Define a path in the Complex Plane:

Now consider $$\int_{C} \frac{e^{iz}}{z}dz=\int_{arc}\frac{e^{iz}}{z}dz+\int_{Arc}\frac{e^{iz}}{z}dz+\int_{-R}^{-r}\frac{e^{iz}}{z}dz+\int_{r}^{R}\frac{e^{iz}}{z}dz$$

By parametizing the integrals over the arcs, and letting $r\to0$ for $arc$ and $R\to\infty$ for $Arc$, we see that $\int_{arc}\to i\int_{\pi}^0d\theta$ and $\int_{Arc}$$\to0$.

So we have $$\int_{C} \frac{e^{iz}}{z}dz=PV\int_{-\infty}^{\infty}\frac{e^{iz}}{z}dz-\pi i$$ where $PV$ denotes the Cauchy Principal Value.

Since the contour does not enclose any poles, the entire contour integral is $0$.

So $$0=PV\int_{-\infty}^{\infty}\frac{e^{iz}}{z}dz-\pi i$$ $$PV\int_{-\infty}^{\infty}\frac{e^{iz}}{z}dz=\pi i$$ Note that due to Euler's Formula $$Im(PV\int_{-\infty}^{\infty}\frac{e^{iz}}{z}dz)=\int_{-\infty}^{\infty}\frac{sin(z)}{z}dz$$

And so $$\int_{-\infty}^{\infty}\frac{sin(z)}{z}dz=\pi$$
Lastly since $\frac{sin(z)}{z}$ is an even function:
$$\int_{0}^{\infty}\frac{sin(z)}{z}dz=\frac{\pi}{2}$$