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 $0<s<r<\infty$ the contour of integration is described by $\gamma=[s,r]+\gamma_r+[-r,-s]-\gamma_s$, with $\gamma_r(t)=re^{it}$ and $\gamma_s(t)=se^{it}$ on $[0,\pi]$. Recall Exercise IV$.4.20.$
Exercise IV$.4.20.$ For $r$ with $0<r<\infty$ let $I(r)=\int_{\gamma_r}z^{-1}e^{iz}dz$ with $\gamma_r(t)=re^{it}$ on $[0,\pi]$. Show that $I(r)\rightarrow 0$ as $r\rightarrow \infty$ and also that $I(r)\rightarrow \pi i$ as $r\rightarrow 0$
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.
 A: 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}$$
A: Here is a real approach for verification:
$$
\begin{align}
\int_0^\infty\frac{\sin(t)}t\,\mathrm{d}t
&=\int_0^\infty\int_0^\infty\sin(t)\,e^{-st}\,\mathrm{d}s\,\mathrm{d}t\\
&=\int_0^\infty\left[-\frac{s\sin(t)+\cos(t)}{1+s^2}\,e^{-st}\right]_0^\infty\,\mathrm{d}s\\
&=\int_0^\infty\frac1{1+s^2}\,\mathrm{d}s\\
&=\left[\tan^{-1}(s)\vphantom{\int}\right]_0^\infty\\[3pt]
&=\frac\pi2
\end{align}
$$
A: This is one of the known problems that can be evaluated using Feynman’s Trick. First, we assume the integrand is a special case of the function$$f(a,x)=\frac {e^{-ax}\sin x}{x}$$Where $f(0,x)$ is your problem. We choose this substitution because the function converges uniformly. Taking the partial derivative we have$$F(a)=\int\limits_0^{\infty}\frac {\partial}{\partial a}\frac {e^{-ax}\sin x}x\, dx=\int\limits_0^{\infty}e^{-ax}\sin x\, dx=:I$$Integration by parts twice gives in terms of $I$$$I=-1-a^2I$$So we have that$$F(a)=-\arctan a+C$$The constant can be found by taking the limit as $a\to\infty$, and the constant is $\tfrac {\pi}2$. Therefore, the general solution is$$\int\limits_0^{\infty}\frac {e^{-ax}\sin x}x\ dx=-\arctan a+\frac {\pi}2$$
