# Confirm that $\int_{0}^{\infty}t^{-1}\sin t dt=\pi/2$ [duplicate]

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.

## marked as duplicate by Guy Fsone, Arnaud D., Rolf Hoyer, kingW3, Aaron MontgomeryNov 22 '17 at 18:51

• There are multiple questions on the stack exchange and elsewhere that show that$$\int\limits_0^{\infty}\frac {\sin t}t\, dt=\frac {\pi}2$$ – Crescendo Nov 5 '17 at 18:13
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 iPV\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}$$Here is a real approach for verification:$$ \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} $$• Are you using Laplace Transform properties? – aleden Nov 5 '17 at 19:48 • No, just using \frac1t=\int_0^\infty e^{-st}\,\mathrm{d}s. – robjohn Nov 5 '17 at 20:00 • +1 Cool trick. I'll be on the lookout for more opportunities to use it. – zahbaz Nov 5 '17 at 20:04 This is one of the known problems that can be evaluated using Feynman’s Trick. First, we assume the integrand is a special case of the function$$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