# Calculate $\int_{-\infty}^{\infty} \frac{\sin(z)}{z}dz$

$\Gamma_1$ is the line connecting $\epsilon$ to $R$.

$\Gamma_2$ is the counterclockwise upper semicircle with radius $R$.

$\Gamma_3$ is the line connecting $-R$ to $-\epsilon$.

$\Gamma_4$ is the clockwise upper semicircle with radius $\epsilon$.

Denote $\Gamma=\Gamma_1+\Gamma_2+\Gamma_3+\Gamma_4$.

I used Cauchy's integral theorem to show that

$\int_{\Gamma}^{} \frac{e^{iz}}{z}dz=0$.

Now I need to evaluate the different parts of the integral $\int_{\Gamma}^{} \frac{e^{iz}}{z}dz$ and show that $\int_{-\infty}^{\infty} \frac{\sin(z)}{z}dz=\pi$ using the fact that $\lim_{\epsilon\rightarrow 0} \int_{\Gamma_4}^{ } \frac{\sin(t)}{t}dt=-i\pi$.

So far I managed to show that $\int_{\Gamma_2}^{ } \frac{e^{iz}}{z}dz=0$.

So I am left with

$\int_{\Gamma_1}^{ } \frac{e^{iz}}{z}dz+\int_{\Gamma_3}^{ } \frac{e^{iz}}{z}dz=-\int_{\Gamma_2}^{ } \frac{e^{iz}}{z}dz \Rightarrow$

$\Rightarrow \int_{-R}^{-\epsilon} \frac{\cos(z)}{z}dz+i\int_{-R}^{-\epsilon} \frac{\sin(z)}{z}dz+\int_{\epsilon}^{R} \frac{\cos(z)}{z}dz+i\int_{\epsilon}^{R} \frac{\sin(z)}{z}dz=-\int_{\Gamma_2}^{ } \frac{\cos(z)}{z}dz-i\int_{\Gamma_2}^{ } \frac{\sin(z)}{z}dz$.

But I can't see how to evaluate these integrals. What am I missing here?

• One could avoid this by showing that $\int_0 ^\infty \frac{\sin x}{x} \ dx = \pi/2$ and use the fact that the integrand is even. – Sean Roberson Apr 27 '18 at 17:08
• Observe that the integral over $\Gamma_2$ is 0 and integrals over $\Gamma_1$and $\Gamma_3$ can be combined together in a multiple of integral of a real even function. – user Apr 27 '18 at 17:09
• Do you mean combine $\int_{\Gamma_1}^{ } \frac{e^{iz}}{z}dz + \int_{\Gamma_3}^{ } \frac{e^{iz}}{z}dz$ to $\int_{-R}^{R} \frac{e^{iz}}{z}dz$? if so, isn't there a problem at $z=0$? how can you do this integral? – bp7070 Apr 27 '18 at 17:39

## 1 Answer

First note that by Jordan lemma: $$\lim_{R\to\infty}\int_{\Gamma_2}\frac{e^{iz}}{z}dz=0.$$

Next $$\lim_{R\to\infty}\left[\int_{\Gamma_1}\frac{e^{iz}}{z}dz+\int_{\Gamma_3}\frac{e^{iz}}{z}dz\right]= \int_{\epsilon}^\infty\frac{e^{ix}}{x}dx+\int_{-\infty}^{-\epsilon}\frac{e^{ix}}{x}dx=\int_{\epsilon}^\infty\frac{e^{ix}}{x}dx+\int_{\epsilon}^\infty\frac{e^{-ix}}{-x}dx\\ =2i\int_{\epsilon}^\infty\frac{\sin x}{x}dx.$$ In the limit $$\epsilon\rightarrow 0$$ we have $$0=\oint\frac{e^{iz}}{z}dz= \underbrace{-\pi}_{\Gamma_4} i+\underbrace{2i\int_{0}^\infty\frac{\sin x}{x}dx}_{\Gamma_1+\Gamma_3}\Rightarrow \int_0^\infty\frac{\sin x}{x}=\frac{\pi}{2}.$$ Finally observing that $$\frac{\sin x}{x}$$ is an even function: $$\int_{-\infty}^\infty\frac{\sin x}{x}=2\int_0^\infty\frac{\sin x}{x}=\pi.$$