# Integrating $\sin(x)/x$, how to treat the pole at the origin? [duplicate]

I want to use residue theory to integrate

$$\int_{-\infty}^{\infty}\frac{\sin(x)}{x}dx$$

What would be a good contour to use?

I plan to take the imaginary part of this integral:

$$\int \frac {e^{iz}}{z} dz$$

The integrand has a simple pole at the origin, so that is a bit problematic.

So, I thought of choosing an upper semicircular contour, with a small semicircular indent going over the origin.

The estimate over the big semicircle is easy to give and the integral on it goes to zero, as the radius $R$ grows to infinity.

The integration over the two half lines are the desired integrals. And the integration should be set equal to zero, since this choice of contour avoids the pole at the origin, and by Cauchy's Theorem, the integration of an analytic function over a closed contour is just zero.

Now the only trouble is evaluating the integral over the small semicircular indent, call it $C_{\epsilon}$.

Parametrizing, we have that $z= \epsilon e^{i\theta}$ for $\pi \ge \theta \ge 0$.

Then the contour integral is (after a bit of simplification):

$$\int_{\pi}^0 e^{\large i\epsilon e^{i\theta}}i d\theta$$

$$=-i\int_0^{\pi} e^{\large i\epsilon e^{i\theta}} d\theta$$

I'm not sure how to evaluate this integral, so any hints or comments are welcome.

Also, if you think a completely different approach would be better, please feel free to comment.

Thanks,

## marked as duplicate by mrf complex-analysis StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Dec 20 '15 at 14:12

• Remember that you will take the limit as $\epsilon \to 0$. It is easy to justify interchanging the limit with the integral. Inasmuch as $\lim_{\epsilon \to 0}e^{i\epsilon e^{i\theta}}=1$, the resulting integral over the semicircle is $-i \pi$. Thus, the Cauchy Principal Value of the integral of interest is $\pi$ (after taking the imaginary part). – Mark Viola Dec 20 '15 at 5:45
• @User001: Since $\left|\,ie^{i\theta}\,\right|=1$ and $\left|\,e^z-1\,\right|=\left|\,\int_0^1e^{tz}z\,\mathrm{d}t\,\right|\le\left| z\,\right|e^{\left|z\right|}$ we have $$\left|\,e^{\large i\epsilon e^{i\theta}}-1\right|\le\epsilon e^\epsilon$$ Does that help? – robjohn Dec 20 '15 at 20:15
Let $\gamma_r$ is the small upper half circle with radius of $r<1$ bypassing $0$. Then $$\int_{\gamma_r}\frac{e^{iz}}{z}dz=\int_{\gamma_r}\left(\frac1{z}+g(z)\right)dz\tag1$$ where $$g(z)=\sum_{n=1}^{\infty}\frac{i^nz^{n-1}}{n!}$$ and is analytic for $|z|<1$. Since $|g(z)|<M$ for $|z|<1$ $$\left|\int_{\gamma_r}g(z)dz\right|<\pi Mr\to0$$ as $r\to0$. So $$\lim_{r\to0}\int_{\gamma_r}g(z)dz=0$$ Thus by $(1)$ $$\lim_{r\to0}\int_{\gamma_r}\frac{e^{iz}}{z}dz=\lim_{r\to0}\int_{\gamma_r}\frac1{z}dz=\lim_{r\to0}\int_{\pi}^{0}\frac{ire^{i\theta}}{re^{i\theta}}d\theta=-\pi i$$