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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,

Eugene Zhang
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User001
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    https://math.stackexchange.com/questions/594641/computing-int-infty-infty-frac-sin-xx-mathrmdx-with-residue-calc – angryavian Dec 20 '15 at 03:56
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    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 05:45
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    @angryavian: I was just about to post a link to this answer, but you beat me with a link to the corresponding question. – robjohn Dec 20 '15 at 09:59
  • Thanks for the link at @angryavian. – User001 Dec 20 '15 at 19:20
  • Thanks so much @Dr.MV. I think that I was able to justify it. The upper bound for the integrand was $e^{\large -\epsilon sin(\theta)}$, which is continuous over $[0,\pi]$, hence integrable over $[0,\pi]$, and so the dominated convergence theorem is applicable here. – User001 Dec 20 '15 at 19:27
  • Although Daniel Fischer claims (on the link mentioned by robjohn) that in fact I can use a weaker assumption and show the uniform convergence of $\large e^{i \epsilon e^{i\theta}} \to 1$, in order to justify swapping the limit with the integral, but I am having some trouble with specifying a suitable $\epsilon_0$ so that for any $\delta >0$, whenever $\epsilon < \epsilon_0$, we have that $|\large e^{i \epsilon e^{i\theta}} - 1| < \delta$. What do you think would be a good choice? Thanks, @Dr. MV – User001 Dec 20 '15 at 19:27
  • Thanks as usual, @robjohn :-) – User001 Dec 20 '15 at 19:36
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    @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
  • One doesn't select $\epsilon$; one must find for any $\epsilon$ a $\delta$. @Robjohn has provided a way forward for you. ;-)) - Mark – Mark Viola Dec 20 '15 at 21:08

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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 $(1)$ $$ \lim_{r\to0}\int_{\gamma_r}\frac{e^{iz}}{z}dz=\lim_{r\to0}\int_{\gamma_r}\frac1{z}dz=\lim_{r\to0}\int_{\pi}^{0}\frac{ire^{i\theta}}{re^{i\theta}}d\theta=-\pi i $$

Eugene Zhang
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