Sin(x)/x via contour integration

[snipez90]

Homework Statement

Evaluate \int_{0}^{\infty} \frac{\sin x}{x}\,dx

Homework Equations

1. Cauchy's integral theorem
2. \int_{|z| = r} \frac{dz}{z} = 2\pi i

The Attempt at a Solution

I want to see if the following works. The hint suggests the use of the contour here: http://upload.wikimedia.org/wikibooks/en/9/9e/ContourSinzz.gif" . We denote by \gamma _R ^+ and \gamma _{\epsilon} ^+ the semicircles of radius R and \epsilon with positive and negative orientations, respectively.

We consider the function
f(z) = \frac{e^{iz}-1}{z}.
By Cauchy's theorem, we have
\int_{-R}^{-\epsilon} + \int_{\gamma _{\epsilon} ^+} + \int_{\epsilon}^{R} + \int_{\gamma _{R} ^+}.

First we note that
\int_{\gamma _{\epsilon} ^+} f(z)\, dz = \int_{\gamma _{\epsilon} ^+} \frac{e^{iz}}{z}\, dz \,-\, \int_{\gamma _{\epsilon} ^+} \frac{dz}{z} = \int_{\pi}^{0} \frac{\exp \left(i \epsilon e^{it} \right)}{\epsilon e^{it}}\cdot i \epsilon e^{it} dt \,+\, \pi i = \int_{\pi}^{0} i\exp \left(i\epsilon e^{it}\right)dt + \pi i.

Since |e^{iz}| = e^{-y}, applying Euler's formula gives
|i\exp{(i\epsilon e^{it})}| \leq e^{-\epsilon \sin t}.
Letting \epsilon \rightarrow 0 and applying Lebesgue's dominated convergence theorem, we have
\lim_{\epsilon \rightarrow 0}\int_{\gamma _{\epsilon} ^+} f(z)\, dz = \lim_{\epsilon \rightarrow 0} \int_{\pi}^{0} i\exp \left(i\epsilon e^{it}\right) dt + \pi i = 0.

Applying the same arguments to the larger semicircle, we have
\left|\int_{\gamma _{\R} ^+} \frac{e^{iz}}{z}\,dz\right| = \left|\int_{0}^{\pi} i\exp \left(iRe^{it} \right) dt \right| \leq \int_{0}^{\pi} \exp(-R \sin t)\,dt.
Clearly this last integrand is dominated over the interval [0,\pi], so applying DCT again shows that this last integral vanishes as R \rightarrow \infty.

Thus the contribution from the integrals over each semicircle is -\pi i as we let R approach infinity and epsilon approach 0. Hence
\int_{-\infty}^{\infty} \frac{e^{ix}-1}{x} dx = \pi i.
Taking imaginary parts and noting parity symmetry gives the desired result.

*EDIT* Hopefully I'm not missing anymore differentials.

 

[Euge]

The solution is correct. Another way to show $\lim\limits_{\epsilon \to 0} \int_{\gamma_{\epsilon}^+} f(z)\, dz = 0$ is to note that since $f(z)$ has a removable singularity at the origin, it is bounded near the origin and hence $\int_{\gamma_{\epsilon}^+} f(z)$ is bounded by a constant times the arclength of $\gamma_\epsilon^+$, which is a constant times $\epsilon$. Therefore $\lim\limits_{\epsilon \to 0} \int_{\gamma_{\epsilon}^+} f(z)\, dz = 0$, as desired.