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.