I am not asking how to calculate it. It is clear that can be done by integrating $\mathrm{e}^{iz}/z$ in complex plane. I am asking the following:
- Since $\,\mathrm{f}\left(z\right) = \sin\left(z\right)/z$ has a removable singularity at $z=0$ we may continue it to the whole complex plane by defining $\lim_{z \to 0}\mathrm{f}\left(z\right) = 1$. Then $\,\mathrm{f}$ is analytic and its integral over any closed curve $\gamma$ should be zero.
So I was trying to verify it by choosing the standard $\gamma$ contour from residue calculation ( though no residue here ) in the upper half plane:
$\gamma_R$ is a semicircle in the upper half plane on which clearly the integral of $\,\mathrm{f}\left(z\right)$ goes to zero as $R \to \infty$ ( no problem there ).
$\gamma_x$ the real line going from $-\infty$ to $+\infty$ and excluding an infinitesimal interval $\left[-\epsilon,+\epsilon\right]$ centered on the origin. The integral over $\gamma_x$ should give us twice the Sine integral which is $\pi$.
$\gamma_{\epsilon}$- A small semicircle which may ( or may not ) exclude the origin. I expect the integral over $\gamma_{\epsilon}$ to go to $-\pi$ as I take the limit $\epsilon\to 0$ so as to cancel the integral on $\gamma_x$ and give me zero total contour integral. But this does not happen. Since $\,\mathrm{f}\left(z\right)$ is regular at the origin the integral over that curve always goes to zero... So, what's happening ?. Am I wrong in assuming that the total contour integral should be zero .?