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:

    1. $\gamma_R$ 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 ).

    2. $\gamma_x$ 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 $\pi$.

    3. $\gamma_{\epsilon}$- 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 .?

  • $\begingroup$ How "the integral over $\;\gamma_x=\left[(-\infty,\,\infty)\right]\;$ should give us twice the Sine integral [sic] which is $\;\pi\;$ ?? $\endgroup$ – DonAntonio Oct 26 '16 at 18:35
  • $\begingroup$ By Sine integral I meant $\int_0^\infty\frac{\sin x}{x}dx=\frac{\pi}{2}$ $\endgroup$ – am301 Oct 26 '16 at 18:54
  • $\begingroup$ @am But then I don't understand: if you already know , what do you need all the things you did in your question for? The function $\;\frac{\sin x}x\;$ is clearly even, so it Principal Value is obviously $\;\pi\;$ ...! $\endgroup$ – DonAntonio Oct 26 '16 at 18:57
  • $\begingroup$ I was just trying to understand why I can't do a simple contour integral in the half plane to get the answer. Of course there are many ways to do it, the simplest being using the residue theorem on the integral of $e^{iz}/z$ over the curve I discussed $\endgroup$ – am301 Oct 26 '16 at 19:09
  • $\begingroup$ "the integral over that curve always goes to zero": why ? $\endgroup$ – Yves Daoust Oct 27 '16 at 7:49

No, the assumption of the contour integral is ok, but the integral over a half-circle in the upper (or the lower) half plane does not go to zero, because $\sin(z) = (e^{iz}-e^{-iz})/2i$ and one of the two exponentials is unbounded. That's probably why people usually takes only one of them to do this calculation.


Since $f(z)=\frac{\sin(z)}z$ is entire, then Cauchy's Integral Theorem guarantees that for any closed rectifiable path $C$ in the complex plane,

$$\begin{align} \oint_C \frac{\sin(z)}z\,dz&=0 \end{align}$$

In particular, if $C$ is the curve that is comprised of (1) the line segment from $-R$ to $R$, for any $R>0$, and (2) the semicircular arc $z=Re^{i\phi}$, for $0\le \phi\le\pi$, we have

$$\begin{align} \oint_C \frac{\sin(z)}{z}\,dz&=\int_{-R}^R \frac{\sin(x)}x\,dx+\int_0^\pi \frac{\sin(Re^{i\phi})}{Re^{i\phi}}\,iRe^{i\phi}\,d\phi\\\\ &=\int_{-R}^R \frac{\sin(x)}x\,dx+i\int_0^\pi \sin(Re^{i\phi})\,d\phi \tag 1\\\\ &=0 \tag 2 \end{align}$$

Although we can let $R\to \infty$ in $(1)$, this process does not provide a way to evaluate either of the integrals on the right-hand side of $(1)$. We can deduce, however, that

$$\int_{-\infty}^\infty \frac{\sin(x)}x\,dx=-\lim_{R\to \infty}\int_0^\pi \sinh(iRe^{i\phi})\,d\phi$$


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