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In Stein's Complex Analysis notes, the following exampleis given.

They then proceed to calculate the integral over the small semicircle.

My question is, why is it necessary to dodge the origin? Afterall, the singularity at $z=0$ is removable?

Spine Feast
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3 Answers3

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The singularity is removable for the function you want to integrate but not for their $f(z).$

Igor Rivin
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  • Could you elaborate? – Spine Feast Dec 05 '13 at 17:44
  • $f(z)$ has a pole at the origin, since the imaginary part is $\sin(z)/z^2$ – Igor Rivin Dec 05 '13 at 17:46
  • By $f(z)$ you mean $\frac{1-e^{iz}}{z^2}$? But then, if one replaces the original integrand by $\frac{1}{2} \frac{\sin^2 (x/2)}{x^2}$, and considers $f(z)= \frac{1}{2} \frac{\sin^2 z}{z^2}$, it is no longer singular, or? – Spine Feast Dec 05 '13 at 18:14
  • I don't know what you are talking about. The function DOES have an actual pole at $0.$ It is conceivable that there is some way to do the contour integral without cutting out zero, but that's not the way Stein et al are doing it... – Igor Rivin Dec 05 '13 at 18:16
  • I don't see why $f(z)= \frac{1}{2} \frac{\sin^2 (z/2)}{z^2}$ has a pole at zero. Isn't it just a removable singularity? – Spine Feast Dec 05 '13 at 18:22
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    It does not have a pole at zero, but that's not the function we are talking about. – Igor Rivin Dec 05 '13 at 18:24
  • Would it not work if we consider that function instead then? – Spine Feast Dec 05 '13 at 18:26
  • Be my guest, if you can compute the integral with it. You had asked why they chose the contour, and it is because they have to with the function they chose. – Igor Rivin Dec 05 '13 at 18:30
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The motivation for the question seems to be: Why does the author integrate $f(z)=(1-e^{iz})/z^2$ over the complex plane, rather than the original integrand $g(z)=(1-\cos z)/z^2$ ? You're right that $g$ is an entire function, and it could be integrated over a semicircle without having to dodge the origin.

However, the integral of $g$ over the arc $\gamma_R^+$ doesn't obey a nice bound. While integrating $f$, the author needs the inequality $$ \left|\frac{1-e^{iz}}{z^2}\right|\leq\frac{2}{|z|^2}, $$ which is justified by the fact that $|e^{iz}|\leq 1$ for $z$ in the upper half-plane. The same can't be said for $\cos z=(e^{iz}+e^{-iz})/2$; instead, the $e^{-iz}$ term is large in the upper half-plane. That's why $g$ is less convenient than $f$.

Chris Culter
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We can evaluate the integral using Feynman's trick:

$$ I = \int_0^\infty \frac{1 - \cos(x)}{x^2}dx = I(1) $$

$$ I(a)= \int_0^\infty \frac{1 - \cos(ax)}{x^2}dx $$

$$ \frac{dI(a)}{da}= \int_0^\infty \frac{\partial}{\partial a} \Bigg(\frac{1 - \cos(ax)}{x^2}\Bigg)dx = \int_0^\infty \frac{\sin(ax)}{x}dx = \frac{\pi}{2}$$

$$ I(a) = \frac{\pi}{2}a+C_1 $$ $$ I(0) = 0 = 0+C_1 \to C_1 = 0 $$ $$ I = I(1) = \frac{\pi}{2} $$

Marc Navarro
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