While i am reading one example in the book, i came across the book teaching me how to evaluate $\int_{-\infty}^{\infty}\dfrac{\sin x}{x}dx$ by using residue theorem.

However, while they say we still construct a semi circle with radius $R$ and centered at $0$, and by letting $\int_{C_R}\dfrac{e^{iz}-1}{z}dz$, we can slowly solve the question.

However, they claimed that $\int_{C_R}\dfrac{e^{iz}-1}{z}dz = 0$ for the closed semi circle $C_R(0,R)$. I have some confusion here as i thought that $z = 0$ is a point where the function is not analytic? I do understand now that if $z=0$ is a removable singularity of the function, then when you integrate, you will get $0$. However, $0$ seems to lie on the boundary of the closed semi circle. So i am not sure if $z=0$ is a singularity or not

Please enlighten

The example is from the book Bak and Newman on application of residue theorem

  • $\begingroup$ What's the residue of $\frac{e^{iz} - 1}{z}$ at $z= 0$? It ends up being zero. More specifically, the Laurent series is actually a power series. Yes, there are special cases for singularities that lie on the boundary. However, this is not a singularity and can be ignored. $\endgroup$ – Kaynex Apr 18 '17 at 12:45
  • $\begingroup$ Yes i think my professor also calculated the residue to be $0$, but then again it begs the question that the theorem says one should use residue theorem when there is a singularity, since $0$ "lies" on the boundary, can it be considered an singularity $\endgroup$ – nan Apr 18 '17 at 12:56
  • $\begingroup$ But it's not a singularity, since it's Laurent series is a power series. $\endgroup$ – Kaynex Apr 18 '17 at 18:18

They key idea is that removable singularities aren't really singularities at all. You can extend $f$ to the point $0$ such that this extension is holomorphic. When the author writes $\frac{e^{iz} -1}{z}$, they really mean the extended version of this function. To be specific, $$f(z) = \frac{e^{iz} -1}{z}$$ can be thought of as the entire (i.e holomorphic on $\mathbb{C}$) function given by the Taylor series $$\sum_{n=0}^{\infty} i^{n} \ \frac{z^{n}}{(n+1)!}.$$

You're correct in thinking that you can't take the line integral of a function $g$, when the curve you're integrating over goes through a singularity of $g$ (that was a mouthful!). However, in this case $0$ isn't really a singularity of $f$ at all - i.e there's no issue calculating the line integral $$\int_C f(z) \ dz,$$ where $C$ is a curve that passes through $0$.

The main point to take away, is that when the author writes $$\frac{e^{iz} -1}{z},$$ they really mean the 'extended', holomorphic everywhere version of this function.


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