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
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.