integrate around non-isolated singularity

hbghlyj

$$f(z)=\csc\frac1z$$

1. FunctionSingularities[1/Sin[1/z], z]

Copy the Code

$\{0\}\cup\Set{\frac1{n\pi}|n\inZ}$
So 0 is a non-isolated singularity.
But $f$ is integrable on the unit circle
$$\int_{\abs z=1}f(z)\rmd z=-1\cdot2\pi i\operatorname*{Res}_{z=\infty}f(z)=\frac\pi3i$$
Here $-1$ is the winding number of unit circle around $\infty$.

1. Chop[NIntegrate[1/Sin[1/z], {z, 1, I, -1, -I, 1}] - Pi/3 I] == 0

Copy the Code

Out:        True

Czhang271828

Absolutely true, you can regard $\{z\mid |z|>1\}$ as the interior part of $\partial \mathbb D$.

hbghlyj

Czhang271828 发表于 2023-6-5 13:21
Absolutely true, you can regard $\{z\mid |z|>1\}$ as the interior part of $\partial \mathbb D$.

$\gamma$ is the (triangle) contour $i\longrightarrow-i\longrightarrow1\longrightarrow i$
How to prove
$$\int_\gamma\frac{1}{\sin \left(\frac{1}{z}\right)}=\frac{i\pi }{6}$$

1. NIntegrate[1/Sin[1/z], {z, I, -I, 1, I}]

Copy the Code

hbghlyj

Sum of residues at $z=\frac1{\pi n},n=1,2,\dots$
$$\sum _{n=1}^{\infty }\frac1{\frac d{dz}\sin(1/z)|_{z=1/\pi n}}=\sum _{n=1}^{\infty }\frac1{-\frac{\cos(1/z)}{z^2}|_{z=1/\pi n}}=\sum _{n=1}^{\infty } \frac{(-1)^{n+1}}{\pi ^2 n^2}=\frac{1}{12}$$
Then
$$\int_\gamma\frac{1}{\sin \left(\frac{1}{z}\right)}\rmd z=2\pi i\cdot\frac{1}{12}=\frac{i\pi }{6}$$
Is this correct
But the function $f(z)=\frac{1}{\sin \left(\frac{1}{z}\right)}$ has non-isolated singularity at 0