integrate around essential singularity

hbghlyj

$f:\mathbb C\to\mathbb C$ is analytic on $\mathbb C\setminus\{0\}$ and has an essential singularity at 0, is $f$ integrable on the unit circle?
For example, $f(z)=\sin(1/z)$ is analytic on $\mathbb C\setminus\{0\}$ and has an essential singularity at 0,
$$\int_{\abs z=1}f(z)\rmd z=-1\cdot2\pi i\operatorname*{Res}_{z=\infty}f(z)=2\pi i$$
Here $-1$ is the winding number of unit circle around $\infty$.

1. Residue[Sin[1/z], {z, Infinity}]

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residue at $\infty$ is $-1$

1. Series[Sin[z^(-1)], {z, Infinity, 6}]

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WolframAlpha

$$\frac{1}{z}-\frac{1}{6 z^3}+\frac{1}{120 z^5}+O\left(\left(\frac{1}{z}\right)^7\right)$$ (Laurent series) (Converges everywhere away from the origin)

Integrate on the piecewise-linear contour $1\to-1+i\to-1-i\to1$

1. Integrate[Sin[z^(-1)],{z,1,-1+I,-1-I,1}]//FullSimplify

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Out:        $2\pi i$
Related: kuing.cjhb.site/forum.php?mod=viewthread&tid=10034

hbghlyj

Residue of an analytic function - Encyclopedia of Mathematics

The residue theorem implies the theorem on the total sum of residues: If $f(z)$ is a single-valued analytic function in the extended complex plane, except for a finite number of singular points, then the sum of all residues of f(z), including the residue at the point at infinity, is zero.

$f(z)=\sin(1/z)$ has residue $1$ at $z=0$ and $-1$ at $z=\infty$. The total sum of residue of $f(z)$ is 0.

hbghlyj

I compute the Laurent expansion at $\infty$ by substituting $1/z$ into the Taylor expansion of $\sin(z)$ $$\sin(1/z)=z^{-1} +O(z^{-3})$$ which is confirmed by WolframAlpha
WolframAlpha says the residue at $\infty$ is $-1$.
The residue is the coefficient of $(z-a)^{−1}$, but $a=\infty$ doesn't make sense.
Also, the coefficient of $z^{-1}$ is 1, not $-1$

hbghlyj

Definition

Given a holomorphic function f on an annulus $ A(0,R,\infty ) $ (centered at 0, with inner radius $ R $ and infinite outer radius), the residue at infinity of the function f can be defined in terms of the usual residue as follows:

$$ \operatorname {Res} (f,\infty )=-\operatorname {Res} \left({1 \over z^{2}}f\left({1 \over z}\right),0\right) $$

Thus, one can transfer the study of $ f(z) $ at infinity to the study of $ f(1/z) $ at the origin.

Note that $ \forall r>R $, we have

$$ \operatorname {Res} (f,\infty )={-1 \over 2\pi i}\int _{C(0,r)}f(z)\,dz $$

hbghlyj

Calculate residue at essential singularity

You're right. Another method is to note that the sum of the residues in $\widehat{\mathbb{C}}$ is $0$, the other answer shows the residue in $\infty$ is $0$, and the residue in $3$ is easily seen to be $\sin \frac{1}{3}$, from which $$\operatorname{Res}\left(\frac{\sin \frac{1}{z}}{z-3}; 0\right) = -\sin \frac{1}{3}$$ follows.

hbghlyj

1. var('z')
2. sin(1/z).residue(z==0)

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Out:        0

1. var('z')
2. (sin(1/z)/(z-3)).residue(z==0)

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Out:        ValueError: power::eval(): division by zero

hbghlyj

Find the residue of $e^{1/z}\sin(z)$ at $z=0$
Found that only $z=0$ is an essential singularity, where
$$a_{-1}=\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{(2n-1)!(2n)!}$$