Find $\int_{\vert z \vert = 1} {dz \over z(e^{1/ z}-1)}$

Question

${1 \over z(e^{1/ z}-1)}$ has a succession of isolated singularities $({1 \over 2k \pi} )_{k \in N}$ which converges to 0, then 0 is a non-isolated singularity.

Edit: There is no residue on 0, what to do instead?

Answer 1

METHODOLOGY $1$:

Exploiting Cauchy's Integral Theorem, the value of the integral is unaltered by integrating over $|z|=R>1$. That is to say,

$$\begin{align} \oint_{|z|=1}\frac{1}{z(e^{1/z}-1)}\,dz&=\oint_{|z|=R>1}\frac{1}{z(e^{1/z}-1)}\,dz\\\\ &=\int_0^{2\pi}\frac{1}{Re^{i\phi}\left(e^{\frac{1}{Re^{i\phi}}}-1\right)}\,iRe^{i\phi}\,d\phi\\\\ &=i\int_0^{2\pi} \frac{1}{\frac{1}{Re^{i\phi}}\left(1+\frac{1}{2Re^{i\phi}}+O\left(\frac{1}{R^2e^{i2\phi}}\right)\right)}\,d\phi\\\\ &=i\int_0^{2\pi}Re^{i\phi}\left(1-\frac{1}{2Re^{i\phi}}+O\left(\frac{1}{R^2e^{i2\phi}}\right)\right)\,d\phi\\\\ &\to -i\pi\,\,\text{as}\,\,R\to \infty \end{align}$$

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METHODOLOGY $2$:

Alternatively, we can use the Residue at Infinity and write

$$\begin{align} \oint_{|z|=1}\frac{1}{z(e^{1/z}-1)}\,dz&=-2\pi i \text{Res}\left(\frac{1}{z(e^{1/z}-1)},z=\infty\right)\\\\ &=-2\pi i \text{Res}\left(-\frac{1}{z^2}\frac{1}{z^{-1}(e^z-1)},z=0 \right)\\\\ &=2\pi i\left(-\frac12\right)\\\\ &=-i\pi \end{align}$$

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NOTES:

We can use Taylor's Theorem to write

$$\begin{align} z(e^{1/z}-1)&=z\left(1+\frac{1}{z}+\frac{1}{2z^2}+O\left(\frac{1}{z^3}\right)\right)-z\\\\ &=\left(1+\frac{1}{2z}+O\left(\frac{1}{z^2}\right)\right) \end{align}$$

Letting $z=Re^{i\phi}$, we obtain

$$\begin{align} \frac{1}{z(e^{1/z}-1)}&=\frac{1}{\left(1+\frac{1}{2Re^{i\phi}}+O\left(\frac{1}{R^2e^{i2\phi}}\right)\right)}\\\\ &=\left(1+\frac{1}{2Re^{i\phi}}+O\left(\frac{1}{R^2e^{i2\phi}}\right)\right)^{-1}\\\\ &=\left(1-\frac{1}{2Re^{i\phi}}+O\left(\frac{1}{R^2e^{i2\phi}}\right)\right) \end{align}$$

Then, with $dz=iRe^{i\phi}$, we arrive at the final lines.

Answer 2

Substituting $z\mapsto\frac1z$, $\frac{\mathrm{d}z}{z}\mapsto-\frac{\mathrm{d}z}{z}$ but the direction of the contour is reversed. Therefore, $$ \begin{align} \int_{|z|=1}\frac{\mathrm{d}z}{z\left(e^{1/z}-1\right)} &=\int_{|z|=1}\frac{\mathrm{d}z}{z\left(e^z-1\right)}\\ &=-\pi i \end{align} $$ since $$ \begin{align} \frac1{z\left(e^z-1\right)} &=\frac1{z^2\left(1+\frac z2+O\!\left(z^2\right)\right)}\\ &=\frac{1-\frac z2+O\!\left(z^2\right)}{z^2}\\ &=\frac1{z^2}\color{#C00000}{-\frac1{2z}}+O(1) \end{align} $$