I have trouble to proove that the residue at inifinity of this function is zero (I found that the sum of the residues at finite is zero). I tried to expand in series at $z=0$ the function $f(1/w^2)/w^2$ but I obtain a combination of three series and I can't operate with this.

  • $\begingroup$ The residue at infinity is equal to the integral over a large circle. You have proven that the integral is zero. $\endgroup$ – Maxim Apr 16 at 18:36

You have $$\frac{1}{w^2}f(\frac{1}{w}) = \frac{1}{w^2}\frac{e^{\frac{2\pi}{w}}-1}{\frac{1}{w}(1+\frac{1}{w^2})^2} = w^3 \frac{1}{(1+w^2)^2} \left(e^{\frac{2\pi}{w}}-1\right)$$

We have $$\frac{1}{(1+u)^2}= -\frac{d}{du}\frac{1}{1+u} = -\frac{d}{du}\sum_{n=0}^\infty (-u)^n = \sum_{n=0}^\infty (n+1)(-1)^n u^n$$ so $$ \frac{1}{(1+w^2)^2} = \sum_{n=0}^\infty (n+1)(-1)^n w^{2n}$$ We have also $$ e^{\frac{2\pi}{w}}-1 = \sum_{m=1}^\infty \frac{(2\pi)^m}{m!} w^{-m}$$ So in total \begin{align} \frac{1}{w^2}f(\frac{1}{w}) =& \sum_{n=0}^\infty \sum_{m=1}^\infty (n+1)(-1)^n \frac{(2\pi)^m}{m!} w^{2n-m+3} = \newline =& \sum_{n=0}^\infty \sum_{N=-\infty}^{2n+2} (n+1)(-1)^n \frac{(2\pi)^{2n+3-N}}{(2n+3-N)!} w^N = \newline =& \sum_{N=-\infty}^\infty \sum_{n=\max (0,\frac{N-2}{2}) }^\infty (n+1)(-1)^n \frac{(2\pi)^{2n+3-N}}{(2n+3-N)!} w^N \end{align} and the residuum at $w=0$ is \begin{align} Res_{w=0} \frac{1}{w^2}f(\frac{1}{w}) =& \sum_{n=0}^\infty (n+1)(-1)^n \frac{(2\pi)^{2n+4}}{(2n+4)!} = \newline =& \sum_{n=0}^\infty \left(\frac12 \frac{1}{(2n+3)!} - \frac{1}{(2n+4)!}\right) (-1)^n(2\pi)^{2n+4} = \newline =& \pi\big(-\sin(2\pi)+2\pi\big) - \big(\cos(2\pi) -1 + 2\pi^2\big) = 0\end{align}

Another method is to use the fact that for $f(z)$: $$ Res_{z=\infty} f(z) = - \big(Res_{z=0} f(z) + Res_{z=i} f(z)+ Res_{z=-i} f(z)\big) $$


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