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I am trying to compute the following integral: $$ F_{k}(\xi) = \int_{\mathbb{S}^1} e^{-\xi \frac{z+1}{z-1}} \dfrac{z^k}{z-1}dz$$ where $k\in \mathbb{Z}$ and $\xi \in \mathbb{R}$.

I only need to show that $$F_k(\xi) = 0 \,\,\,\text{for all}\,\,\, \xi > 0,\,\, k \geq 0.$$ $$\text{ For all }\, k < 0 \text{ there exists } \xi_k > 0 \text{ such that } F_k(\xi_k) \neq 0$$

Any ideas/hints?

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  • $\begingroup$ first you can try to simplify the fractions multiplying by the conjugate of the denominator. $\endgroup$ – Masacroso Nov 25 '17 at 11:49
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By enforcing the substitution $z=\frac{t+1}{t-1}$ (notice that the function $g(z)=\frac{z+1}{z-1}$ is an involution bringing $S^1$ into the imaginary line) we get $$ F_k(\xi) = \int_{-i\infty}^{+i\infty}e^{-\xi t}\frac{(t+1)^k}{(t-1)^{k+1}}\,dt$$ and if $k\geq 0$ the meromorphic function $e^{-\xi t}\frac{(t+1)^k}{(t-1)^{k+1}}\,dt$ has a unique singularity, i.e. a pole at $t=1$. If $\xi>0$, by moving the integration line towards the right we get that $F_k(\xi)$ only depends on the value of the residue of $e^{-\xi t}\frac{(t+1)^k}{(t-1)^{k+1}}$ at $t=1$. If $\xi>0$ but $k<0$, the only singularity of the integrand function is at $t=-1$, hence be moving the integration line towards the right we get that $F_k(\xi)$ simply equals zero. If $\xi<0$ we can perform the same analysis, but the integration line has to be moved towards the left, so we get $F_k(\xi)=0$ if $k\geq 0$ and $$ F_k(\xi)=2\pi i\operatorname*{Res}_{t=-1}\left(e^{-\xi t}\frac{(t+1)^k}{(t-1)^{k+1}}\right) $$ if $k<0$.

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  • $\begingroup$ I don't know if there is any sign error, but your result would contradict the result.. But I'm starting to convince myself it was them who made a mistake! $\endgroup$ – Ignatius Nov 27 '17 at 19:39
  • $\begingroup$ @Ignatius: I noticed there was some incongruence, but I think that the substitutions and signs in my answer are correct. Please let me know if you spot a mistake somewhere, I will fix it. $\endgroup$ – Jack D'Aurizio Nov 27 '17 at 19:48
  • $\begingroup$ You were right, Jack! Thank you for your help. $\endgroup$ – Ignatius Dec 2 '17 at 16:26

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