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Let $a = \sqrt{\pi}e^{\pi i/4} = (1+i)\sqrt{\pi/2} $

Consider the function $$ f(z) = \frac{e^{-z^2}}{1+e^{-2az}} $$ I have already shown that $f(z) - f(z+a) = e^{-z^2}$ if that helps at all.

The rest of the question states:

i) For some (large) R > 0, locate the pole(s) of f inside the parallelogram with corners -R, R, R+a, -R+a, and calculate the residue.

I know there will be a pole when $e^{-2az} = -1$ and solving for z I get $z=\frac{\pi i(2n+1)}{2a}$ for n is any integer. How do I go about finding this residue and what is the purpose of giving me the parallelogram (if there is one)?

ii) (haven't look at this one much) Estimate the integral along the short sides of the parallelogram, and using residue theorem, deduce that $$\int^\infty_\infty e^{-x^2}dx = \sqrt{\pi}$$ I assume its simple once I get the first part, but any help is appreciated. I am a little stuck at the moment.

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To find the residue at that pole, take the derivative of the denomiator, evaluated at the pole. If the integrand is $a(z)/b(z)$, the the residue at the pole $z_k$ is $a(z_k)/b'(z_k)$. Then the residue at the above pole is

$$\frac{e^{\pi^2 (2 n+1)^2/(4 a^2)}}{2 a}$$

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  • $\begingroup$ Is that it? Wouldn't summing over all integers just give me zero than? and does this relate to part ii) at all? I don't know but this just doesnt seem satisfying $\endgroup$ – Pauli Mar 20 '13 at 20:30
  • $\begingroup$ I was just trying to get across the residue for a single pole. Your parallelogram to me was unclear, so I held off from going further than this. Also, summing all the residues over $n$ looks like a nonconvergent series. I have a much simpler way to evaluate the gaussian using Cauchy's theorem if you are interested. $\endgroup$ – Ron Gordon Mar 20 '13 at 20:39
  • $\begingroup$ It wouldn't hurt to show it $\endgroup$ – Pauli Mar 20 '13 at 21:53
  • $\begingroup$ @RonGordon are you going to show you simpler method? $\endgroup$ – dustin Mar 11 '15 at 5:29

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