I am asked to evaluate the integral

$$\int_\gamma \frac{e^{2z^2}}{z^{77}}\,{\rm d}z$$

where $\gamma$ is a circle centre $0$ traversed once anti-clockwise.

Clearly the integrand has a pole of order $77$ at $z=0$ which is enclosed within the interior of $\gamma$. So how would one evaluate this? If we write the integrand as a laurent series

$$f(z) = \frac{e^{2z^2}}{z^{77}} = \sum_{n=0}^\infty \frac{2^nz^{2n-77}}{n!}$$

the residue of this function is the coefficient of the $1/z$ term, this occurs when $n=38$. So the residue is

$${\rm Res}_{z=a} f(z) = \frac{2^{38}}{38!}$$

Now this is ridiculously small so I'm not too sure that I will find the correct answer via this method. What would be a better way to proceed?

  • $\begingroup$ ??? Why in the world would you conclude that this is wrong just because the residue appears to be small? $\endgroup$ – David C. Ullrich Apr 30 '16 at 12:51
  • $\begingroup$ @DavidC.Ullrich because it's an exam question as usually if you get silly numbers it's wrong. $\endgroup$ – user2850514 Apr 30 '16 at 12:52
  • $\begingroup$ By the standards of these things, $77$ is a ridiculously large order for a pole. $\endgroup$ – Travis Apr 30 '16 at 12:56
  • 1
    $\begingroup$ Your result is correct. I don't think that there is a simpler way to proceed. $\endgroup$ – Christian Blatter Apr 30 '16 at 17:43

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