Is there a way to perform a contour integral around zero of something like f(z)/z e^(1/z), where f is holomorphic at 0? If you expand you get something like:(adsbygoogle = window.adsbygoogle || []).push({});

[tex] \frac{1}{z} \left( f(0) + z f'(0) + \frac{1}{2!} z^2 f''(0) + ... \right) \left( 1 + \frac{1}{z} + \frac{1}{2!} \frac{1}{z^2} + ... \right) [/tex]

[tex] = \frac{1}{z} \left( f(0) + f'(0) + \frac{1}{2!^2} f''(0) + ... \right) + ...[/tex]

So that the result is [itex]2\pi i ( f(0) + f'(0) + \frac{1}{2!^2} f''(0) + ... )[/tex]. Is this correct, and is there a simpler way to write the final answer? If not for the factorials being squared, the term inside the parentheses would just be f(1), but as it is I don't see what I can do with it in general. By the way, the f I'm interested in has complicated poles and branch cuts away from zero, so deforming the contour probably won't help.

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# Contour integral around essential singularity

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