# Does $e^{az}/\Gamma(z)dz$ have a nice indefinite integral? Definite integral?

Now obviously we can just expand $\frac{1}{\Gamma(z)}$ into a power series and then integrate with $e^{az}$. But the coefficients of $\frac{1}{\Gamma(z)}$ are way too ugly.

We can represent $\frac{1}{\Gamma(z)}$ using the contour integral due to Hankel $$\frac{1}{\Gamma(z)} = \frac{i}{2\pi} \oint_C (-t)^{-z} e^{-t}dt$$ where $C$ is a circle around 0 that comes from and goes back to positive infinity. But even if we exchange the order of integration, we don't really get anything nice, not mentioning the question of whether the exchange is warranted.

An indefinite integral might not be necessary. I'm at the moment only concerned with the integral from $c-i\infty$ to $c+i\infty$ for some real $c$, or in other words a vertical line on the complex plane.

So hopefully someone can point me the right way, or outline a good solution. Thanks!

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The integral over the contour you specified is divergent. –  Alexandre Eremenko Oct 22 '12 at 22:01