Can Euler's Reflection Formula Simplify the POTW #178 Gamma Function Problem?

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Here is this week's POTW:

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Show that

$$\frac{1}{\Gamma(s)} = -\frac{1}{2\pi i} \int_\gamma (-z)^{-s}e^{-z}\, dz,$$

where $\gamma$ is a contour which is the union of a line from $+\infty$ to $\epsilon$ (where $\epsilon > 0$), an $\epsilon$-circle about the origin, and a line from $\epsilon$ to $+\infty$.

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Note: If you like, you may use Euler's reflection formula.

Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

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No one answered this week's problem. You can read my solution below.

Spoiler

For $m = 1,2,3,\ldots$, the integrals $\int_{\gamma, |z| \le m} (-z)^{s-1}e^{-z}\, dz$ are entire functions of $s$, and converge uniformly on compact disks, so the limit function

$$\int_{\gamma} (-z)^{s-1}e^{-z}\, dz$$

is entire. This integral does not depend on $\epsilon$, by Cauchy's theorem. Now the integral along the arc of $\gamma$ is negligible as $\epsilon \to 0$. In the upper half of $\gamma$, $\arg(-z) = -\pi$, and in the lower half, $\arg(-z) = \pi$. So the contribution of the rays of $\gamma$ is $$e^{(s-1)\pi i} \int_\epsilon^\infty x^{s-1}e^{-x}\, dx + e^{-(s-1)\pi i} \int_\epsilon^\infty x^{s-1}e^{-x}\, dx = 2i\sin[\pi(s-1)]\int_\epsilon^\infty x^{s-1}e^{-x}\, dx = -2i\sin(\pi s)\int_\epsilon^\infty x^{s-1}e^{-x}\, dx.$$ Since $\int_\epsilon x^{s-1}e^{-x}\, dx \to \Gamma(s)$ as $\epsilon \to 0$ for $\operatorname{Re}(s) > 0$, it follows that

$$\Gamma(s) = -\frac{1}{2i\sin \pi s}\int_\gamma (-z)^{s-1}e^{-z}\, dz$$

for $\operatorname{Re}(s) > 0$. But by the principle of analytic continuation, the above formula holds for all $s\in \Bbb C \setminus \Bbb Z$. Using Euler's reflection formula $\Gamma(s)\Gamma(1 - s) = \pi/\sin(\pi s)$, we deduce

$$\frac{1}{\Gamma(s)} = \frac{i}{2\pi} \int_\gamma (-z)^{-s}e^{-z}\, dz.$$