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I was wondering is there anyone know how to show that the gamma function $\Gamma(z)$ satisfies the conditions of Watson's Lemma, where z is on the right half plane. After I changed variable t=xv $$\Gamma(z)=\frac{1}{z} \int_{0}^{\infty}e^{-t}t^{z}dt=z^{z}\int_{0}^{\infty}e^{-z(v-\log{v})}dv$$

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and those conditions are... – Jonathan Sep 9 '12 at 5:36
I believe you need a more general method, like Laplace's method. – Fabian Sep 9 '12 at 5:46
Well, in complex plane, Laplace's method does not work. Even if it works, it won't generate complete asymptotic series. – user39846 Sep 9 '12 at 22:18
conditions are when you generate the standard form that can apply Watson's lemma, $\int_{0}^{\infty}e^{-zt}f(t)dt$ you have to show f(t) is analytic from 0 to infinity, and f(t) is bounded by $Ke^{bt}$, where K,b are some positive constants. – user39846 Sep 9 '12 at 22:23

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