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 $$\Gamma(z)=\frac{1}{z} \int_{0}^{\infty}e^{-t}t^{z}dt=z^{z}\int_{0}^{\infty}e^{-z(v-\log{v})}dv$$
Apply Watson's lemma to Gamma function
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numerical-methods
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0and those conditions are... – 2012-09-09
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0I believe you need a more general method, like [Laplace's method](http://en.wikipedia.org/wiki/Laplace%27s_method). – 2012-09-09
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0Well, in complex plane, Laplace's method does not work. Even if it works, it won't generate complete asymptotic series. – 2012-09-09
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0conditions 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. – 2012-09-09