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How can I prove that

$$\displaystyle \Gamma(z)=\lim_{n \to \infty} \displaystyle \int_0^n \left( 1-\frac{t}{n}\right)^n t^{z-1}\ \text{d} t\;=\displaystyle \int_0^{\infty} e^{-t} t^{z-1}\ \text{d} t\;$$

Issue is how can I prove that the order of the limit and the integral can be changed.

I know about the dominated convergence theorem and the monotone convergence theorem, but the additional problem here is that the integration limit itself depends on n.

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3  
Integrate to infinity, and multiply the integrand by $\chi_{[0,n]}$. –  David Mitra Feb 12 '12 at 1:15
1  
shouldn't the integration be with respect to $t$? –  Stefan Smith Feb 12 '12 at 1:24
    
@user20520 Indeed, I edited my question. –  wnvl Feb 12 '12 at 1:29
    
@wnvl You're from here right? I'm Weiler. –  Pedro Tamaroff Feb 12 '12 at 3:47
    
@DavidMitra I think you've hit the nail right on the head. Maybe put it as an answer? –  Ragib Zaman Feb 12 '12 at 4:15

1 Answer 1

up vote 2 down vote accepted

Put $t = n u$ so $dt = n du $

You get

$${n^z}\int\limits_0^1 {{{\left( {1 - u} \right)}^n}{u^{z - 1}}du} $$

$$\int\limits_0^1 {{{\left( {1 - u} \right)}^n}{u^{z - 1}}du} = B\left( {z,n + 1} \right) = \frac{{\Gamma \left( {n + 1} \right)\Gamma \left( z \right)}}{{\Gamma \left( {z + n + 1} \right)}}$$

But then you have

$$\eqalign{ & \Gamma \left( {z + n + 1} \right) = \left( {z + n} \right)\Gamma \left( {z + n} \right) \cr & \Gamma \left( {z + n + 1} \right) = \left( {z + n} \right)\left( {z + n - 1} \right) \cdots \left( {z + 1} \right)z\Gamma \left( z \right) \cr} $$

$$\mathop {\lim }\limits_{n \to \infty } \int\limits_0^n {{{\left( {1 - \frac{t}{n}} \right)}^n}{t^z}\frac{{dt}}{t}} = \mathop {\lim }\limits_{n \to \infty } \frac{{n!{n^z}}}{{\left( {z + n} \right)\left( {z + n - 1} \right) \cdots \left( {z + 1} \right)z}}$$

Which is by (Gauss') definition the Gamma function.

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