# Convergence of this integral [duplicate]

Possible Duplicate:
Some questions about the gamma function

My statistics text book prescribed by my school states that the integral $$\Gamma(n)=\int_{0}^{\infty}e^{-x}x^{n-1}dx$$ is convergent for $n>0$.It does not prove the statement.So can anyone please help me prove it?Thanks again!

• Thank you.I flagged this question as well.Looks like I need some background in analysis before I enter deeper water. – Amitabh Udayiman May 6 '12 at 15:05
• I think that this question is at a lower level than the question Some questions about the gamma function. – Américo Tavares May 6 '12 at 17:33

I assume that $n$ is a real number. Split the gamma improper integral $$\Gamma(n)=\int_{0}^{\infty}e^{-x}x^{n-1}dx\tag{0}$$ into $I_1+I_2$, where $$I_1=\int_{0}^{1}e^{-x}x^{n-1}dx\tag{1}$$ and $$I_2=\int_{1}^{\infty}e^{-x}x^{n-1}dx\tag{2}$$
1. To prove that the integral $I_2$ is always convergent use the fact that for any real number $\alpha$ the integral $$\int_{1}^{\infty }e^{-x}x^{\alpha }dx\tag{3}$$ is convergent, by the limit comparison test $$\lim_{x\rightarrow \infty }\frac{e^{-x}x^{\alpha }}{x^{-2}}=0\tag{4}$$ with the convergent integral $$\displaystyle\int_{1}^{\infty }\dfrac{dx}{x^{2}}\tag{5}.$$
2. As for $I_1$ consider two cases. (a) If $n\geq 1$ observe that $\lim_{x\rightarrow 0}e^{-x}x^{n-1}=0$, so $I_1$ is a proper integral. (b) If $0<n<1$, the integrand $e^{-x}x^{n-1}$ behaves like $x^{n-1}$ near $n=0$, because $e^{-x}\rightarrow 1$ as $x\rightarrow 0$. Since $$\displaystyle\int_{0}^{1}\dfrac{dx}{x^{1-n}}\tag{6}$$ is convergent if and only if $1-n<1$, i.e. $n>0$, so is $I_1$.
It follows that $\Gamma(n)=I_1+I_2$ is convergent for $n>0.$