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I have something to ask regarding convergence of gamma function. I have done the proof as below. Please tell me if it is correct.

$\int_0^\infty e^{-x}x^{n-1}dx$ is convergent for $n>0$

Proof: For $n\in(0,1],~\int_0^\infty e^{-x}x^{n-1}dx$ is convergent since $\int_0^\infty e^{-x}x^{n-1}dx=\int_0^1 e^{-x}x^{n-1}dx+\int_1^\infty e^{-x}x^{n-1}dx\le\int_0^1 x^{n-1}dx+\int_1^\infty e^{-x}dx.$

On the other hand, $\int e^{-x}x^{(n+1)-1}dx=-x^n.e^{-x}+n\int e^{-x}x^{n-1}dx\cdots(1)$

Since, $\lim\limits_{x\to\infty}x^n.e^{-x}=0,$ by successive application of $(1)$ $n$ can be put in $(0,1],$ whence the result follows.

TonyK
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Jave
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1 Answers1

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The proof in the OP fails when $n>1$ since for $x>1$, $x^{n-1}>1$ or $n>1$.

But we can assert that $e^x\ge \frac{x^{\lfloor n\rfloor+1}}{(\lfloor n\rfloor +1)!}$ for $x \ge 1$. So, for $L\ge1$

$$\begin{align} \left|\int_1^L e^{-x}x^{n-1}\,dx\right|&\le \left(\lfloor n\rfloor+1\right) ! \int_1^L x^{n-\lfloor n\rfloor -2}\,dx\\\\ &=\frac{\left(\lfloor n\rfloor+1\right) !}{\left(n-\lfloor n\rfloor-1\right) }\left(L^{\left(n-\lfloor n\rfloor-1\right) }-1\right) \end{align}$$

Since $n-\lfloor n\rfloor -1<0$, $\lim_{L\to\infty}\left(\frac{\left(\lfloor n\rfloor+1\right) !}{\left(n-\lfloor n\rfloor-1\right) }\left(L^{\left(n-\lfloor n\rfloor-1\right) }-1\right) \right)=\frac{\left(\lfloor n\rfloor+1\right) !}{1-\left(n-\lfloor n\rfloor\right) }$, and the integral on the left-hand side of $(1)$ converges. And we are done!

Mark Viola
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