# Integration By Parts of Gamma Function [duplicate]

A textbook I'm self-studying - Introduction to Mathematical Statistics by Hogg - has the following text:

T(a) = $$\int_{0}^{\infty} y^{\alpha-1}e^{-y} dy$$ [gamma function]

If $$\alpha > 1$$, an integration by parts shows that $$T(\alpha) = (\alpha - 1)\int_{0}^{\infty} y^{\alpha-2}e^{-y} dy$$.

I'm having trouble deriving this for myself. How does an Integration by Parts lead to this result?

Thank you.

Let $dv=e^{-x}\ dx$ and $u=x^{\alpha-1}$. It thus follows that

$$v=-e^{-x},\quad du=(\alpha-1)x^{\alpha-2}\ dx$$

Thus,

\begin{align}\int x^{\alpha-1}e^{-x}\ dx&=\int u\ dv\\&=uv-\int v\ du\\&=-x^{\alpha-1}e^{-x}+(\alpha-1)\int x^{\alpha-2}e^{-x}\ dx\end{align}

And as the bounds go to $[0,\infty)$, we find that $uv\to0$, hence

$$\int_0^\infty x^{\alpha-1}e^{-x}\ dx=(\alpha-1)\int_0^\infty x^{\alpha-2}e^{-x}\ dx$$

• Thank you! This makes sense. For the bounds part of the proof: my understanding is that when the bounds approach zero, $-x^{\alpha - 1}$ is zero, making the $uv$ term zero. When the bounds approach $\infty$, $e^{-x}$ is zero, making the $uv$ term also zero. Is this correct? Mar 10 '17 at 19:55
• @Daniel Particularly, as $x\to+\infty$, $e^{-x}$ goes to $0$ faster than the polynomial term goes to infinity. And I assume $\alpha>1$. Mar 10 '17 at 19:56
• Thank you. How can one prove that $e^{-x}$ goes to 0 faster? Mar 10 '17 at 19:58
• @Daniel L'Hospital's rule or similar methods. Mar 10 '17 at 19:59