Understanding derivation of Gamma function using integration by parts.

Question

I am currently trying to understand the derivation of the Gamma function, namely, this step

$$ \begin{align}\int ^\infty _0 (- \exp(-y))'& y^{\alpha -1} \,\mathrm dy \\&= - \exp(-y) y^{\alpha -1} | ^ \infty _ 0 + (\alpha -1 )\int ^\infty _0 (- \exp(-y))' y^{\alpha -2} \,\mathrm dy \\&= (\alpha -1) \Gamma (\alpha -1)\end{align} $$

Namely, in the second step, how does one conclude that $ \exp(-y) y^{\alpha -1} | ^ \infty _ 0 = 0$ ?

Answer 1

Based on what you wrote, you tacitly assumed $\alpha > 1$, therefore your problem boils down to show that $$\lim_{y \to \infty} \frac{y^{\alpha - 1}}{e^y} = 0,$$ which is a classic result. There are multiple ways to prove this, one way is to notice that $$ e^{y} = 1 + y + \frac{y^2}{2!} + \cdots > \frac{y^{\lfloor \alpha \rfloor}}{\lfloor \alpha \rfloor!},$$ then apply the squeeze principle.

Answer 2

$$ 0 \le \frac{y^{\alpha-1}}{e^y} \le \frac{y^n}{e^y} \text{ where $n$ is the least integer that is $\ge \alpha-1$}. $$ Note that $n\ge0.$

If $n=0$ then it is trivial that $y^n/e^y\to0$ as $y\to\infty.$

Otherwise, L'Hopitals rule tells us that the limit of that last expression as $y\to\infty$ is the same as that of $\displaystyle \frac{ny^{n-1}}{e^y},$ if that exists. So we have a proof by mathematical induction that the limit is $0.$

L'Hopital's rule often doesn't give much insight even though it gets to the bottom line very efficiently. I think of $y^n/e^y$ like this: suppose $y$ is incremented from $1\text{ million}$ to $1\text{ million}+1.$ Then $e^y$ gets multiplied by $e$, which is quite a bit more than $2,$ thereby cutting the fraction down to less than half what it was. But the numberator $y^n$ increases by only a tiny fraction of what it was: $$ (1\text{ million}+1)^n = (1\text{ million})^n + n(1\text{ million})^{n-1} + \text{lower-degree terms} $$ If $n$ is small compared to $1\text{ million}$ then those later terms can be seen to be small compared to $y^n.$

Therefore $y^n/e^y$ becomes less than half what it was each time $y$ is incremented by $1.$ And if $n$ is not small compared to a million, just use a trillion, etc.