Proving the gamma function integral converges.

Question

I am trying to prove that the integral $\int_0^{\infty}x^{t-1}e^{-x}dx$ converges for all $t>0$.

My first thought is to bound the function above by some other function which converges, however I could not think up of a function which does so.

I know the function has a local maximum on $[0,\infty)$ at $x = t-1$, so I thought about separating the integral about that point and then bounding each one separately. However I could not, again, think of a function which bounds the curve for any $t$.

My final approach was to use the power series of $e$ and then integrate power by power.

Are any of my approaches correct or am I completely on the wrong path?

Answer 1

According to subadditivity of definite integrals, we split the integral into two as follows $$\int_0^{\infty}x^{t-1}e^{-x}dx = \underbrace{\int_0^1 x^{t-1}e^{-x}dx}_{(1)} + \underbrace{\int_1^\infty x^{t-1}e^{-x}dx}_{(2)}$$

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• For the first integral, we note that since $t > 0$, we can find a segment $[a, A]$ such that $t \in [a,A]$. Then, for $t \in [a,A]$ and $x \in [0, +\infty)$ we have $$x^{t-1}e^{-x} = \frac{1}{x^{1-t}e^{x}} \le \frac{1}{x^{1-A}\cdot 1}$$ By comparison test, we observe that the integral $(1)$ converges.

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• For the second integral we note that $$\lim_{x\to\infty}\frac{x^{t-1}e^{-x}}{\frac{1}{x^2}} = \lim_{x\to\infty}x^{t+1}e^{-x} = 0$$ and again by comparison test, the integral $(2)$ converges.

Answer 2

Indeed, split it up into an integral away from $0$ and an integral near $0$, say $(1,\infty)$ and $[0,1]$. First consider $(1,\infty)$. In these situations it is common to "borrow" from the dominating term as follows. Write $x^{t-1} e^{-x}$ as $$ \frac{x^{t-1}}{e^{\epsilon x}} e^{-(1-\epsilon)x} $$ for $\epsilon >0$ very small ($<1$ is enough). Next you can show using limits that the term on the left is bounded as $x \to \infty$. Thus an integral away from $0$ will converge because $\int_1^\infty e^{-(1-\epsilon)x} dx$ does. For the part close to $0$, the $e^{-x}$ term is bounded and the integral converges because $\int_0^1 x^{t-1} dx$ converges.

Answer 3

First note that $$\int_0^1x^{t-1}dx$$ converges and then use that $|x^{t-1}e^{-x}|\leq x^{t-1}$ for $x\in[0,1]$. Then see that $$\int_1^{\infty}x^{-2}dx$$ converges and use that $$\frac{x^{t-1}e^{-x}}{x^{-2}}=\frac{x^{t+1}}{e^x}\rightarrow 0,x\rightarrow \infty.$$