# Prove the next integral converges for any $x>0$: $\int_0^{\infty}t^{x-1}e^{-t}dt$

Prove the next integral converges for any $x>0$: $$\int_0^{\infty}t^{x-1}e^{-t}dt$$ I can't find a proper way to prove that, But what i did so far was:

1. integration by parts: $$\frac{e^{-t}t^x}{x}|_0^{\infty}+ \frac1x\int_0^{\infty}t^{x}e^{-t}dt$$ (My idea behind this part was maybe finding an Indefinite integral i can use)
2. $\frac{e^{-t}t^x}{x}|_0^{\infty}=0$ so i'm left only with: $\frac1x\int_0^{\infty}t^{x}e^{-t}dt$
3. The interesting thing about the above, is that it's some-kind of recursion where $x\int(x)=\int(x+1)$
4. Although it's interesting it doesn't help in my converging proving, Any ideas?
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Property #3 is because this is the Gamma function. See en.wikipedia.org/wiki/Gamma_function. It's a generalization to the factorial. – Suugaku Apr 12 '13 at 22:06
Actually i found that out after playing around with this function. – StationaryTraveller Apr 12 '13 at 22:09

There is no need to do an integration by parts, though we mention later how it can be useful.

Break up the integral into (say) the integral from $0$ to $1$ and the integral from $1$ to $\infty$.

For the integral from $0$ to $1$, note that the integrand is $\le et^{x-1}$ over the interval. Then use Comparison. There is no trouble at all if $x-1\ge 0$. For $-1\lt x-1\lt 0$, use the known fact that $\int \frac{dt}{t^p}$ converges if $p\lt 1$. This is a basic fact that probably has already been proved in your course. If it hasn't, use the Integral Test.

For the integral from $1$ to $\infty$, rewrite $e^{-t}$ as $e^{-t/2}e^{-t/2}$, and use the fact that $\lim_{t\to\infty}t^{x-1}e^{-t/2}=0$ (you only need the fact the expression is bounded). Then we do a comparison with $\int_1^\infty e^{-t/2}$, which clearly converges.

Integration by parts can be used to bypass the need to separate the integral into two parts, since $t^xe^{-t}$ is "well-behaved" at $0$.

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I still can't see how rewriting and using the limits help, Cause i get the next thing: $\lim_{x\to\infty}{\frac{t^{x-1}e^{-t}}{e^{-\frac t2}}}$=$\lim_{x\to\infty}{\frac{t^{x-1}e^\frac{t}{2}}{e^\frac{t}{2}e^\frac{t}{2}‌​}}=0$. And does $0$ mean anything when it comes to improper integral behavior? – StationaryTraveller Apr 12 '13 at 22:40
You get $\int_1^\infty \frac{t^{x-1}}{e^{t/2}}e^{-t/2}\,dt$. The term $\frac{t^{x-1}}{e^{t/2}}$ approaches $0$, so in particular is bounded by some number $B$. And $\int_1^\infty Be^{-t/2}\,dt$ certainly converges. – André Nicolas Apr 12 '13 at 22:50

Simply we have $$t^{x-1}e^{-t}\sim_0 t^{x-1}$$ and $$\int_0^1t^{x-1}dt \,\,\,\text{is convergent}\iff x>0$$ moreover $$t^{x-1}e^{-t}=_\infty o\left(\frac{1}{t^2}\right)$$ so $$\text{the integral}\,\,\,\int_1^\infty t^{x-1}e^{-t}dt \,\,\,\ \text{is convergent}\,\,\forall x\in\mathbb{R}$$ hence we have $$\text{the integral}\,\,\,\int_0^\infty t^{x-1}e^{-t}dt \,\,\,\ \text{is convergent if}\,\, x>0$$

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