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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
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Actually i found that out after playing around with this function. –  StationaryTraveller Apr 12 '13 at 22:09

2 Answers 2

up vote 1 down vote accepted

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
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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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