Gamma as indefinite integral

**fobos3** · November 8th 2008, 12:46 PM

Let
$\int x^{a-1}e^{-x}dx=F(x)$

Mathematica returns
$F(x)=-\Gamma (a,x)$

We have
$-\Gamma (a,x)=-(F(\infty)-F(x))$
$F(x)=F(x)-F(\infty)$
$F(\infty)=0$

So how do you prove the last line?

**Mathstud28** · November 8th 2008, 01:14 PM

Originally Posted by fobos3

Let
$\int x^{a-1}e^{-x}dx=F(x)$

Mathematica returns
$F(x)=-\Gamma (a,x)$

We have
$-\Gamma (a,x)=-(F(\infty)-F(x))$
$F(x)=F(x)-F(\infty)$
$F(\infty)=0$

So how do you prove the last line?

Assuming I understand what you are saying correctly, since this integral is uniformly convergent (I leave that to you to prove) then we have that

$\begin{aligned}F\left(\infty\right)&=\lim_{x\to\in fty}\int_0^{\infty}x^{a-1}e^{-x}dx\\ &=\int_0^{\infty}\lim_{x\to\infty}x^{a-1}e^{-x}dx\\ &=\int_0^{\infty}0dx\\ &=0\quad\blacksquare \end{aligned}$

For proof of the limit we see that our limit is infinity over infinity, an undeterminate form for which L'hopital's may be applied. But we see that this yields another indeterminate form. So we may keep taking limits until

$\lim_{x\to\infty}\frac{x^{a-1}}{e^x}=\lim_{x\to\infty}\frac{(a-1)x^{a-2}}{e^x}=\underbrace{\cdots}_{a-3\text{ number of times}}=\lim_{x\to\infty}\frac{(a-1)!}{e^x}=0$

**Laurent** · November 9th 2008, 05:21 AM

Originally Posted by Mathstud28

Assuming I understand what you are saying correctly, since this integral is uniformly convergent (I leave that to you to prove) then we have that

$\begin{aligned}F\left(\infty\right)&=\lim_{x\to\in fty}\int_0^{\infty}x^{a-1}e^{-x}dx\\ &=\int_0^{\infty}\lim_{x\to\infty}x^{a-1}e^{-x}dx\\ &=\int_0^{\infty}0dx\\ &=0\quad\blacksquare \end{aligned}$

For proof of the limit we see that our limit is infinity over infinity, an undeterminate form for which L'hopital's may be applied. But we see that this yields another indeterminate form. So we may keep taking limits until

$\lim_{x\to\infty}\frac{x^{a-1}}{e^x}=\lim_{x\to\infty}\frac{(a-1)x^{a-2}}{e^x}=\underbrace{\cdots}_{a-3\text{ number of times}}=\lim_{x\to\infty}\frac{(a-1)!}{e^x}=0$

I understand nothing of the initial post, so your answer is very meritorious. However, it may have inheritated some confusion from the way the question was asked. Plainly written, $F(x)=\int_{x_0}^x t^{a-1} e^{-t} dt$ for some $x_0>0$ . For instance, $x_0=0$ is fine if $a>0$ . This is a function of $x$ where $x$ is acting only as a limit of the integration interval. Hence $\lim_{x\to\infty} F(x)=\int_{x_0}^\infty t^{a-1} e^{-t}dt$ , which is $\Gamma(a)$ if we chose $x_0=0$ . No limit under the $\int$ symbol involved. The formula $\lim_{x\to\infty}\int_0^{\infty}x^{a-1}e^{-x}dx$ makes by the way no sense at all.

**Laurent** · November 9th 2008, 05:31 AM

Originally Posted by fobos3

Let
$\int x^{a-1}e^{-x}dx=F(x)$

Mathematica returns
$F(x)=-\Gamma (a,x)$

We have
$-\Gamma (a,x)=-(F(\infty)-F(x))$
$F(x)=F(x)-F(\infty)$
$F(\infty)=0$

So how do you prove the last line?

I think I finally understood it. From what you write, I guess that $\Gamma(a,x)=-\int_x^\infty t^{a-1}e^{-t}dt$ . This is a primitive for $x^{a-1}e^{-x}$ which tends to 0 at $+\infty$ . The reason why it tends to 0 is because the integral converges on $[1,+\infty)$ : $\Gamma(a,x)=-\int_1^\infty t^{a-1}e^{-t}dt + \int_1^x t^{a-1}e^{-t}dt$ and the second integral converges to the first one as $x$ tends to $+\infty$ , by definition of the integral on $[1,+\infty)$ . Why is the integral convergent? For instance because, for $t$ large enough, $t^{a-1}e^{-t}\leq \frac{C}{t^2}$ for some positive $C$ , as results from $t^2 t^{a-1}e^{-t}\to_{t\to\infty} 0$ (which is obtained using MathStud28's argument for instance, if you like L'Hospital's rule).

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