Show for large x, Γ(x+1) = exp(-x)*x^(x+1/2)*√(2*pi)

  • #1
3,003
2

Homework Statement



I would like to figure out how my book came up with the relation "for large values of x":

[tex]\Gamma(x+1) = e^{-x}x^{x+\frac{1}{2}}\sqrt{2\pi}\qquad(1)[/tex]



Homework Equations



Definition of gamma function:

[tex]\Gamma(z) = \int_0^\infty t^{z-1}e^{-t}\,dt\qquad(2)[/tex]

The Attempt at a Solution



When I see that (1) has a multiple of [itex]\pi[/itex] in it, it makes me think of "something squared" in the exponent. I had initially thought of using a Taylor series expansion method. For example, it can be shown that (1) is also equivalent to

[tex]

\Gamma(x+1) = e^{-x}x^{x+1/2}\int_{-\sqrt{x}}^\infty e^{-\sqrt{x}t}\left (1 +\frac{t}{\sqrt{x}}\right )^x\,dt \qquad(3)

[/tex]


After some manipulation, (3) can be put in the form:

[tex]

\Gamma(x+1) = e^{-x}x^{x+1/2}\int_{-\sqrt{x}}^\infty \exp\left [-\sqrt{x}t + x\left (\frac{t}{\sqrt{x}} - \frac{t^2}{2x} + \frac{t^3}{3x^{3/2}} \dots \right )\right ] \qquad(4)[/tex]

or

[tex]\Gamma(x+1) = e^{-x}x^{x+1/2}\int_{-\sqrt{x}}^\infty \exp\left [-\sqrt{x}t + \left (\sqrt{x}t - \frac{t^2}{2} + \frac{t^3}{3x^{3/2}} \dots \right )\right ] \qquad(5)[/tex]

We can see from (5) that for large values of x, any terms with an x in the denominator can be neglected, hence

[tex]\Gamma(x+1) = e^{-x}x^{x+1/2}\int_{-\sqrt{x}}^\infty e^{ - \frac{t^2}{2}} \qquad(6)
[/tex]


But (6) seems to be a dead end for me. I tried breaking it into two integrals, one running from -√x to 0 and another from 0 to ∞, the latter of which is the gaussian integral. However, the forst part is unintegrable.

Any thoughts on another approach or a continuation of this one?

Thanks :smile:
 

Answers and Replies

  • #2
fzero
Science Advisor
Homework Helper
Gold Member
3,119
289
Use

[tex]
\lim_{x-\rightarrow \infty} \int_{-\sqrt{x}}^\infty e^{ - \frac{t^2}{2}}dt = \int_{-\infty}^\infty e^{ - \frac{t^2}{2}}dt .
[/tex]

You could find the error terms by expanding in a series using the fundamental theorem of calculus.
 
  • #3
3,003
2
Hello again fzero :smile: I am not sure what the notation in your limit means (i.e. what does x- -->∞ mean) ? Also, why do i want to let x-->∞ ? Wouldn't that cause the e^(-x) --> 0 and hence it would render (1) equal to 0 ?
 
Last edited:
  • #4
fzero
Science Advisor
Homework Helper
Gold Member
3,119
289
Hello again fzero :smile: I am not sure what the notation in your limit means (i.e. what does x- -->∞ mean) ? [/tex]

I meant x -->∞, the large x limit.

Also, why do i want to let x-->∞ ? Wouldn't that cause the e^(-x) --> 0 and hence it would render (1) equal to 0 ?
The idea of the calculation is that you evalute the integral in the large x limit by expanding in powers of [tex]x^{-1/2}[/tex], leaving the prefactors alone. Your expansion

[tex]
\int_{-\sqrt{x}}^\infty e^{-\sqrt{x}t}\left (1 +\frac{t}{\sqrt{x}}\right )^x\,dt = \int_{-\sqrt{x}}^\infty \exp\left [-\sqrt{x}t + \left (\sqrt{x}t - \frac{t^2}{2} + \frac{t^3}{3x^{3/2}} \dots \right )\right ]
[/tex]

is incomplete because you haven't dealt with the x dependence of the integration limits. However, as [tex]a\rightarrow \infty[/tex], we have a Maclaurin expansion

[tex]\int_{-a}^\infty f(t) dt = \int_{-\infty}^\infty f(t) dt + \lim_{a\rightarrow \infty} \frac{d}{d(1/a)} \int_{-a}^\infty f(t) dt + \cdots [/tex]

If we're only interested in the leading order term, then we only need to set the integration variable to [tex]-\infty[/tex]. If we want to compute subleading terms, then we need to be more careful.
 

Related Threads on Show for large x, Γ(x+1) = exp(-x)*x^(x+1/2)*√(2*pi)

Replies
16
Views
3K
Replies
2
Views
5K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
17
Views
15K
Replies
7
Views
2K
Top