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

[Saladsamurai]

Homework Statement

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

$$\Gamma(x+1) = e^{-x}x^{x+\frac{1}{2}}\sqrt{2\pi}\qquad(1)$$

Homework Equations

Definition of gamma function:

$$\Gamma(z) = \int_0^\infty t^{z-1}e^{-t}\,dt\qquad(2)$$

The Attempt at a Solution

When I see that (1) has a multiple of $\pi$ 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

$$<br /> \Gamma(x+1) = e^{-x}x^{x+1/2}\int_{-\sqrt{x}}^\infty e^{-\sqrt{x}t}\left (1 +\frac{t}{\sqrt{x}}\right )^x\,dt \qquad(3)<br />$$After some manipulation, (3) can be put in the form:

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

or

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

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

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

 

[fzero]

Use

$$\lim_{x-\rightarrow \infty} \int_{-\sqrt{x}}^\infty e^{ - \frac{t^2}{2}}dt = \int_{-\infty}^\infty e^{ - \frac{t^2}{2}}dt .$$

You could find the error terms by expanding in a series using the fundamental theorem of calculus.

 

[Saladsamurai]

Hello again fzero 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 ?

 

[fzero]

Hello again fzero 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 $$x^{-1/2}$$, leaving the prefactors alone. Your expansion

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

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

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

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