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 /> <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 /> <br />After some manipulation, (3) can be put in the form:

<br /> <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)<br />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

<br /> \lim_{x-\rightarrow \infty} \int_{-\sqrt{x}}^\infty e^{ - \frac{t^2}{2}}dt = \int_{-\infty}^\infty e^{ - \frac{t^2}{2}}dt .<br />

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

<br /> \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 ] <br />

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.