- #1

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## 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