# Gamma Function

1. Oct 30, 2005

### amcavoy

I am wondering how the following statement holds true:

$$\Gamma\left(\frac{1}{2}\right)=\int_0^{\infty}e^{-x}x^{-\frac{1}{2}}\,dx=\sqrt{\pi}$$

I know how to show that:

$$\int_0^{\infty}e^{-x^2}\,dx=\frac{\sqrt{\pi}}{2}$$

But I can't seem to apply that method (converting to a double-integral) to the gamma function. Any ideas?

Thanks.

Last edited: Oct 30, 2005
2. Oct 30, 2005

### Hurkyl

Staff Emeritus
$$\int_0^{+\infty} e^{-x^2} \, dx$$

and

$$\int_0^{+\infty} e^{-x} x^{1/2} \, dx$$

look a lot different... after all, one has $-x^2$ in the exponent, and the other has $-x$...

3. Oct 30, 2005

### amcavoy

Yeah, I just figured that since the results look similar, there may be a way to rearrange that gamma function to make it look like the other. --I guess not.

Would you suggest integration by parts to solve this? That is the only way I can think of, but it doesn't seem too promising because I can't get rid of that fractional exponent. Is there some sort of trick to this?

Thanks again.

4. Oct 30, 2005

### Hurkyl

Staff Emeritus
I never said you couldn't. :tongue: My post was a hint!

5. Oct 30, 2005

### saltydog

One method is to use Laplace Transforms:
First calculate:
$$\Gamma(x+1)=\int_0^{\infty}e^{-\beta}\beta^{x}d\beta$$
let:
$$\beta=st$$
$$\Gamma(x+1)=s^{x+1}\int_0^{\infty}e^{-st}t^x dt$$
rearrange, equate to the Laplace transform of $t^{x}$ and then let x=-1/2. Can you finish it?
Also, shouldn't your integral have $x^{-1/2}$?

6. Oct 30, 2005

### benorin

yep

Yes, it should, for $\Gamma(z)=\int_{t=0}^{\infty}e^{-t}t^{z-1}dt$ for $z\in\mathbb{C}$ such that $\Re(z)>0$.
Hence $\Gamma\left(\frac{1}{2}\right)=\int_{t=0}^{\infty}e^{-t}t^{-\frac{1}{2}}dt$, apply the substitution $x=\sqrt t \Rightarrow dx=\frac{1}{2}t^{-\frac{1}{2}}dt$ to get
$\Gamma\left(\frac{1}{2}\right)=2\int_{t=0}^{\infty}e^{-x^{2}}dx = \sqrt\pi$

Last edited: Oct 30, 2005
7. Oct 30, 2005

### amcavoy

Thanks a lot everyone for the help. I like that last substitution to change it into the form I am familiar with. Thank you

8. Oct 31, 2005

### Hurkyl

Staff Emeritus
Do you understand why he thought to use that substitution in the first place? (Hint: my hint is why, though he probably knew it without my hint) Will you think to look at that next time you get stuck on an integral?

9. Oct 31, 2005

### amcavoy

Yes, thank you both very much.