Proving the Gamma Function Integral for $\frac{1}{2}$

[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.

 

[Hurkyl]

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

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 $-x$...

 

[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.

 

[Hurkyl]

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.

I never said you couldn't. My post was a hint!

 

[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 $x^{-1/2}$?

 

[benorin]

yep

One method is to use Laplace Transforms:
Also, shouldn't your integral have $x^{-1/2}$?

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
$\Gamma\left(\frac{1}{2}\right)=2\int_{t=0}^{\infty}e^{-x^{2}}dx = \sqrt\pi$

 

[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

 

[Hurkyl]

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?

 

[amcavoy]

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?

Yes, thank you both very much.