gamma and e^(-x^2)

galactus · #1 $Old$ December 31st, 2008, 06:23 PM

You know, I was just playing around and noticed something. I am sure you all

have thought of this before, but I just 'discovered this'. I always integrated

$e^{-x^{2}}$ by the polar method.

But if we use gamma it is easy.

Knowing that ${\Gamma}(\frac{1}{2})=\sqrt{\pi}$

we can use the definition:

${\Gamma}(n)=\int_{0}^{\infty}t^{n-1}e^{-t}dt=2\int_{0}^{\infty}x^{2n-1}e^{-x^{2}}dx$

Therefore,

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

Since $e^{-x^{2}}$ is even, we get

$\int_{-\infty}^{\infty}e^{-x^{2}}dx=\sqrt{\pi}$

I know, I know, it's obvious to you all and older than Moses. But I had not

thought about using gamma before for this famous integral and it just

dawned on me. I always just used the old polar thing and never worried

about another method. Stuck in that rut, I reckon. But, if there are those

out there who like this, here it is.

Krizalid · #2 $Old$ December 31st, 2008, 06:31 PM

Quote:

Originally Posted by galactus $View Post$

But if we use gamma it is easy.

Knowing that ${\Gamma}(\frac{1}{2})=\sqrt{\pi}$

How'd ya prove this little fact?

galactus · #3 $Old$ December 31st, 2008, 06:55 PM

You want me to prove that?. That is a famous gamma identity, so I did not bother proving every aspect.

But, if you would like to see it, let me get back. Okey-doke. I always just used it and took it for granted and never bothered proving it.

But, I am thinking we can play on the old polar thing to show it.

To be lazy, I would just use ${\Gamma}(n){\Gamma}(1-n)=\frac{\pi}{sin({\pi}n)}$ and let n=1/2.

I know, you want something more tangible.

Let's see, we could start with $\int_{0}^{\infty}t^{\frac{-1}{2}}e^{-t}dt$

Then, sub in $t=y^{2}, \;\ dt=2ydy$ .

I will finish later. Got to go for a second.

galactus · #4 $Old$ December 31st, 2008, 07:38 PM

Well, Kriz, it appears I ended up using the polar thing to do that. But I think you knew that, didn't you?.

To conclude:

$2\int_{0}^{\infty}y^{-1}e^{-y^{2}}ydy=2\int_{0}^{\infty}e^{-y^{2}}dy$

or $2\int_{0}^{\infty}e^{-x^{2}}dx$

$\left[{\Gamma}(\frac{1}{2})\right]^{2}=4\int_{0}^{\infty}\int_{0}^{\infty}e^{-(x^{2}+y^{2})}dxdy$

Now, we use the polar thing. So, I learned myself something with a little prodding from the Krizmeister. Cheers.

I suppose that infernal polar was hidden there afterall.

I can not believe I had never bothered doing these before. Oh well, I derived something I should have a long time ago. Especially, since I like the Gamma function so much.

Krizalid · #5 $Old$ December 31st, 2008, 08:14 PM

Ohhh, I'm sorry, I misread the question, it's actually calculating again the gaussian integral.

Cheers and Happy New Year!!!!

galactus · #6 $Old$ December 31st, 2008, 09:00 PM

Since we're on Gamma, derive this one:

${\Gamma}(2n)=\frac{1}{\sqrt{\pi}}\cdot 2^{2n-1}\cdot {\Gamma}(n)\cdot {\Gamma}(n+\frac{1}{2})$

This is the duplication formula for Gamma.

Try using the identity $2sin{\theta}cos{\theta}=sin2{\theta}$ and put $2{\theta}={\phi}$

Happy New Years

Krizalid · #7 $Old$ December 31st, 2008, 09:07 PM

Don't worry, I'm also a good friend of Gamma function.

But it's actually a bit interesting in finding a closed form for $\Gamma\left(n+\frac12\right).$

galactus · #8 $Old$ December 31st, 2008, 09:12 PM

You mean this one:

${\Gamma}(n+\frac{1}{2})=\frac{1\cdot 3\cdot 5\cdots (\not{2}n-1)}{2^{n}}=\frac{(2n)!}{4^{n}n!}\sqrt{\pi}$ ?.

Krizalid · #9 $Old$ December 31st, 2008, 09:12 PM

Yes.

galactus · #10 $Old$ December 31st, 2008, 09:55 PM

Well, Kriz, let' see. I have been working on it while I was waiting on your reponnse and gong to the bathroom.

It is getting late, Kriz. I am going to skip some major steps and join you tomorrow. Gotta go. But I have this so far. Need to explain some steps further if need be.

From $B(p,q)=\frac{{\Gamma}(p){\Gamma}(q)}{{\Gamma}(p+q)}$

Then,

${\Gamma}(p+\frac{1}{2})=\frac{{\Gamma}(p)\cdot {\Gamma}(\frac{1}{2})}{B(p,1/2)}$

$=\frac{(p-1)!\sqrt{\pi}}{\frac{4^{p}(p!)^{2}}{p(2p)!}}$

$\boxed{=\frac{(2p)!\sqrt{\pi}}{4^{p}\cdot p!}}$

We need to show that $B(p,1/2)=\frac{4^{n}(n!)^{2}}{n(2n)!}$ which I think can be gotten from $\frac{4^{p}}{2}\cdot B(p,p)$

because $B(p,p)=\frac{{\Gamma}(p){\Gamma}(p)}{{\Gamma}(2p)}=\frac{(p-1)!(p-1)!}{(2p-1)!}$

Therefore, we get $=\frac{(p!)^{2}\cdot 2p}{p^{2}\cdot (2p)!}$

$=\frac{2(p!)^{2}}{p(2p)!}$

$B(p,1/2)=\frac{4^{p}}{2}\cdot \frac{2(p!)^{2}}{p(2p)!}$

Plugging this in the above gives:

$\frac{(n-1)!\sqrt{\pi}}{\frac{(p!)^{2}4^{p}}{p(2p)!}}=\frac{(2p)!\sqrt{\pi}}{4^{p}p!}$

Hope you can follow that. I will touch it up tomorrow.

It's getting late and my brain is beginning to see p's and q's everywhere.

Cheers, Kriz

Moo · #11 $Old$ January 2nd, 2009, 01:22 PM

The method I really love for calculating this integral is considering the function :

$f(x)=\left(\int_0^x e^{-t^2} ~dt\right)^2+\int_0^1 \frac{e^{-x^2(t^2+1)}}{t^2+1} ~dt$