# Special Gamma Value

• May 18th 2008, 06:49 PM
Chris L T521
Special Gamma Value
I'm just wondering, does anybody know of an analytic proof that shows that :

$\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$?

Or do we just accept that this is true?? (Sun)
• May 18th 2008, 06:59 PM
Mathstud28
Quote:

Originally Posted by Chris L T521
I'm just wondering, does anybody know of an analytic proof that shows that :

$\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$?

Or do we just accept that this is true?? (Sun)

Ok...I will give this a go...see if ti works

$\Gamma\bigg(\frac{1}{2}\bigg)=\int_0^1\frac{1}{\sq rt{-\ln(x)}}dx$

Let $u=\sqrt{-\ln(x)}\Rightarrow{x=e^{-u^2}}$

so $dx=-2ue^{-u^2}$

and $u(1)=\sqrt{-ln(1)}=0$

and $u(0)=\sqrt{-\ln(0)}=\infty$

So we have

$\int_{\infty}^{0}\frac{-2ue^{-u^2}}{u}du=2\int_0^{\infty}e^{-u^2}du$

Now look at your signature (Rofl)
• May 18th 2008, 07:01 PM
Mathstud28
Quote:

Originally Posted by Mathstud28
Ok...I will give this a go...see if ti works

$\Gamma\bigg(\frac{1}{2}\bigg)=\int_0^1\frac{1}{\sq rt{-\ln(x)}}dx$

Let $u=\sqrt{-\ln(x)}\Rightarrow{x=e^{-u^2}}$

so $dx=-2ue^{-u^2}$

and $u(1)=\sqrt{-ln(1)}=0$

and $u(0)=\sqrt{-\ln(0}=\infty$

So we have

$\int_{\infty}^{0}\frac{-2ue^{-u^2}}{u}du=2\int_0^{\infty}e^{-u^2}du$

Now look at your signature (Rofl)

I have never done that before...that is hilarious it came out to be that (Rofl)(Rofl)
• May 18th 2008, 07:01 PM
Chris L T521
Quote:

Originally Posted by Mathstud28
Ok...I will give this a go...see if ti works

$\Gamma\bigg(\frac{1}{2}\bigg)=\int_0^1\frac{1}{\sq rt{-\ln(x)}}dx$

Let $u=\sqrt{-\ln(x)}\Rightarrow{x=e^{-u^2}}$

so $dx=-2ue^{-u^2}$

and $u(1)=\sqrt{-ln(1)}=0$

and $u(0)=\sqrt{-\ln(0}=\infty$

So we have

$\int_{\infty}^{0}\frac{-2ue^{-u^2}}{u}du=2\int_0^{\infty}e^{-u^2}du$

Now look at your signature (Rofl)

hahaha. Thanks for that! :D
• May 18th 2008, 07:03 PM
mr fantastic
Quote:

Originally Posted by Chris L T521
I'm just wondering, does anybody know of an analytic proof that shows that :

$\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$?

Or do we just accept that this is true?? (Sun)

If you run

gamma function

through a famous search engine, you will find more proofs than you've had hot breakfasts.
• May 18th 2008, 07:03 PM
TheEmptySet
Quote:

Originally Posted by Chris L T521
I'm just wondering, does anybody know of an analytic proof that shows that :

$\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$?

Or do we just accept that this is true?? (Sun)

$\Gamma\left(\frac{1}{2}\right)=\int_{0}^{\infty}t^ {-1/2}e^{-t}dt$

let $t=u^2 \to dt =2udu$

$\Gamma\left(\frac{1}{2}\right)=\int_{0}^{\infty}u^ {-1}e^{-u^2}(2udu)=2\int_{0}^{\infty}e^{-u^2}du$

do the same sub with v gives

$\left[\Gamma\left(\frac{1}{2}\right) \right]^2=4\int_{0}^{\infty}\int_{0}^{\infty}e^{-(u^2+v^2)}dudv$

converting to polar coordinates gives

$4\int_{0}^{\pi/2}\int_{0}^{\infty}re^{-r^2}drd\theta=\pi$

$\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$
• May 18th 2008, 08:04 PM
Krizalid
Quote:

Originally Posted by Chris L T521

I'm just wondering, does anybody know of an analytic proof that shows that :

$\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$?

See here.

Quote:

Originally Posted by Chris L T521

Or do we just accept that this is true?? (Sun)

Of course not.

Here's another proof:

Start by showing for $n\ge0$ that $\int_{0}^{1}{x^{n}(1-x)^{n}\,dx}=\frac{(n!)^{2}}{(2n+1)!}.$ Let $n=\frac12;$ the integral is equal to find the area of a semicircle with radius $\frac12,$ hence $\frac{1}{2}\left[ \left( \frac{1}{2} \right)! \right]^{2}=\frac{\pi }{8}.$

Now rearrange this to get what you want.

----

Oops, I realized that you want to find $\Gamma \left( \frac{1}{2} \right)$ and I computed $\left( \frac{1}{2} \right)!$ Anyway, the link above I gave you contains another proof.
• May 18th 2008, 08:56 PM
NonCommAlg
here's my favorite proof of the so-called Gaussian integral:

let $f(x)=\left(\int_0^x e^{-t^2}dt \right)^2 + \int_0^1 \frac{e^{-x^2(1+t^2)}}{t^2+1} dt, \ x \geq 0.$ then: $f'(x)=2e^{-x^2}\int_0^x e^{-t^2} dt - 2x \int_0^1 e^{-x^2(1+t^2)}dt.$

in the first integral let $t=xu$ to get: $f'(x)=2x\int_0^1e^{-x^2(1+ u^2)}du-2x \int_0^1 e^{-x^2(1 + t^2)}dt=0,$ i.e. $f$ is constant.

hence $f(x)=f(0)=\int_0^1 \frac{1}{1+t^2}dt=\frac{\pi}{4}.$ thus $\lim_{x\to\infty}f(x)=\frac{\pi}{4}.$ call this result (1). on the other hand, from the

definition of $f$ it's easily seen that $\lim_{x\to\infty}f(x)=\left(\int_0^{\infty} e^{-t^2} dt \right)^2.$ call this (2). now (1) and (2) complete the proof.