I'm just wondering, does anybody know of an analytic proof that shows that :
$\displaystyle \Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$?
Or do we just accept that this is true??
Ok...I will give this a go...see if ti works
$\displaystyle \Gamma\bigg(\frac{1}{2}\bigg)=\int_0^1\frac{1}{\sq rt{-\ln(x)}}dx$
Let $\displaystyle u=\sqrt{-\ln(x)}\Rightarrow{x=e^{-u^2}}$
so $\displaystyle dx=-2ue^{-u^2}$
and $\displaystyle u(1)=\sqrt{-ln(1)}=0$
and $\displaystyle u(0)=\sqrt{-\ln(0)}=\infty$
So we have
$\displaystyle \int_{\infty}^{0}\frac{-2ue^{-u^2}}{u}du=2\int_0^{\infty}e^{-u^2}du$
Now look at your signature
$\displaystyle \Gamma\left(\frac{1}{2}\right)=\int_{0}^{\infty}t^ {-1/2}e^{-t}dt$
let $\displaystyle t=u^2 \to dt =2udu$
$\displaystyle \Gamma\left(\frac{1}{2}\right)=\int_{0}^{\infty}u^ {-1}e^{-u^2}(2udu)=2\int_{0}^{\infty}e^{-u^2}du$
do the same sub with v gives
$\displaystyle \left[\Gamma\left(\frac{1}{2}\right) \right]^2=4\int_{0}^{\infty}\int_{0}^{\infty}e^{-(u^2+v^2)}dudv$
converting to polar coordinates gives
$\displaystyle 4\int_{0}^{\pi/2}\int_{0}^{\infty}re^{-r^2}drd\theta=\pi$
$\displaystyle \Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$
See here.
Of course not.
Here's another proof:
Start by showing for $\displaystyle n\ge0$ that $\displaystyle \int_{0}^{1}{x^{n}(1-x)^{n}\,dx}=\frac{(n!)^{2}}{(2n+1)!}.$ Let $\displaystyle n=\frac12;$ the integral is equal to find the area of a semicircle with radius $\displaystyle \frac12,$ hence $\displaystyle \frac{1}{2}\left[ \left( \frac{1}{2} \right)! \right]^{2}=\frac{\pi }{8}.$
Now rearrange this to get what you want.
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Oops, I realized that you want to find $\displaystyle \Gamma \left( \frac{1}{2} \right)$ and I computed $\displaystyle \left( \frac{1}{2} \right)!$ Anyway, the link above I gave you contains another proof.
here's my favorite proof of the so-called Gaussian integral:
let $\displaystyle f(x)=\left(\int_0^x e^{-t^2}dt \right)^2 + \int_0^1 \frac{e^{-x^2(1+t^2)}}{t^2+1} dt, \ x \geq 0.$ then: $\displaystyle f'(x)=2e^{-x^2}\int_0^x e^{-t^2} dt - 2x \int_0^1 e^{-x^2(1+t^2)}dt.$
in the first integral let $\displaystyle t=xu$ to get: $\displaystyle f'(x)=2x\int_0^1e^{-x^2(1+ u^2)}du-2x \int_0^1 e^{-x^2(1 + t^2)}dt=0,$ i.e. $\displaystyle f$ is constant.
hence $\displaystyle f(x)=f(0)=\int_0^1 \frac{1}{1+t^2}dt=\frac{\pi}{4}.$ thus $\displaystyle \lim_{x\to\infty}f(x)=\frac{\pi}{4}.$ call this result (1). on the other hand, from the
definition of $\displaystyle f$ it's easily seen that $\displaystyle \lim_{x\to\infty}f(x)=\left(\int_0^{\infty} e^{-t^2} dt \right)^2.$ call this (2). now (1) and (2) complete the proof.