It seems as if no one has asked this here before, unless I don't know how to search.
The Gamma function is $$ \Gamma(\alpha)=\int_0^\infty x^{\alpha-1} e^{-x}\,dx. $$ Why is $$ \Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}\text{ ?} $$ (I'll post my own answer, but I know there are many ways to show this, so post your own!)
One can use the trick with spherical coordinates: $$ \Gamma\left(\frac{1}{2}\right) = \int_0^\infty \mathrm{e}^{-x} \frac{\mathrm{d} x}{\sqrt{x}} = \int_0^\infty \mathrm{e}^{-x} \mathrm{d} (2 \sqrt{x}) = \int_{-\infty}^\infty \mathrm{e}^{-u^2} \, \mathrm{d} u $$ Then: $$ \Gamma\left(\frac{1}{2}\right) \Gamma\left(\frac{1}{2}\right) = \int_{-\infty}^\infty \int_{-\infty}^\infty \mathrm{e}^{-x^2} \mathrm{e}^{-y^2} \, \mathrm{d}x \, \mathrm{d} y = \underbrace{\int_0^\infty r \mathrm{e}^{-r^2} \, \mathrm{d} r}_{\frac{1}{2}} \cdot \underbrace{\int_{0}^{2\pi} \, \mathrm{d} \phi}_{2 \pi} = \pi $$
Alternatively one can use the second Euler's integral: $$ \frac{\Gamma\left(\frac{1}{2}\right) \Gamma\left(\frac{1}{2}\right) }{\Gamma(1)} = \int_0^1 t^{-1/2} (1-t)^{-1/2} \mathrm{d}t = \int_0^1 \mathrm{d} \left(2\arcsin\left(\sqrt{t}\right)\right) = \pi $$ Now, using $\Gamma(1) = 1$ the result follows.
Yet another method is to use the duplication identity: $$ \Gamma(2s) = \frac{2^{2s-1}}{\sqrt{\pi}} \Gamma(s) \Gamma\left(s+\frac{1}{2}\right) $$ at $ s= \frac{1}{2}$.
We only need Euler's formula:
$$\Gamma(1-z) \Gamma(z) = \frac{\pi}{\sin \pi z} \Longrightarrow \Gamma^2\left(\frac{1}{2}\right ) = \pi $$
If there's any justice in the universe, someone must have asked here how to show that $$ \int_{-\infty}^\infty e^{-x^2/2}\,dx = \sqrt{2\pi}. $$ Let's suppose that has been answered here. Let (capital) $X$ be a random variable whose probability distribution is $$ \frac{e^{-x^2/2}}{\sqrt{2\pi}}\,dx. $$ Consider the problem of finding $\operatorname{E}(X^2)$. It is $$ \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty x^2 e^{-x^2/2}\,dx = \text{(by symmetry)} \frac{2}{\sqrt{2\pi}} \int_0^\infty x^2 e^{-x^2/2}\,dx $$ $$ \sqrt{\frac2\pi}\int_0^\infty xe^{-x^2/2}\Big(x\,dx\Big) = \sqrt{\frac2\pi}\int_0^\infty \sqrt{u}\ e^{-u}\,du = \sqrt{\frac2\pi}\ \Gamma\left(\frac32\right) = \frac12\sqrt{\frac2\pi} \Gamma\left(\frac12\right). $$ So it is enough to show that this expected value is $1$. That is true if the sum of two independent copies of it has expected value $2$. So: $$ \Pr\left(X^2+Y^2<w\right) = \frac{1}{2\pi}\iint\limits_\mathrm{disk} e^{-(x^2+y^2)/2}\,dx\,dy $$ where the disk has radius $\sqrt{w}$. This equals $$ \frac{1}{2\pi}\int_0^{2\pi}\int_0^{\sqrt{w}} e^{-\rho^2/2} \,\rho\,d\rho\,d\theta = \int_0^{\sqrt{w}} e^{-\rho^2/2} \,\rho\,d\rho. $$ This last equality holds because we are integrating with respect to $\theta$ something not depending on $\theta$. Differentiating this with respect to $w$ gives the probability density function of the random variable $X^2+Y^2$: $$ e^{-w/2}\sqrt{w}\frac{1}{2\sqrt{w}} = \frac{e^{-w/2}}{2}\text{ for }w>0. $$ So $$ \operatorname{E}(X^2+Y^2) = \int_0^\infty w \frac{e^{-w/2}}{2}\,dw =2. $$
Method 1: $$\Gamma(1/2) = \int_0^{\infty} x^{-1/2} \exp(-x) dx$$ Set $x = t^2$, to get $$\Gamma(1/2) = \int_0^{\infty} 2\exp(-t^2) dt = \sqrt{\pi}$$
Method 2: $$\beta(x,y) = \dfrac{\Gamma(x) \Gamma(y)}{\Gamma(x+y)}$$ Take $y=x=1/2$, to get $$\beta(1/2,1/2) = \dfrac{\Gamma^2(1/2)}{\Gamma(1)} \implies \Gamma(1/2) = \sqrt{\beta(1/2,1/2)}$$
$$\beta(x,y) = 2 \int_0^{\pi/2} \sin^{2x-1}(\theta) \cos^{2y-1}(\theta) d \theta$$ Hence, $$\beta(1/2,1/2) = 2 \int_0^{\pi/2} d \theta = \pi$$ $$\Gamma(1/2) = \sqrt{\pi}$$
I'm surprised that no one has mentioned that this $\sqrt{\pi}$ is also "the same one" as the $\sqrt{\pi}$ in Stirling's formula.
It is a known fact that $\Gamma$ is uniquely characterized as the function that satisfies the equation $\Gamma(z+1) = z \Gamma(z)$ and whose logarithm "has nice asymptotics at $+\infty$". The conventional way to explain what "nice asymptotics" should mean in this context involves logarithmic convexity, but I prefer to leave it like this: any function that satisfies that functional equation must coincide with $\Gamma(z)$ up to a periodic factor, which, if nontrivial, would prevent it from having an asymptotic expansion at $+\infty$ in terms of powers and logs.
Now let's turn to Stirling's formula and assume that it holds both for integer and half-integer values of $\Gamma$. Clearly,
$$\Gamma(n+1/2) = \Gamma(1/2) \cdot \frac{1}{2} \cdot \frac{3}{2} \cdot \dots \cdot \frac{2n-1}{2} = \Gamma(1/2) \cdot \frac{(2n)!}{2^{2n} n!},$$
which is basically the duplication formula. Now if we plug it into the Stirling's formula, we will find out that $\Gamma(1/2) = \sqrt{\pi}$.
It also follows from the famous Riemann functional equation with $s=1/2$: $$ \zeta(s) = 2^s\pi^{s-1}\ \sin\left(\frac{\pi s}{2}\right)\ \Gamma(1-s)\ \zeta(1-s) $$ but Euler's reflection formula is probably more basic.
This is a "proof". We know that the surface area of the $n-1$ dimensional unit sphere is $$ |S^{n-1}| = \frac{2\pi^{\frac{n}2}}{\Gamma(\frac{n}2)}. $$ On the other hand, we know that $|S^2|=4\pi$, which gives $$ 4\pi = \frac{2\pi^{\frac32}}{\Gamma(\frac32)} = \frac{2\pi^{\frac32}}{\frac12\Gamma(\frac12)}. $$
There are a few details missing (showing the integral converges, justifying the differentiation under the integral sign, and so forth), but the main idea should be clear.
Consider the Fourier transform of $f(x)=e^{-x^2}$.
$$\hat f(\xi)= \int_{-\infty}^{\infty} e^{-x^2} e^{-2\pi i x\xi} \ dx =\int_{-\infty}^{\infty} e^{-x^2} (\cos(2\pi x \xi) - i \sin(2\pi x \xi)) dx =\int_{-\infty}^{\infty} e^{-x^2} \cos(2\pi x \xi) \ dx$$
Differentiate under the integral sign and integrate by parts.
$$\hat f'(\xi) =-2\pi \int e^{-x^2}\sin(2\pi \xi x) \ dx = -2\pi^2 \xi \int e^{-x^2} \cos(2\pi x \xi) \ dx = -2\pi^2\xi \hat f(\xi)$$
Separate variables.
$$\frac{\hat f'(\xi)}{\hat f(\xi) }= -2\pi^2 \xi \Rightarrow \hat f(\xi) =Ce^{-\pi^2 \xi^2}.$$
Use the inversion formula and change variables .
$$e^{-x^2} = C\int e^{-\pi^2 \xi^2} \cos(2\pi x \xi) \ d\xi = \frac{C}{\pi} \int e^{- \xi^2} \cos(2 x \xi) \ d\xi $$
$$e^{-\pi^2 x^2}\frac{C}{\pi} =\int e^{- \xi^2} \cos(2 x\pi \xi) \ d\xi =\frac{C^2}{\pi}e^{-\pi^2x^2}\Rightarrow C=\sqrt{\pi}$$
$$\hat f (0) = \sqrt{\pi} = \int e^{-x^2} \cos(0) \ dx = \int e^{-x^2} \ dx.$$
$$B\left(\frac{1}{2},\frac{1}{2}\right)=\frac{\left[\Gamma\left(\frac{1}{2}\right)\right]^{2}}{\Gamma{(1)}}=\left[\Gamma\left(\frac{1}{2}\right)\right]^{2}$$ $$B\left(\frac{1}{2},\frac{1}{2}\right)=\frac{\pi}{\sin{\frac{\pi}{2}}}=\pi$$
Chris.
The Gamma function is $$\Gamma(\alpha)=\int_0^\infty x^{\alpha-1} e^{-x}\,dx$$
Therefore
$$\Gamma\left( \frac{1}{2}\right) =\int_0^\infty \frac{1}{\sqrt{x}} e^{-x}\,dx$$
Thus, after the change of variable $t=\sqrt{x}$, this turns into the Euler integral $$t=\sqrt{x} \implies x = t^2$$ $$\frac{dt}{dx} = \frac{1}{2\sqrt{x}} = \frac{1}{2t} \iff dx = 2t\,dt$$
We have $$\int_0^\infty \frac{1}{t} e^{-t^2}\,2t\,dt = 2\int_0^\infty e^{-t^2}\,dt $$
And following holds: $$\Gamma\left( \frac{1}{2}\right) = 2\int_0^\infty e^{-t^2}\,dt = \int_{-\infty}^\infty e^{-t^2}\,dt$$
Which $\int_{-\infty}^\infty e^{-t^2}\,dt$ is the Gaussian integral. It follows:
$$\left[ \Gamma\left( \frac{1}{2}\right)\right]^2 = \bigg(\int_{-\infty}^{+\infty} e^{-x^2} dx\bigg)\bigg(\int_{-\infty}^{+\infty} e^{-y^2} dy\bigg)$$ $$\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}e^{-(x^2+y^2)} \,dx\,dy$$
Now change to polar coordinates
$$\int_0^{+2 \pi}\int_0^{+\infty}e^{-r^2} r\,dr\,d\theta$$
The $\theta$ integral just gives $2\pi$, while the $r$ integral succumbs to the substitution $u=r^2$
$$\left[ \Gamma\left( \frac{1}{2}\right)\right]^2 = 2\pi\int_{0}^{+\infty}e^{-u} \,du/2=\pi$$
$$\Gamma\left( \frac{1}{2}\right)=\sqrt{\pi}$$
By the Bohr-Mollerup theorem,
$$\Gamma(1/2)=\lim_{n\to\infty}\frac{\sqrt n(n!)}{(1/2)(-1/2)\dots(1/2-n)}=\lim_{n\to\infty}\frac{\sqrt n4^{n+1}(n!)^2}{(2n)!(n+\frac12)}$$
Apply the Stirling approximation and watch almost everything simplify!
$$\Gamma(1/2)=\sqrt\pi\lim_{n\to\infty}\frac n{n+\frac12}=\sqrt\pi$$
