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

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!)

Solution 1:

We only need Euler's formula:

$$\Gamma(1-z) \Gamma(z) = \frac{\pi}{\sin \pi z} \Longrightarrow \Gamma^2\left(\frac{1}{2}\right ) = \pi $$

Solution 2:

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}$.