gamma function for half integers

Question

How can one prove the gamma function for positive half-integers?

Here is my attempt for $\frac{3}{2}$:

How do I solve the last integral?

Answer 1

Hint: $$ \begin{align} \Gamma(\alpha+1) &=\int_0^\infty x^\alpha e^{-x}\,\mathrm{d}x\\ &=-\int_0^\infty x^\alpha\,\mathrm{d}e^{-x}\\ &=\alpha\int_0^\infty x^{\alpha-1}e^{-x}\,\mathrm{d}x\\[3pt] &=\alpha\,\Gamma(\alpha) \end{align} $$ and $$ \begin{align} \Gamma\left(\frac12\right)^2 &=\int_0^\infty x^{-1/2}e^{-x}\,\mathrm{d}x\int_0^\infty y^{-1/2}e^{-y}\,\mathrm{d}y\\ &=4\int_0^\infty e^{-x^2}\,\mathrm{d}x\int_0^\infty e^{-y^2}\,\mathrm{d}y\\ &=\int_{-\infty}^\infty e^{-x^2}\,\mathrm{d}x\int_{-\infty}^\infty e^{-y^2}\,\mathrm{d}y\\ &=\int_{-\infty}^\infty\int_{-\infty}^\infty e^{-\left(x^2+y^2\right)} \,\mathrm{d}x\,\mathrm{d}y\\ &=\int_0^{2\pi}\int_0^\infty e^{-r^2}r\,\mathrm{d}r\,\mathrm{d}\theta\\ &=\frac12\int_0^{2\pi}\int_0^\infty e^{-r}\,\mathrm{d}r\,\mathrm{d}\theta\\[8pt] &=\pi \end{align} $$