How can i prove that \begin{align*} \Gamma\left(\frac{n}{2}\right) &= \frac{\sqrt{\pi} \left(n-1\right)!}{2^{n-1} \left(\frac{n-1}{2}\right)! } \end{align*}
I try this \begin{align*} \Gamma\left(\frac{n}{2}\right) &= \int\limits_0^\infty x^{\frac{n}{2}-1}e^{-x} \mathrm{d}x\qquad\qquad\mid\text{integration by parts }\int fg^\prime= fg-\int f^\prime g\\ &=-x^{\frac{n}{2}-1}e^{-x}\mathop{\big|}\limits_0^\infty+\int\limits_0^\infty \left(\frac{n}{2}-1\right) x^{\frac{n}{2}-2}e^{-x}\mathrm{d}x\\ &=-\lim\limits_{x\rightarrow\infty}x^{\frac{n}{2}-1}e^{-x}+0+\int\limits_0^\infty \left(\frac{n}{2}-1\right) x^{\frac{n}{2}-2}e^{-x}\mathrm{d}x\qquad\qquad\mid\lim\limits_{x\rightarrow\infty}x^{\frac{n}{2}-1}e^{-x}\rightarrow 0\\ &=\left(\frac{n}{2}-1\right)\int\limits_0^\infty x^{\frac{n}{2}-2}e^{-x}\mathrm{d}x\qquad\qquad\mid\text{integration by parts }\\ &=\left(\frac{n}{2}-1\right)\left(-x^{\frac{n}{2}-2}e^{-x}\mathop{\big|}\limits_0^\infty+\int\limits_0^\infty \left(\frac{n}{2}-2\right) x^{\frac{n}{2}-3}e^{-x}\mathrm{d}x\right)\\ &=\left(\frac{n}{2}-1\right)\left(\frac{n}{2}-2\right)\int\limits_0^\infty x^{\frac{n}{2}-3}e^{-x}\mathrm{d}x\\ &=\left(\frac{n}{2}-1\right)\left(\frac{n}{2}-2\right)\Gamma\left(\frac{n}{2}-2\right)\\ &=\left(\frac{n}{2}-1\right)\left(\frac{n}{2}-2\right)\left(\frac{n}{2}-3\right)\Gamma\left(\frac{n}{2}-3\right)\\ \end{align*} but I don't know what to do now. Any ideas?
First prove that $\Gamma\Big(\frac{1}{2}\big)=\sqrt{\pi}$
Proof:
Given the known result
$$\int_{-\infty}^{+\infty}e^{-\frac{x^2}{2}}dx=\sqrt{2\pi}$$
set $\frac{x^2}{2}=t$ and get
$$\int_{-\infty}^{+\infty}e^{-\frac{x^2}{2}}dx=2\int_0^{\infty}\frac{\sqrt{2}}{2}t^{-\frac{1}{2}}e^{-t}dt=\sqrt{2}\int_0^{\infty}t^{\frac{1}{2}-1}e^{-t}dt=\sqrt{2}\Gamma\Big(\frac{1}{2}\Big)\rightarrow \Gamma\Big(\frac{1}{2}\Big)=\sqrt{\pi}$$
Second prove that
$\Gamma(n)=(n-1)\Gamma(n-1)$
Proof:
simply integrating by parts you get
$$\Gamma(n)=\int_0^{\infty}x^{n-1}e^{-x}dx=(n-1)\int_0^{\infty}x^{n-2}e^{-x}dx=(n-1)\Gamma(n-1)$$
Finally: put the two result together and get you proof recursively