to show $\int_{-\infty}^{\infty} x^{2n}e^{-x^2}dx = (2n)!{\sqrt{\pi}}/4^nn!$ I was trying to show  the following 
$\int_{-\infty}^{\infty} x^{2n}e^{-x^2}dx = (2n)!{\sqrt{\pi}}/4^nn!$ by using $\int_{-\infty}^{\infty} e^{-tx^2}dx = \sqrt{\pi/t}$ 
thus
I differentiated this exponential integral n times to get the following. 
$\int_{-\infty}^{\infty} \frac{d^ne^{-tx^2}}{dt^n}dx $$=\frac{2^{n}\times \sqrt{\pi}t^{\frac{2n-1}{2}}} {1\times 3\times 5 \times ... \times (2n-1)}$
after applying limit for $t\rightarrow 1} I am not getting the desired result. Where am I going wrong ? 
Thanks  
 A: The $n$th derivative of $\sqrt{\pi/t} = t^{-1/2}\sqrt{\pi}$ is 
$$(-1)^n \left(\frac{1}{2}\right)\left(\frac{3}{2}\right)\cdots \cdot \left(\frac{2n-1}{2}\right)t^{-1/2 - n}\sqrt{\pi},$$
which can be written
$$\frac{(-1)^n(1)(3)(5)\cdots (2n-1) t^{-1/2 - n}\sqrt{\pi}}{2^n}.$$
This is the same as 
$$\frac{(-1)^n (1)(2)(3)\cdots (2n-1)(2n)t^{-1/2 - n}\sqrt{\pi}}{2^n\cdot 2^n n!} = \frac{(-1)^n(2n)!t^{-1/2 - n}\sqrt{\pi}}{4^n n!}.$$
So since 
$$\frac{d^n}{dt^n} e^{-tx^2} = (-1)^n x^{2n}e^{-tx^2}$$
we have
$$\int_{-\infty}^\infty (-1)^n x^{2n}e^{-tx^2}\, dx = \frac{(-1)^n(2n)!t^{-1/2 - n}\sqrt{\pi}}{4^n n!}.$$
Cancelling the $(-1)^n$ on both sides, then substituting $t = 1$, we deduce
$$\int_{-\infty}^\infty x^{2n}e^{-tx^2}\, dx = \frac{(2n)!\sqrt{\pi}}{4^nn!}.$$
A: The first thing is your answer should be
$$\int_{-\infty}^{\infty} \frac{d^ne^{-tx^2}}{dt^n}dx=\frac{1\times 3\times 5 \times ... \times (2n-1)\times \sqrt{\pi}t^{-\frac{2n-1}{2}}}{2^{n}} $$
based on the repeated differentiation of $t^{-1/2}\sqrt{\pi}$
Then, if you multiply both numerator and denominator by $2\times4\times\cdots\times(2n)$, you will end up with
$$\int_{-\infty}^{\infty} \frac{d^ne^{-tx^2}}{dt^n}dx=\frac{(2n)!\times \sqrt{\pi}t^{\frac{2n-1}{2}}}{2^{n}\times2^n\times n!}=\frac{(2n)!\times \sqrt{\pi}t^{-\frac{2n-1}{2}}}{4^{n}\times n!}$$
Setting $t=1$ will give you the desired answer.
A: By exploiting a change of variable and integration by parts:
$$I=\int_{\mathbb{R}}x^{2n}e^{-x^2}\,dx = \int_{0}^{+\infty}z^{n-\frac{1}{2}}e^{-z}\,dz = \int_{0}^{+\infty}\frac{d^n}{dz^n}\left(z^{n-\frac{1}{2}}\right)e^{-z}\,dz$$
hence:
$$ I = \frac{(2n-1)!!}{2^n}\int_{0}^{+\infty}z^{-1/2}e^{-z}\,dz = \sqrt{\pi}\,\frac{(2n-1)!!}{2^n}=\color{red}{\frac{(2n)!}{4^n n!}\sqrt{\pi}}$$
as wanted.
