Please show $\int_0^\infty x^{2n} e^{-x^2}\mathrm dx=\frac{(2n)!}{2^{2n}n!}\frac{\sqrt{\pi}}{2}$  without gamma function?  Prove:
$$\int_0^\infty x^{2n} e^{-x^2}\mathrm dx=\frac{(2n)!}{2^{2n}n!}\frac{\sqrt{\pi}}{2}$$
Thanks!
 A: Alternatively, set $$I(\alpha) = \int_0^\infty e^{-\alpha x^2}\mathrm{d}x,$$ differentiate $n$ times with respect to $\alpha$ and evaluate at $\alpha = 1$.
EDIT:
To spell things a little more out, this technique is known as Differentiation under the integral sign.
Using the fact that $I(\alpha) =\frac12\sqrt{\frac{\pi}{\alpha}}$ and differentiating to obtain $$\frac{\mathrm{d}^n}{\mathrm{d}\alpha^n} I(\alpha) = (-1)^n\int_0^\infty x^{2n} e^{-\alpha x^2}\mathrm{d}x, $$
some algebraic manipulation and evaluating at $\alpha = 1$ will yield the wanted identity.
A: Alternatively, integration by parts works immediately.
Let $$a_n=\int_0^\infty x^{2n}e^{-x^2}.$$  Consider $U=x^{2n-1}$ so that $du=(2n-1)x^{2n-2}$, and $dv=xe^{-x^{2}}$ so that $V=-\frac{1}{2}e^{-x^{2}}$. 
Then
$$\int_{0}^{\infty}x^{2n}e^{-x^{2}}dx=\frac{1}{2}e^{-x^{2}}x^{2n-1}\biggr|_{0}^{\infty}-\int_{0}^{\infty}(2n-1)x^{2n-2}\frac{-1}{2}e^{-x^{2}}dx$$ 
$$=\frac{(2n-1)}{2}\int_{0}^{\infty}x^{2n-2}e^{-x^{2}}dx=\frac{(2n-1)2n}{2^{2}n}\int_{0}^{\infty}x^{2n-2}e^{-x^{2}}dx$$
Hence $$a_n=\frac{(2n)(2n-1)}{2^2n}a_{n-1}$$ and since $a_0=\frac{\sqrt{\pi}}{2}$ we conclude $$a_n=\int_{0}^{\infty}x^{2n}e^{-x^{2}}dx=\frac{(2n)!}{2^{2n}n!}\frac{\sqrt{\pi}}{2}$$ by induction.
Hope that helps,
A: Making a change of variable $u=x^2$ gives
$$
\int_0^\infty  {x^{2n} e^{ - x^2 } dx}  = \frac{1}{2}\int_0^\infty  {u^{n - 1/2} e^{ - u} du}  = \frac{1}{2}\Gamma (n + 1/2).
$$
Then from the well-known formula for the gamma function
$$
\Gamma (n + 1/2) = \frac{{(2n)!}}{{4^n n!}}\sqrt \pi  
$$
we get
$$
\int_0^\infty  {x^{2n} e^{ - x^2 } dx}  = \frac{{(2n)!}}{{2^{2n} n!}}\frac{{\sqrt \pi  }}{2}.
$$
Second approach. Writing
$$
\int_0^\infty  {x^{2n} e^{ - x^2 } dx}  = \frac{1}{2} \frac{{\sqrt {2\pi (1/2)} }}{{\sqrt {2\pi (1/2)} }} \int_{ - \infty }^\infty  {x^{2n} \exp \bigg( - \frac{{x^2 }}{{2(1/2)}}\bigg)dx} 
$$
shows that
$$
\int_0^\infty  {x^{2n} e^{ - x^2 } dx}  = \frac{{\sqrt \pi  }}{2}{\rm E}[X^{2n} ],
$$
where ${\rm E}[X^{2n} ]$ is the $2n$-th moment of the Normal$(0,1/2)$ distribution.
Now you can see how others, here for example, find ${\rm E}[X^{2n} ]$ (for a Normal$(0,\sigma^2)$ distribution).
The simplest approach may be to use integration by parts. I'll leave it to you.
EDIT (in light of the OP's edit): Integration by parts gives the result, using induction, as follows:
$$
\int_0^\infty  {x^{2(n + 1)} e^{ - x^2 } dx}  = \frac{{2n + 1}}{2}\int_0^\infty  {x^{2n} e^{ - x^2 } dx}  = \frac{{[2(n + 1)]!}}{{2^{2(n + 1)} (n + 1)!}}\frac{{\sqrt \pi  }}{2}.
$$
For the base case $n=0$, note that $\int_0^\infty  {e^{ - x^2 } dx}  = \frac{{\sqrt \pi  }}{2}$.
A: Let's suppose that, one way or another, you know that $ \displaystyle \int_{-\infty}^{\infty} e^{-x^2} \, dx = \sqrt{\pi}$. Then
$$\int_{-\infty}^{\infty} e^{2tx - x^2} dx= e^{t^2} \int_{-\infty}^{\infty} e^{-(t - x)^2} \, dx = e^{t^2} \sqrt{\pi}.$$
On the other hand,
$$\int_{-\infty}^{\infty} e^{2tx - x^2} dx = \int_{-\infty}^{\infty} \left( \sum_{n \ge 0} \frac{2^n t^n x^n}{n!} \right) e^{-x^2} dx = \sum_{n \ge 0} \frac{2^n t^n}{n!} \int_{-\infty}^{\infty} x^n e^{-x^2} \, dx.$$
Finally, note that by evenness, 
$$\int_{-\infty}^{\infty} x^{2n} e^{-x^2} \, dx = 2 \int_0^{\infty} x^{2n} e^{-x^2} \, dx.$$ 
