Need help with the integral $\int_{0}^\infty e^{-x^{2}}x^{2n+1}dx $ Problem: Prove that $$I = \int_{0}^\infty  e^{-x^{2}}x^{2n+1}dx = \frac{n!}{2}     $$
Source: A problem I found on an integral test. The problem bugged me for long and I did end up leaving it on the exam. Back home I decided to tackle it again. Here's my go on it.
My try: I have never encountered such a problem before, not even in my assignments and workbooks. Here's how I tried it
We can first take an indefinite integral and then we can work up towards a reduction formula.We can plug in the limits later. I don't know if it's any good but I did write this up on the exam: $$I_{2n+1} = \int{e^{x^{-2}}}{x^{2n+1}}dx$$
On applying Integration by parts:
$$I_{2n+1} = e^{-x^{2}}\frac{x^{2n+2}}{2n+2} +\frac{2}{2n+2} \int{e^{-x^{2}}}{x^{2n+3}}dx$$
or $$I_{2n+1} = e^{-x^{2}}\frac{x^{2n+2}}{2n+2} +\frac{I_{2n+3}}{n+1} $$
or
$$I_{2n+1} = {e^{-x^{2}}\over 2}\Bigl(\frac{x^{2n+2}}{n+1}\Bigl) +\frac{I_{2n+3}}{n+1} $$
What can I do next? Am I doing it incorrectly? Is this the wrong way? Please help me as I can't stop thinking about this problem! Thanks!
Edit 1: As suggested (already tried by me) lets put $u = -x^2$
$$ du = -2xdx$$
plug the above expression in the problem
$$I_{n} = -{1 \over 2}\int{e^{u}}{u^{n}}du$$
$$I_{n} = -{1 \over 2}e^{-u}\frac{u^{n+1}}{n+1} + \frac{I_(n+1)}{n+1}$$
Certainly helps. Now what?
Edit 2: Use the gamma function to get the result. Now unfortunately I didn't read it for the test that's why I didn't get it. Thank you all for pointing out the right direction!
 A: 
I thought it might be instructive to present a way forward that relies on differentiating under the integral.  To that end, we proceed.


Let $I(a)$ be given by the integral 
$$I(a)=\int_0^\infty e^{-ax^2}\,x\,dx=\frac1{2a} \tag1$$
Differentiating $(1)$ $n$ times reveals
$$\begin{align}
I^{(n)}(a)&=(-1)^n\int_0^\infty e^{-ax^2}\,x^{2n+1}\,dx\\\\
&=\frac12 \frac{d^n}{da^n}\left(\frac1a\right)\\\\
&=\frac1{2a^{n+1}}(-1)^n\,n!\tag2
\end{align}$$
Finally, setting $a=1$ in $(2)$, we find that
$$\bbox[5px,border:2px solid #C0A000]{\int_0^\infty e^{-x^2}\,x^{2n+1}\,dx=\frac12\,n!}$$
And we are done!
A: $n\geq 0$ integer.
$\displaystyle J=\int_{0}^\infty e^{-x^{2}}x^{2n+1}dx$
Perform the change of variable $y=x^2$,
$\begin{align}J&=\frac{1}{2}\int_{0}^\infty e^{-x}x^{n}dx\\
&=\frac{1}{2}\Gamma(n+1)\\
&=\boxed{\frac{1}{2} n!}\\
\end{align}$
A: Let's pretend that we have no idea about the Gamma function. Apply a substitution, 
$$t = x^2$$ 
From which,
$$I_n = \int_0^\infty e^{-t} (x^2)^n \cdot \frac x {2x} \mathrm d t$$
$$I_n = \frac 1 2 \int_0^\infty e^{-t} t^n \mathrm dt$$
As we are trying to prove that $2I_n  = n!$, it is sufficient to prove that, from the recursive definition of the factorial,
$$2I_1 = 1$$
$$I_n = nI_{n-1}$$
Can you take it from here with IBP?
A: Let us start from your initial formula
$$ I_{2n+1} = \int{e^{x^{-2}}}{x^{2n+1}}dx $$
and define the definite integral
$$ J_{2n+1} = \int_0^{\infty}{e^{x^{-2}}}{x^{2n+1}}dx$$
If we take the formula you got via integration by parts:
$$I_{2n+1} = e^{-x^{2}}\frac{x^{2n+2}}{2n+2} +\frac{2}{2n+2} \int{e^{-x^{2}}}{x^{2n+3}}dx$$
and evaluate it at it's limits, it reduces to
$$J_{2n+1} = \frac{1}{n+1} \int_0^{\infty}{e^{-x^{2}}}{x^{2n+3}}dx=\frac{J_{n+3}}{n+1}.$$
Your formula now follows from evaluating $J_1$ and induction.  For this purpose, it is useful to rewrite the equation as $J_{2k+1}=kJ_{2(k-1)+1}$.
