How to prove this integral? I was confused about an integral showing on my teacher's slide, could anyone tell me how is the following integral derived? 
$$
\int^\infty_{-\infty} x^{2k} e^{-\frac{x^2}{2\sigma^2}} \; \mathrm{d}x = 1 \cdot 3 \; \ldots \; (2k-1) \sigma^{2k+2} \sqrt{2\pi}
$$
Thanks!
 A: Starting with the value of the Gaussian integral, deduce by using the simple substitution $x=t\sqrt a$, 
that $~I(a)=\displaystyle\int_{-\infty}^\infty e^{-ax^2}~dx=\sqrt{\frac\pi a},~$ then differentiate both sides with regard to a, k times. Lastly, 
replace a by $\dfrac1{2\sigma^2},$ and you're done!
A: Hint: Express the integrand as:
$$x^{2k-1}\cdot (x e^{-\frac{x^2}{2\sigma^2}})$$
and apply integration by parts repeatedly. Induction will help a lot.
A: Let $I_k$ be the integral, and write
$$I_k^2 = \int_{-\infty}^{\infty} x^{2k} e^{-\frac{x^2}{2 \sigma^2}} \,dx \int_{-\infty}^{\infty} y^{2k} e^{-\frac{y^2}{2 \sigma^2}} \,dy = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} (x^{2k} + y^{2k}) e^{-\frac{x^2 + y^2}{2 \sigma^2}} \,dx \,dy.$$
In polar coordinates this is
$$\int_0^{2 \pi} (\cos^{2k} \theta + \sin^{2k} \theta)\,d\theta \int_0^{\infty} r^{2k + 1} e^{-\frac{r^2}{2 \sigma^2}} dr.$$
The first integral can be handled with trigonometric identities, and the second be handled with the $u$-substitution $u = \frac{r}{\sqrt{2} \sigma}$ and then $k$ applications of integration by parts (induction makes quicker work of this).
Alternatively, you can apply this trick find $I_0$ and then apply integration by parts $k$ times to write $I_k$ in terms of $I_0$.
A: Hint
To simplify the antiderivative, start writing $x=\sqrt{2} \sigma  y$. This makes $$I_k=\int x^{2k} e^{-\frac{x^2}{2\sigma^2}} \; \mathrm{d}x =2^{k+\frac{1}{2}} \sigma ^{2 k+1} \int y^{2k}e^{-y^2}\; \mathrm{d}y$$ and let $$J_k=\int y^{2k}e^{-y^2}\; \mathrm{d}y$$
Integrating by parts with $u=e^{-y^2}$, $v'=y^{2 k}$, so $u'=-2 e^{-y^2} y$, $v=\frac{y^{2 k+1}}{2 k+1}$, we obtain
 $$J_k=\int y^{2k}e^{-y^2}\; \mathrm{d}y=\frac{e^{-y^2} y^{2 k+1}}{2 k+1}+\int\frac{2 e^{-y^2} y^{2 k+2}}{2 k+1}\; \mathrm{d}y$$ Using the infinite   bounds, the term in the middle disappears and we are left with $$J_k=\frac{2}{2k+1}J_{k+1}$$ that is to say $$J_{k+1}=\frac{2k+1}{2}J_k$$ with $J_0=\sqrt{\pi }$. So, still using the bounds and continuing that way, you will arrive to $$I_k=(2 k-1) \sigma ^2 I_{k-1}$$ with $I_0=\sigma\sqrt{2 \pi}$.
I am sure that you can take from here.
A: Consider the Gaussian integral $\displaystyle F(t):=\frac{1}{\sqrt{2\pi \sigma^2}}\int_{-\infty}^\infty e^{-(1-t)x^2/2\sigma^2}\,dx$. For $t=0$ we have the usual Gaussian integral and so $F(0)=1$. Consequently $F(t)$ can then evaluated via a scaling substitution to obtain yielding $$F(t)=\frac{1}{\sqrt{1-t}}=\sum_{k=0}^\infty \binom{k-1/2}{k}t^k=\sum\limits_{k=0}^\infty \frac{(k-1/2)_{k}}{k!}t^k$$ where we have expressed the Taylor series expansion of the reciprocal square root by means of the falling factorial $(n)_{k}=n(n-1)\cdots (n-k+1)$. At the same time, we can express $F(t)$ as
$$F(t)=\frac{1}{\sqrt{2\pi \sigma^2}}\int_{-\infty}^\infty e^{t x^2/2\sigma^2}e^{-x^2/2\sigma^2}\,dx
=\sum_{k=0}^\infty \frac{t^k}{k!} \cdot \frac{1}{\sqrt{2\pi \sigma^2}}\int_{-\infty}^\infty\,\left(\frac{x^2}{2\sigma^2}\right)^k e^{-x^2/2\sigma^2}\,dx$$
where we have Taylor-expanded the first Gaussian. Identifying coefficients of the two expressions for $F(t)$ then yields the desired conclusion:
$$\int_{-\infty}^\infty\,x^{2k} e^{-x^2/2\sigma^2}=(2\sigma^2)^k(k-1/2)_{k}\sqrt{2\pi \sigma^2}=1\cdot 3\cdots (2k-1)\sigma^{2k+2}\sqrt{2\pi}$$
To clarify the last equality, note that $$2^k (k-1/2)_k=2(k-1/2)\cdot 2(k-3/2)\cdots 2(1/2)=1\cdot 3\cdots (2k-1).$$
