Limit as n-> infinity of Integral of Gaussian Multiplied by Polynomial of Order n

Question

Can anyone tell me the result of the following limit or, at least, if it converges (n integer)? $$\lim_{n\to\infty} \int_{-\infty}^{\infty} x^{2n}e^{({-1/2})ax^2}dx$$ I believe the integral evaluates to $\sqrt{(2\pi/a)}\dfrac{(2n)!}{(2a)^nn!}$

Thanks.

Answer 1

With the change of variable $ax^2/2=t$ we have that

$$\int_{-\infty}^\infty x^{2n}e^{-\frac{ax^2}2}\mathrm dx=2\left(\frac2a\right)^{n-1/2}\int_0^\infty t^{n-1/2}e^{-t}\mathrm dt=2\left(\frac2a\right)^{n-1/2}\Gamma(n+1/n)$$

Now, by the reflection formula $\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin(\pi z)}$ and the formula $\Gamma(z+1)=z\Gamma(z)$ we knows that $\Gamma(1/2)=\sqrt\pi$, and by induction

$$\Gamma(n+1/2)=\sqrt{\pi}\frac{(2n-1)!!}{2^n}$$

where $(2n-1)!!:=1\cdot 3\cdot 5\cdots(2n-1)$. Then, rearranging all, we finally have that

$$I_n(a):=\int_{-\infty}^\infty x^{2n}e^{-\frac{ax^2}{2}}\mathrm dx=\sqrt{\frac{2\pi}a}\cdot\frac{(2n-1)!!}{a^n}$$

but observe, by the archimedean principle of $\Bbb R$, that exist some $N\in\Bbb N$ such that $N>a$ for any (fixed) $a\in\Bbb R$. Then

$$\frac{(2m-1)!!}{a^{m+1}}=\frac{(2(m-1)-1)!!}{a^m}\cdot\frac{2m-1}{a}$$

where $(2m-1)/a>1$ if $m\ge N$. Then, by an induction argument, we can see that $$|I(a)|:=\lim_{n\to\infty}|I_n(a)|=\infty,\quad a\in\Bbb R\setminus\{0\}$$

For $a\in\Bbb C\setminus\Bbb R$ the limit is not defined, notice that if we define $a=re^{i\theta}$ then $a^n=r^n\cdot e^{in\theta}$, and the limit $$\lim_{n\to\infty}r^n\cdot e^{in\theta}=\lim_{n\to\infty}r^n\cdot (\cos(n\theta)+i\sin(n\theta))$$ doesnt exists.

Answer 2

Let $I_n$ be the integral. Then

$$I_n > \int_2^\infty 2^{2n} e^{(-1/2) a x^2} dx\to \infty. $$

Answer 3

There are a multitude of ways to show that the limit does not exist, such as the one provided by @Masacroso . Here is another... Let $W$ be a standard normal random variable. Then the given integral can be written as
\begin{align} I_n = \sqrt\frac{2\pi}{a} \;\;\mathbb{E}[W^{2n}] \end{align} Using Markov's inequality, for any $k >0$, and in particular for $k=2$, we have $$\mathbb{E}[W^{2n}] \geq k^{2n}\; \mathbb{P}(|W| \geq k ) = 4^{n}\; \mathbb{P}(|W| \geq 2 ) \longrightarrow \infty \textrm{ as } n \to \infty$$ since $\mathbb{P}(|W| \geq 2 ) \in (0,1)$.