Compute integral $\int_{-\infty}^{\infty} x^{4n} e^{-x^2/2} \, dx$

Question

While solving a more general problem, I came across the following integral

$$I= \int_{-\infty}^{\infty} x^{4n} e^{-x^2/2} \, dx$$

To start off, I defined $t:=x^2/2$ to get

\begin{align*} I= \int_{-\infty}^\infty x^{4n} e^{-x^2/2} \, dx &= 2\int_0^\infty x^{4n} e^{-x^2/2} \, dx \\ &= 2 \int_0^\infty (\sqrt{2t})^{4n} e^{-t} \frac{dt}{\sqrt{2t}} \\ &= 2 (\sqrt{2})^{4n - 1} \int_0^\infty t^{2n - 1/2} e^{-t} \, dt \end{align*}

This seems to be a gamma function integral. I would start integrating by parts

\begin{align*} \int_0^{+\infty} t^{2n-\frac{1}{2}}e^{-t} \, dt&=[ -t^{2n-\frac{1}{2}}e^{-t}]_0^{+\infty} - (-)(2n-\frac{1}{2})\int_0^{+\infty}t^{2n-\frac{1}{2}-1}e^{-t} \, dt \\ &=(2n-\frac{1}{2})\int_0^{+\infty}t^{2n-\frac{1}{2}-1}e^{-t} \, dt \end{align*}

But how to proceed further?

I looked up in a table and the result should be

$$I=2 (\sqrt{2})^{4n - 1} \Gamma\left(2n+ \frac 1 2 \right)$$

PS: Alternative methods avoiding the gamma function are very welcomed! :)

EDIT: I think that what Yves suggested was the following

$$I= \int_{-\infty}^\infty x^{4n} e^{-x^2/2}= (2n-\frac{1}{2}) (2n-\frac{3}{2}) (2n-\frac{5}{2})\cdots\frac{3}{2}\frac{1}{2} \int_0^{+\infty}t^{-\frac{1}{2}}e^{-t}\,dt=(2n-\frac{1}{2}) (2n-\frac{3}{2}) (2n-\frac{5}{2})\cdots\frac{3}{2}\frac{1}{2}\ \ 2\int_0^{+\infty}e^{-t} \, d(\sqrt{t})$$

Alrigtht! I see that $(2n-\frac{1}{2}) (2n-\frac{3}{2}) (2n-\frac{5}{2})\cdots\frac{3}{2}\frac{1}{2} = \Gamma\left(2n+ \frac 1 2 \right)$ NAIVE question though: how to see that $\int_0^{+\infty}e^{-t} \, d(\sqrt{t})=2 (\sqrt{2})^{4n - 1}$?

Answer 1

You're really done after the first integration by parts: as $\displaystyle \Gamma(z) = \int_0^\infty x^{z-1} e^{-x}\,dx,$

$\displaystyle 2 (\sqrt{2})^{4n - 1} \int_{0}^{\infty} t^{2n -\frac{1}{2}} e^{-t} dt =2 (\sqrt{2})^{4n - 1} \int_{0}^{\infty} t^{2n + \frac{1}{2}-1} e^{-t} dt =2 (\sqrt{2})^{4n - 1} \Gamma\left(2n+ \frac 1 2 \right).$

In other words, your integral is the gamma function with $z = 2n+\frac{1}{2}$.

Regarding the other method, you have:

You've shown that if $\displaystyle I_n = \int_0^{\infty} t^{2n-\frac{1}{2}}\,{dt} $ then $I_n = \left(2n-\frac{1}{2}\right)I_{n-1}$ so that $$I_n =(2n-\frac{1}{2}) \cdot (2n-\frac{3}{2}) \cdot (2n-\frac{5}{2})\cdots\frac{3}{2} \cdot I_1 $$

Where $\displaystyle I_1 = \int_0^{+\infty}t^{\frac{1}{2}}e^{-t}\,dt$, then $t = x^2$ shows that
$$\displaystyle I_1 = 2 \int_0^{\infty} x^2 e^{-x^2}\,{dx}. $$

Which evaluates to $\displaystyle \frac{1}{2}\sqrt{\pi}$ (see here). Therefore

$$\begin{aligned} I &= \sqrt{\pi} \cdot (2n-\frac{1}{2}) \cdot (2n-\frac{3}{2}) \cdot (2n-\frac{5}{2})\cdots\frac{3}{2} \cdot \frac{1}{2}\\& =\frac{\sqrt{\pi}}{2^{2n}}(4n-1) (4n-3) (4n-5)\cdots 3 \cdot 1 \\& =\frac{\sqrt{\pi}}{2^{2n}}\frac{(4n-1)(4n-2) (4n-3)(4n-4) (4n-5)\cdots 1}{(4n-2) (4n-4) \cdots 2 } \\& =\frac{\sqrt{\pi}}{2^{2n}}\frac{(4n-1)(4n-2) (2n-3)(2n-4) (2n-5)\cdots 3 \cdot \cdot 2 \cdot 1}{2^{2n} (2n-1) (2n-2) \cdots 1 } \\& = \frac{(4n)!\sqrt{\pi}}{2^{4n}(2n)!}.\end{aligned} $$

Therefore your original integral is $\displaystyle I =2 (\sqrt{2})^{4n - 1} \frac{(4n)!\sqrt{\pi}}{2^{4n}(2n)!}.$

This agrees with $\displaystyle I = 2 (\sqrt{2})^{4n - 1} \Gamma\left(2n+ \frac 1 2 \right)$ as $\displaystyle \Gamma\left(2n+ \frac 1 2 \right) = \frac{(4n)!\sqrt{\pi}}{2^{4n}(2n)!}$ (see here).

Answer 2

you can use the known formula (in any case very easy to prove) of the odd simple moments of the Standard Gaussian

$$\mathbb{E}[X^{2n}]=\frac{(2n)!}{2^n\cdot n!}$$

Thus you integral results to me

$$\frac{\sqrt{2\pi}(4n)!}{4^n\cdot(2n)!}$$

To prove that this result match with the one you expected, just observe that

$$\frac{\sqrt{2\pi}(4n)!}{4^n\cdot(2n)!}=2^{2n+1/2}\cdot\Gamma\left(2n+\frac{1}{2}\right)$$

---

Your integral is the following

$$I=\sqrt{2\pi}\int_{-\infty}^{\infty} x^{2m}\phi(x)dx$$

where

• $m=2n$

• $\phi(x)$ is the density of a Standard Gaussian, that is a Gaussian distribution with mean zero and variance 1.

• now you have only to apply the known formula for the simple odd moments I showed. The proof is available in many basic Statistics Textbook and it is very easy: it is enough to calculate the Taylor expansion of MGF and get the derivatives in $t=0$

Answer 3

Note that

$$I(a)= \int_{-\infty}^{\infty} e^{-a x^2} \, dx=\sqrt{\frac \pi a} $$

Then

$$\int_{-\infty}^{\infty} x^{4n} e^{-x^2/2} \, dx = \frac{d^{2n} I(a)}{da^{2n}}\bigg|_{a=\frac12}= \sqrt{\pi}\cdot \frac{d^{2n} (a^{-\frac12})}{da^{2n}}\bigg|_{a=\frac12} = \frac{(4n-1)! \sqrt{2\pi}}{2^{2n-1}(2n-1)!} $$