# Integral of x^n e^(-x^2)

1. Oct 9, 2011

### lol_nl

1. The problem statement, all variables and given/known data
Doing calculations for quantum mechanics, I have stumbled upon several integrals which I cannot solve by hand.
All of these have the form:
$$\int x^n exp[-x^2] dx$$
For instance, when calculating the expectation value for x^2, the integral reduces to
$$\int_{-\infty}^{\infty} x^2 exp[-x^2] dx$$
multiplied by some constant factors.

2. Relevant equations
There is a standard integral given in the back of my book (Griffiths):
$$\int_{0}^{\infty} exp[-\frac{x^2}{a^2}] dx = n! a^{n-1}$$
However, when I apply this to n=2, I arrive at a different answer than when I use Mathematica to compute it.

Also, when I try WolframAlpha, it returns a function that involves the error function without explaining how it got there.
3. The attempt at a solution

There seems to be no way of simplifying or rearranging the integral. Integration by parts only works for taking the derivative of the exponential part, but that leaves higher and higher orders of n. Making substitution for x^2 still leaves a factor x in the denominator since u = x^2 implies du = 2 x dx. The result then would give a square root of u in the integral, which I cannot solve by integration by parts.
Is there any way to find an explicit formula for this integral?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Oct 9, 2011

### SteamKing

Staff Emeritus
For the case where n = 2, I would recommend u = x and dv = x * EXP (-x^2) for integration by parts.

IIRC, some of the integrals, like for n = 2, have closed form solutions. Others, however, must be evaluated numerically.

3. Oct 9, 2011

### lol_nl

Thanks. Hence I arrive at the answer for n=2:
$$\int_{-\infty}^{\infty} x^2 exp(-x^2) dx = 1/2 \int_{-\infty}^{\infty} exp(-x^2) dx - 1/2 [exp(-x^2)]_{-\infty}^{\infty} = \sqrt{\pi}/2$$

4. Oct 9, 2011

### kloptok

Integrals of the form you describe are particularly simple when n is odd. You can then use symmetry arguments to solve the integral by inspection. Do you follow me? (Hint: think of the integrand, is it even/odd?)

5. Oct 9, 2011

### lol_nl

The exponential factor is the gaussian distribution and is even. If n is odd, x^n is odd and therefore the integrand is odd. Integrating from -infinity to infinity will then always give 0.

6. Oct 9, 2011

### lol_nl

Also, I have noticed that for higher powers of n, in principle you van always use integration by parts to reduce the power by 2 each step, so n=4 is easily computed.
Side question: why is
$$[x^n exp(-x^2)]_{-\infty}^{\infty} = 0$$?
Or with other words, why is
$$\lim_{x \rightarrow \infty} x^n exp(-x^2) = 0$$?

7. Oct 9, 2011

### vela

Staff Emeritus
The exponential dies off faster than any polynomial grows, so it always wins in the limit. You can show the limit is 0 by simply applying the Hospital rule repeatedly to xn/e-x2. When you differentiate enough times, the numerator will become 0, but the exponential will still be around.

8. Oct 9, 2011

### vela

Staff Emeritus
By the way, another trick you can use to evaluate integrals like this is to note that
$$x^{2n}e^{-\alpha x^2} = (-1)^n\frac{d^n}{d\alpha^n}e^{-\alpha x^2}$$so that
$$\int_{-\infty}^\infty x^{2n}e^{-\alpha x^2}\,dx = \int_{-\infty}^\infty (-1)^n\frac{d^n}{d\alpha^n}e^{-\alpha x^2}\,dx = (-1)^n\frac{d^n}{d\alpha^n}\int_{-\infty}^\infty e^{-\alpha x^2}\,dx$$At the end, set $\alpha = 0$ to get the result you want.