# Integrate (x^2)exp(-x^2)dx from - to + infinity

Hello, I'm doing a quantum mechanics problem from Griffiths chapter 1 where I'm trying to find the expectation value of x^2 given the function ρ(x) = Aexp[-λ(x-μ)^2].
The problem is in the title, but I'll rewrite it here simplified:

∫(x^2)exp(-x^2)dx

Subbing u = x^2 <=> x = sqrt(u) => dx = -0.5(u^(-1/2))du which becomes

-0.5∫(u/sqrt(u))exp(u)du = -0.5∫ sqrt(u)exp(u)du

So from here I tried integration by parts, u = sqrt(u), du = -0.5u^(-3/2)du,
dv = exp(u)du, v = exp(u)

uv - ∫vdu = sqrt(u)exp(u) + 0.5∫u^(-3/2)exp(u)du

From here I equatted this with -0.5∫sqrt(u)exp(u)du and noted

u^(-3/2) = (u^(-1/2))*(u^-1) = (u^(-1/2))/u

so then I did the algebra and got:

∫u^(-1/2)exp(u)du = u^(-1/2)(u/(u-1))exp(u) = sqrt(u)exp(u)/(u-1)

Am I on the right track? I am aware I can do this with the complementary error function or the gamma function but I'm trying to avoid using them. Any help is appreciated thank you.

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Ray Vickson
Homework Helper
Dearly Missed
Hello, I'm doing a quantum mechanics problem from Griffiths chapter 1 where I'm trying to find the expectation value of x^2 given the function ρ(x) = Aexp[-λ(x-μ)^2].
The problem is in the title, but I'll rewrite it here simplified:

∫(x^2)exp(-x^2)dx

Subbing u = x^2 <=> x = sqrt(u) => dx = -0.5(u^(-1/2))du which becomes

-0.5∫(u/sqrt(u))exp(u)du = -0.5∫ sqrt(u)exp(u)du

So from here I tried integration by parts, u = sqrt(u), du = -0.5u^(-3/2)du,
dv = exp(u)du, v = exp(u)

uv - ∫vdu = sqrt(u)exp(u) + 0.5∫u^(-3/2)exp(u)du

From here I equatted this with -0.5∫sqrt(u)exp(u)du and noted

u^(-3/2) = (u^(-1/2))*(u^-1) = (u^(-1/2))/u

so then I did the algebra and got:

∫u^(-1/2)exp(u)du = u^(-1/2)(u/(u-1))exp(u) = sqrt(u)exp(u)/(u-1)

Am I on the right track? I am aware I can do this with the complementary error function or the gamma function but I'm trying to avoid using them. Any help is appreciated thank you.
You cannot avoid the error function, as it occurs in one term of the answer. Put it another way: if you could find a closed-form for the integral, involving only elementary functions, you could also find a closed, elementary form for the error function, and this is provably impossible.

RGV

gabbagabbahey
Homework Helper
Gold Member
You cannot avoid the error function, as it occurs in one term of the answer.
Sure you can. The definite integral $\int_{-\infty}^{\infty} x^2 e^{-x^2} dx$ (which from the thread title is what is actually desired) can be computed by applying integration by parts once with $u=x$ and $dv=x e^{-x^2} dx$, and then computing $\int_{-\infty}^{\infty} e^{-x^2} dx$ in the usual way (multiplying by $\int_{-\infty}^{\infty} e^{-y^2} dy$ to give the square of the integral, converting this to a 2D integral in polar coordinates to easily compute it, and then taking the square root)

Ray Vickson
Homework Helper
Dearly Missed
Sure you can. The definite integral $\int_{-\infty}^{\infty} x^2 e^{-x^2} dx$ (which from the thread title is what is actually desired) can be computed by applying integration by parts once with $u=x$ and $dv=x e^{-x^2} dx$, and then computing $\int_{-\infty}^{\infty} e^{-x^2} dx$ in the usual way (multiplying by $\int_{-\infty}^{\infty} e^{-y^2} dy$ to give the square of the integral, converting this to a 2D integral in polar coordinates to easily compute it, and then taking the square root)
Of course the definite integral is doable---but not by the methods the OP posted. In the main body of the post the OP seemed to be wanting the indefinite integral, and I was responding to that.

RGV

Mute
Homework Helper
Do you know the result

$$I(\alpha) = \int_{-\infty}^\infty dx~e^{-\alpha x^2} = \sqrt{\frac{\pi}{\alpha}}?$$

If you are allowed to use that result, you can find the values of the averages of $x^{2n}$ by supposing $\alpha$ is actually a variable parameter and taking derivatives with respect to it. e.g.,

$$I'(\alpha) = -\int_{-\infty}^\infty dx~x^2e^{-\alpha x^2} = -\langle x^2 \rangle_\alpha,$$
where I've used a subscript $\alpha$ to denote that it's still a variable, and you have to set $\alpha$ to be whatever it is supposed to be in your original integral.

This is by far the slickest way to get Gaussian averages, in my opinion.

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