- #1
clope023
- 992
- 131
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.
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.