# Awkward integral?

1. Dec 23, 2008

### philip041

I can't understand what my lecturer has done, it looks like he has replaced d^2/dx^2 with an identity but I'm not sure.

$$A^2\int^{\infty}_{-\infty}{dxe^{-1/2\alpha^2x^2}}\cdot\left(\frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\right)e^{-1/2\alpha^2x^2}$$

he gets it to equal this, (but the stuff in brackets I have no idea where it came from!)

$$A^2\cdot\frac{\hbar^2}{2m}\int^{\infty}_{-\infty}{dx\left(\alpha^2} - \alpha^4 x^2\right)e^{-\alpha^2x^2}$$

he then goes on to do this but I think if I understood the first step this bit would be ok, but here you go anyway..

$$A^2\cdot\frac{\hbar^2}{2m}\left(\alpha^2\frac{\pi^{1/2}}{\alpha} - \alpha^4\frac{\pi^{1/2}}{2\alpha^3}\right)$$

is there an identity im missing? also where has the minus which was in front of hbar^2 at the beginning gone?

2. Dec 23, 2008

### George Jones

Staff Emeritus
What does

$$e^{-\frac{1}{2} \alpha^2x^2}} \left(\frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\right)e^{-\frac{1}{2} \alpha^2x^2}$$

equal?

3. Dec 23, 2008

### HallsofIvy

Staff Emeritus
He hasn't "replaced d^2/dx^2 with an identity", he has done the indicated second derivative.