# Calculating the expectation value for kinetic energy $\langle E_k \rangle$ for a known wave function

I have a wavefunction ($a=1nm$):

$$\psi=Ax\exp\left[\tfrac{-x^2}{2a}\right]$$

for which I already calculated the normalisation factor (in my other topic):

$$A = \sqrt{\frac{2}{a\sqrt{\pi a}}} = 1.06\frac{1}{nm\sqrt{nm}}$$

What I want to know is how to calculate the expectation value for a kinetic energy. I have tried to calculate it analyticaly but i get lost in the integration:

\begin{align} \langle E_k \rangle &= \int\limits_{-\infty}^{\infty} \overline\psi\hat{T}\psi \,dx = \int\limits_{-\infty}^{\infty} Ax \exp \left[{-\tfrac{x^2}{2a}}\right]\left(-\tfrac{\hbar^2}{2m}\tfrac{d^2}{dx^2}Ax \exp \left[{-\tfrac{x^2}{2a}}\right]\right)\,dx =\dots \end{align}

At this point I go and solve the second derivative and will continue after this:

\begin{align} &\phantom{=}\tfrac{d^2}{dx^2}Ax \exp \left[{-\tfrac{x^2}{2a}}\right] = A\tfrac{d^2}{dx^2}x \exp \left[{-\tfrac{x^2}{2a}}\right]= A\tfrac{d}{dx}\left(\exp \left[{-\tfrac{x^2}{2a}}\right]-\tfrac{2x^2}{2a}\exp \left[{-\tfrac{x^2}{2a}}\right]\right)= \\ &=A \left(-\tfrac{2x}{2a}\exp \left[{-\tfrac{x^2}{2a}}\right] - \tfrac{1}{a}\tfrac{d}{dx}x^2\exp \left[{-\tfrac{x^2}{2a}}\right]\right) = \\ &=A \left(-\tfrac{x}{a}\exp \left[{-\tfrac{x^2}{2a}}\right] - \tfrac{2x}{a}\exp \left[{-\tfrac{x^2}{2a}}\right] + \tfrac{x^3}{a^2}\exp \left[{-\tfrac{x^2}{2a}}\right]\right) = \\ &= A \left(-\tfrac{3x}{a}\exp \left[{-\tfrac{x^2}{2a}}\right] + \tfrac{x^3}{a^2}\exp \left[{-\tfrac{x^2}{2a}}\right]\right) \end{align}

Ok so now I can continue the integration:

\begin{align} \dots &= \int\limits_{-\infty}^{\infty} Ax \exp \left[{-\tfrac{x^2}{2a}}\right]\left(-\tfrac{\hbar^2}{2m} A \left(-\tfrac{3x}{a}\exp \left[{-\tfrac{x^2}{2a}}\right] + \tfrac{x^3}{a^2}\exp \left[{-\tfrac{x^2}{2a}}\right]\right)\right)\,dx = \\ &= \int\limits_{-\infty}^{\infty} -\frac{A^2\hbar^2}{2m}x\exp\left[-\tfrac{x^2}{2a}\right] \left(-\tfrac{3x}{a}\exp \left[{-\tfrac{x^2}{2a}}\right] + \tfrac{x^3}{a^2}\exp \left[{-\tfrac{x^2}{2a}}\right]\right) \,dx\\ &= \int\limits_{-\infty}^{\infty} \frac{A^2\hbar^2}{2m}\left(\tfrac{3x^2}{a}\exp \left[{-\tfrac{x^2}{a}}\right] - \tfrac{x^4}{a^2}\exp \left[{-\tfrac{x^2}{a}}\right]\right) \,dx\\ &= \frac{A^2\hbar^2}{2m} \underbrace{\int\limits_{-\infty}^{\infty}\left(\tfrac{3x^2}{a}\exp \left[{-\tfrac{x^2}{a}}\right] - \tfrac{x^4}{a^2}\exp \left[{-\tfrac{x^2}{a}}\right]\right) \,dx}_{\text{How do i solve this?}}=\dots\\ \end{align}

This is the point where I admited to myself that I was lost in an integral and used the WolframAlpha to help myself. Well I got a weird result. My professor somehow got this ($m$ is a mass of an electron) but I don't know how:

\begin{align} \dots = \frac{\hbar^2}{2m}\cdot\frac{3}{2a} = \frac{3\hbar^2}{4ma} = 0.058eV \end{align}

Can anyone help me to understand the last integral? How can I solve it? Is it possible analyticaly (it looks like professor did it, but i am not sure about it)?