I recently started a quantum mechanics course after a long time with no serious maths and I'm having some problems with the most basic maths operations.
Please, help me solve this triple integral (it's a non-graded book exercise and I know that the result should be: $1 \over \sqrt{2\pi}$)
Data: $\begin{align*} \Psi_{2p1}(r,\theta,\phi) &= \sqrt{1 \over{64\pi a^5}}re^{-r \over 2a} \sin\theta ·e^{i\phi}\\ \Psi_{2px}(r,\theta,\phi) &= \sqrt{1 \over{32\pi a^5}}re^{-r \over 2a} \sin\theta \cos\phi\end{align*}$ Demonstrate that: $\left\langle\Psi_{2p1} | \Psi_{2px}\right\rangle= \frac{1}{\sqrt{2\pi}}$
The actual integral to solve is: $ \int\limits_{0}^\infty \int\limits_{0}^\pi \int\limits_{0}^{2\pi}{\Psi_{2p1}^*\Psi_{2px}r^2\sin\theta\,\mathrm d\phi\,\mathrm d\theta\,\mathrm dr}$
Thanks!
My try: $\left\langle\Psi_{2p1} | \Psi_{2px}\right\rangle = \sqrt{1 \over{2^{11}\pi^2 a^{10}}}\int\limits_{0}^\infty \int\limits_{0}^\pi \int\limits_{0}^{2\pi}r^4e^{-r \over a} \sin^2\theta \cos\phi e^{i\phi} \,\mathrm d\phi\,\mathrm d\theta\,\mathrm dr $ $ ={1 \over{2^{5}\pi a^{5}\sqrt{2}}}\int\limits_{0}^\pi\int\limits_{0}^{2\pi} [(-a /5r)r^5e^{-r \over a}]_o^\infty \sin^2\theta \cos\phi e^{i\phi} \,\mathrm d\phi\,\mathrm d\theta$
Is it right so far? How do I continue?