integration - triple integral (quantum mechanics)

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?

Answer

Start from

$$\left\langle\Psi_{2p1} | \Psi_{2px}\right\rangle = \sqrt{1 \over{2^{11}\pi^2 a^{10}}}\int\limits_{0}^\infty \int\limits_{0}^\pi \int\limits_{0}^{2\pi}r^4e^{-r \over a} \sin^3\theta \cos\phi e^{i\phi}~ \mathrm d\phi~\mathrm d\theta~\mathrm dr $$
(which, note, is not the same as the expression in the question statement; thanks @joriki.)

1. Expand the $e^{i\phi}$ term using Euler's formula to be $\cos\phi + i \sin\phi$. If you multiply this against the $\cos\phi$ factor already in the expression, and integrate from 0 to $2\pi$, you see that the term $\sin\phi\cos\phi$ integrate to zero (why?) and that $\cos^2\phi$ integrate to some constant (why? And I'll leave it to you to compute that constant yourself).





2. Now the innermost integral is taken care of, you can integrate the term $\sin^3\theta$ from $0$ to $\pi$. This gives you another constant (what is it?).




3. Lastly, you need to evaluate the $r$ integral. Note that after the previous two steps you are left with something that looks like
   $$ \text{Constant}\cdot \frac{1}{a^5} \int_0^\infty r^4 e^{-\frac{r}{a}} \mathrm{d}r $$
   Now, you can rewrite it as
   $$ \text{Constant}\cdot \int_0^\infty \left(\frac{r}{a}\right)^4 e^{-\frac{r}{a}} \mathrm d \left(\frac{r}{a}\right) $$
   so doing the change of variables $\rho = r/a$, your integral becomes
   $$ \text{Constant}\cdot \int_0^\infty \rho^4 e^{-\rho} \mathrm{d}\rho $$
   This you can solve simply by repeated integration by parts, or by appealing to the Gamma function (whose values at positive integers are explicitly known).