Normalizing a 3D Wavefunction

We have that $\displaystyle \phi_{(1,0,0)}(\overrightarrow{r}) = C \exp{ \left( \dfrac{-Zr}{a_B} \right) } \ \implies \int_{-\infty}^{\infty} {|\phi_{(1,0,0)}{(\overrightarrow{r})|^2} dr} = 1$ .

Hence, we proceed with spherical coordinates to get the following mess: $\displaystyle \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} C^2 \exp{\left( \dfrac{ -2Z\sqrt{x^2+y^2+z^2}}{a_B} \right)} dxdydz = \int_0^\pi \int_0^{2\pi} \int_0^{\infty} C^2 \exp{\left( \dfrac{-2Z\sqrt{\rho^2\sin^2{\phi}\cos^2{\theta} + \rho^2\sin^2{\phi}\sin^2{\theta} + \rho^2\cos^2{\phi}}}{a_B} \right)} \rho^2 \sin{\phi} d\rho d\theta d\phi.$ Factoring and the Pythagorean identity yields:

$\ \displaystyle C^2 \int_0^\pi \int_0^{2\pi} \int_0^{\infty} \exp{\left( -\dfrac{2Z\rho}{a_B} \right) }\rho^2 \sin{\phi} d\rho d\theta d\phi = \ C^2\pi \left( \dfrac{a_B}{Z} \right)^3 = \ 1$ .

Since $C = \alpha \left( \dfrac{Z}{a_B} \right)^{3/2}$ , we see that $\alpha = \sqrt{\dfrac{1}{\pi}} \approx \ \boxed{0.564}$ .