I was trying to solve the quantum harmonic oscillator problem. It is almost done, just the normalization of the wave function is left.
$Ψ_n = β_n\mathcal{H}_n(\frac{x}{\alpha})e^{\frac{-x^2}{2\alpha^2}}$;
$\mathcal{H}_n$ is the physicist's hermite polynomial, $\alpha$ is a constant.
For normalizing, I need to find the constant $β_n$.
Using the law of total probability;
$\int_{-\infty}^{+\infty}\psi^*\psi dx = 1$
$\Rightarrow \int_{-\infty}^{+\infty} [β_n\mathcal{H}_n(\frac{x}{\alpha})e^{\frac{-x^2}{2\alpha^2}}]^2 dx= 1$
$\Rightarrow \beta_n^2 \int_{-\infty}^{+\infty} \alpha[\mathcal{H}_n(\frac{x}{\alpha})e^{\frac{-x^2}{2\alpha^2}}]^2 \frac{dx}{\alpha} = 1$
$\Rightarrow \beta_n = \frac{1}{\sqrt{\int_{-\infty}^{+\infty} \alpha[\mathcal{H}_n(\frac{x}{\alpha})e^{\frac{-x^2}{2\alpha^2}}]^2 \frac{dx}{\alpha}}}$
I don't know how to start with the solving of the integral in the equation.
But, I am really interested to know how the integral is solved and the value of $\beta_n$ is derived.
From the orthogonality relation of said polynomials, $$ \beta_n = \frac{1}{\sqrt{{\alpha}\sqrt{\pi} 2^n n! }}~~.$$