# Closed form for the integral of a squared Legendre function

Is there a closed form for the integral $$\int_0^{\pi/2}(P_\nu^\mu(\cos\theta))^2\,\mathrm d\theta,\quad\mu>\nu\gt-\frac12$$ where $$P_\nu^\mu(x)$$ is the associated Legendre function of the first kind?

I encountered this integral while trying to derive explicit solutions for a certain Sturm-Liouville problem. I am primarily interested in $$\mu,\nu$$ being nonnegative integers, but a result that is valid for real $$\mu,\nu$$ (subject to the above restriction) is very much welcome.

Neither Maple nor Mathematica seem to be able to make a dent on this integral, but I was able to at least confirm that for $$\mu,\nu$$ an integer, I get results that are rational multiples of $$\pi$$, which makes me believe there ought to be a (simple?) closed form, perhaps involving gamma functions.

I wasn't able to find anything in G&R or the DLMF that resembles this integral, so I am really stuck, and would appreciate any ideas on resolving this.

• Using its recurrence formulas it boils down to evaluating the overlap $\int_{-1}^1P_{\nu}^\mu(x) P_{\nu-1}^{\mu\pm1}(x) \mathrm{d} x$ of two associated Legendre polynomials. Perhaps this source helps? – Timothy Budd Jul 20 '20 at 17:16
• Your link refers to formula 14.3.6, which holds for $x > 1$ (while $\cos\theta\leq 1$). Do you actually mean formula 14.3.1, that is Ferrers rather than Legendre functions? – Max Alekseyev Jul 21 '20 at 2:50
• Also, for integer parameters $\nu,\mu$, associated Legendre function $P_\nu^\mu(x)$ turns into the associated Legendre polynomial. However, such polynomial is zero when $\mu>\nu$. So, please clarify the meaning of $P_\nu^\mu(x)$ in your question. – Max Alekseyev Jul 21 '20 at 3:27
• The measure seems odd - $\sin \theta d\theta$ would look a lot more natural. It might be useful to have more details on how this integral comes about. – Michael Engelhardt Jul 24 '20 at 17:30

## 1 Answer

The best evaluation, a single sum, that I can derive is

$$\int_0^{\pi/2} \big( P_n^m(\cos{\theta}) \big)^2\ d\theta = \frac{\pi}{2} \frac{(2m)!}{m!^4} \Big( \frac{(n+m)!}{(n-m)!} \Big)^2 {}_4F_3 \bigl( \begin{smallmatrix} m+1/2, & m+1/2, & m-n, &m+n+1 \\ m+1 & m+1 & 2m+1 \end{smallmatrix} | 1\bigr)$$

I derived it from the answer given in Math Overflow 291481, which expresses the product of the associated Legendre polynomials as a single sum with powers of $$\sin^2{\theta}$$ within the summand. The integration is then easy. I then simplified the formula to that above.

It is doubtful that the generalized hypergeometric $${}_4F_3$$ simplifies to a ratio of gamma functions. I have a reference that has some $${}_4F_3$$ evaluated at 1, but the 'numerator' parameters start off as $$a, 1+a/2...$$ and that's not the form of the answer. Furthermore, I calculated it a few for small $$m$$ and $$n,$$ and if a ratio of gammas was in fact true, I wouldn't expect to get large primes in my answer.