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.

  • $\begingroup$ 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? $\endgroup$ Jul 20, 2020 at 17:16
  • $\begingroup$ 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? $\endgroup$ Jul 21, 2020 at 2:50
  • $\begingroup$ 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. $\endgroup$ Jul 21, 2020 at 3:27
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    $\begingroup$ 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. $\endgroup$ Jul 24, 2020 at 17:30

1 Answer 1


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.


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