I need to compute the following integral
$$
I\_{n,m} := \int\_0^1 P\_n(x) P\_m(x) \; \mathrm{d}x
$$
where $P\_n$ is the [Legendre polynomial](https://en.wikipedia.org/wiki/Legendre_polynomials).


For an even sum $n+m=2l$ it is easy to show that
$$
I\_{n,m} = \frac{1}{2} \int\_{-1}^1 P\_n(x) P\_m(x) \; \mathrm{d}x
= \delta\_{n,m} \frac{1}{2n+1} \,.
$$


A length calculation arises for an *odd* sum $n+m=2l+1$. Has someone finished the calculation?



Legendre Polynomial Integral over half space
Integration of Equation (34) in [MathWorld](http://mathworld.wolfram.com/LegendrePolynomial.html) gives the integral $I\_{nm}$ as a sum
$$I\_{nm}=\sum \_{q=0}^m \frac{2^{-q}}{q+1} \binom{-m-1}{q} \binom{m}{q} \, \_3F\_2\left(-n,n+1,q+1;1,q+2;\tfrac{1}{2}\right).$$


As noted by the OP, Mathworld also gives the explicit expression:


![](https://ilorentz.org/beenakker/MO/LegendreP.png)


