Legendre's Equation I'm given two solutions to Legendre's equation:
$$P_1=x$$
$$Q_0=\frac{1}{2} \ln\left(\frac{1+x}{1-x}\right)$$
I'm trying to explain why their overlap integral (i.e. $\int_{-1}^{1} P_1 Q_0 dx$) is non-zero.  I computed it and it is indeed non-zero, but I'm having a difficult time justifying why that is. I'm thinking it has something to do with that fact that the $P_n$ and $Q_n$ solutions are constructed w.r.t different weight functions. Or perhaps it has something to do with the completeness of solutions. Any thoughts?
 A: As has been shown, the integral is not zero. 
This is okay.
Since $P$ and $Q$ obey different boundary conditions they are eigenfunctions of different Sturm-Liouville systems, so we should not expect them to be orthogonal. 

Consider a more familiar example, 
$$\begin{array}{l}
y'' + n^2 y = 0 \\
y(0) = y(\pi) = 0.
\end{array}$$
The unnormalized eigenfunctions are $f_{n} = \sin n x$, 
where $n \in \mathbb{N}$. 
Sturm-Liouville theory tells us the eigenfunctions must be orthogonal, and of course they are. 
The related system 
$$\begin{array}{l}
y'' + n^2 y = 0 \\ 
y'(0) = y'(\pi) = 0
\end{array}$$
has eigenfunctions $g_{n} = \cos n x$. 
Again, the eigenfunctions are orthogonal. 
However, Sturm-Liouville theory has nothing to say about whether $f_m$ and $g_n$ are orthogonal, and in fact they are not in general. 
For example, 
$$\int_0^\pi dx\, \sin x \cos 2x = -\frac{2}{3}.$$ 
A: Since $Q_0(x)$ is pointymmetric
$$
Q_0(x)=\frac{1}{2} \ln\left(\frac{1+x}{1-x}\right)=-\frac{1}{2} \ln\left(\frac{1+(-x)}{1-(-x)}\right)=-Q_0(-x)
$$
as is $P_1(x)=-P_1(-x)$, there product is symmetric 
$$
P_1(x)Q_0(x)=(-1)^2P_1(-x)Q_0(-x).
$$
Since your limits are also symmetric we'll get
$$
\int_{-1}^{1} P_1(x) Q_0(x) dx=\int_{-1}^{0} P_1(x) Q_0(x) dx+\int_{0}^{1} P_1(x) Q_0(x) dx=2\int_{0}^{1} P_1(x) Q_0(x) dx
$$
and since $P_1(x)$ and $ Q_0(x)$ are both positive on $[0,1]$, your integral is non-zero.
