Can order of summation and integral be interchanged in : $\int_{-1}^1 ( \sum_{n=0}^\infty P_n(\xi)P_n(\xi^\prime)Q_l(\xi^\prime))\text{d}\xi^\prime$? I am wanting to know if there is a proof that the order of summation and integration can be interchanged in $$\int_{-1}^1 \left( \sum_{n=0}^\infty P_n(\xi)P_n(\xi^\prime)Q_l(\xi^\prime)\right)\text{d}\xi^\prime ,$$ where $\xi,\xi^\prime$ $\in[-1,1]$, and $P_n$ is the Legendre polynomial of order $n$, ($Q_l$ is arbitrary but is assumed to be nicely behaved - i.e. no singularities etc.).
I know the following, namely that:  $$\sum_{n=0}^\infty P_n(\xi)P_n(\xi^\prime)=\cfrac{2}{\pi}\cfrac{K(k)}{\sqrt{2(1-\xi\xi^\prime)+2\sqrt{1-\xi^2}\sqrt{1-\xi^{\prime2}}}}\overset{\text{def}}{=}f(\xi,\xi^\prime)\ ,$$ with $K$ the complete elliptic integral of the first kind, and $$k=\sqrt{\cfrac{4\sqrt{1-\xi^2}\sqrt{1-\xi^{\prime2}}}{2(1-\xi\xi^\prime)+2\sqrt{1-\xi^2}\sqrt{1-\xi^{\prime2}}}}$$ (see https://www.sciencedirect.com/science/article/pii/S0021999121007579 , with error in $k$ definition from paper corrected).
Plotting the partial sums of the infinite Legendre sum seems to indicate there may be a bounding function (which seems potentially to be something like $10\cdot f(\xi,\xi^\prime) \cdot Q_l(\xi^\prime)$ or something like that), which would of course if true let me use the dominated convergence theorem (if I'm reasoning this right).  But how to show this rigorously??
Thanks.
 A: So I think that I might have solved this question, at least to my satisfaction.
So let's define $$f_N=\sum_{n=0}^N P_n(\xi)P_n(\xi^\prime)\ ,$$ where $f_N$ is the $N$'th partial sum.   I already know that the series converges pointwise almost everywhere to $f(\xi,\xi^\prime)$, so it only remains to show that the absolute value of each $N$'th partial sum is bounded by an integrable function (to use the dominated convergence theorem).
To show that that the absolute value of each $N$'th partial sum is bounded by some integrable function, we first observe $(|f_N|-f)$.  We have that $$(|f_N|-f)\leq\underset{\xi^\prime}{\max}(|f_N|-f)\overset{\text{def}}{=}g(N,\xi)\ .$$ This maximum exists, first because $f$ being a positive function implies that $-f$ is bounded from above, and for a fixed $N$, $|f_N|$ is a bounded function as well.  Next we observe that $$g(N,\xi)\leq \underset{N}{\max}g(N,\xi)\overset{\text{def}}{=}h(\xi)\ .$$  This max exists because for each $N$, $g(N,\xi)$ is finite, and as $N\rightarrow\infty$, $g(N,\xi)\rightarrow 0$ almost everywhere (because $f_N\rightarrow f$ almost everywhere as $N\rightarrow \infty$). But $h(\xi)$ is definitely an integrable function of $\xi^\prime$ over $[-1,1]$, because it is constant with respect to $\xi^\prime$.
So this gives finally that $$|f_N|\leq f(\xi,\xi^\prime)+h(\xi)\ .$$ $f$ is an integrable function over $\xi^\prime$ because it has only a logarithmic singularity (due to the asymptotic behavior of $K(k)$ as $k\rightarrow 1$).
Including $Q_l(\xi^\prime)$ into the analysis is now straighforward.  We have $$|f_NQ_l(\xi^\prime)|=|f_N||Q_l(\xi^\prime)|\leq |Q_l(\xi^\prime)|(f(\xi,\xi^\prime)+h(\xi))\ .$$
Assuming $Q_l(\xi^\prime)$ does not contain itself any new singularities or non-integrable behavior (which is what we are assuming), then we have shown that each $N$'th partial sum is bounded by an integrable function.  So the dominated convergence theorem lets us move the integral over $\xi^\prime$ inside the infinite sum.
