Swapping limits when simplifying a hypergeometric function I have been trying to find a simplified expression for ${}_3F_2\left(\begin{smallmatrix}1,1,\frac{3}{2}\\2,3 \end{smallmatrix}; x\right)$. After manipulating the Pochhammers in the definition, I arrive at $${}_3F_2\left(\begin{smallmatrix}1,1,\frac{3}{2}\\2,3 \end{smallmatrix}; x\right) = -8\sum_{n=0}^\infty \frac{\left(-\frac{1}{2}\right)_{n{+}2}}{(n{+}2)!}\frac{x^n}{n{+}1}.$$ This is very close to a binomial series, apart from the extra factor $1/(n+1)$. To remedy this I use $\frac{1}{n+1}=\int_0^1dt\;t^n$ to write that $${}_3F_2\left(\begin{smallmatrix}1,1,\frac{3}{2}\\2,3 \end{smallmatrix}; x\right) = -8\sum_{n=0}^\infty \int_0^1\frac{dt}{(tx)^2}\frac{\left(-\frac{1}{2}\right)_{n{+}2}}{(n{+}2)!}(tx)^{n+2}.$$ It's now straightforward to evaluate the sum$$\sum_{n=0}^\infty \frac{\left(-\frac{1}{2}\right)_{n{+}2}}{(n{+}2)!}(tx)^{n+2} = \frac{1}{\sqrt{1-tx}} + \frac{1}{2}tx -1,$$ which gives that$${}_3F_2\left(\begin{smallmatrix}1,1,\frac{3}{2}\\2,3 \end{smallmatrix}; x\right) = -\frac{8}{x^2}\int_0^1\frac{dt}{t^2}\left(\frac{1}{\sqrt{1-tx}} + \frac{1}{2}tx -1\right).$$ This integral does not converge, so I have not been able to simplify ${}_3F_2\left(\begin{smallmatrix}1,1,\frac{3}{2}\\2,3 \end{smallmatrix}; x\right)$ this way.
I assume that this is probably because I should not have swapped around the integral and the sum when I evaluated the sum using the binomial series. Is it true that the limits in the integral and sum do not commute in the case? How can you check whether it's valid to swap an integral and an infinite sum?
Is it possible to define something like a commutator $$\left[\sum_{n=0}^\infty, \int_0^1dt\right]:= \sum_{n=0}^\infty \int_0^1dt -\int_0^1dt\sum_{n=0}^\infty$$and then compute this quantity to give an extra piece which adds to my final integral to give something that converges?
 A: If your goal is simply to reduce your hypergeometric function, one way you can approach this is working with integral representations as opposed to directly working with the series expansion. Here is a CAS (Mathematica) assisted approach to reducing your hypergeometric function.  You may go back and fill in the details of how the integrals are evaluated if you're looking for a more detailed proof. This is simply meant to sketch out the approach for you.

Using DLMF 16.6.2 and assuming $|x|<1$ we may succesively reduce your ${_3F_2}(\cdots)$ down to ${_1F_0}(\cdot)$ by writing
$$
\begin{align}
{_3F_2}\left({1,1,3/2\atop 2,3};x\right)
&=\int_0^1 \mathrm du\,{_2F_1}\left({1,3/2\atop 3};xu\right)\\
&=2\int_0^1 \mathrm du\,\int_0^1\mathrm dv\,(1-v){_1F_0}\left({3/2\atop -};xuv\right)\\
&=2\int_0^1 \mathrm du\,\int_0^1\mathrm dv\,(1-v)(1-xuv)^{-3/2}.
\end{align}
$$
Evaluating the integral in $v$ (presumably by parts) one has
$$
{_3F_2}\left({1,1,3/2\atop 2,3};x\right)=
\frac{4}{x^2}\int_0^1 \mathrm du\,\frac{2-2\sqrt{1-x u}-x u}{u^2}.
$$
Evaluating over $u$ then gives the final result
$$
{_3F_2}\left({1,1,3/2\atop 2,3};x\right)=
\frac{4}{x^2}\left(-2 x\operatorname{arctanh}\sqrt{1-x}-x\log(x/4)+x+2\sqrt{1-x}-2\right).
$$
