How to compute this integral? Dominated convergence theorem doesn't apply I've met a physics problem and I need to evaluate:
$$\int_0^{\infty}y \sum_{n=1}^{\infty}\frac{(-1)^n}{(n^2r^2 + y^2)^{3/2}} dy$$
If we pretend we can interchange the sum and the integral we get the alternate harmonic series. But I'm not sure there is a theorem that garantees we can interchange them in this case.
And I'm not sure if Taylor expansion would help either, both the integral and the sum of the absolute values are not finite.
Can one show that it diverges?
 A: Since terms are alternating and decreasing in absolute value with respect to $n$, we have
$$\left|\sum_{n=1}^N \frac{(-1)^ny}{(n^2r^2 + y^2)^{\frac32}}\right| \leqslant  \frac{y}{(r^2 + y^2)^{\frac32}} < \frac{1}{y^2},$$
and since $y^{-2}$ is integrable on $[1,\infty)$, by the dominated convergence theorem,
$$\int_1^{\infty} \sum_{n=1}^{\infty}\frac{(-1)^ny}{(n^2r^2 + y^2)^{\frac32}}\, dy =  \sum_{n=1}^{\infty}\int_1^{\infty}\frac{(-1)^ny}{(n^2r^2 + y^2)^{\frac32}}\, dy $$
Note also that
$$\frac{y}{(n^2r^2 + y^2)^{\frac32}} =\frac{1}{2nr}\frac{2nry}{n^2r^2 + y^2} \frac{1}{\sqrt{n^2r^2 + y^2}} \leqslant \frac{1}{2nr} \cdot 1 \cdot \frac{1}{nr} = \frac{1}{2n^2r^2},$$
and $\frac{y}{(n^2r^2 + y^2)^{\frac32}}\to 0 $ uniformly and monotonically with respect to $n$. By the Dirichlet test, the series is uniformly convergent for $y \in [0,1]$ and we can switch the sum and the integral over the finite interval $[0,1]$, as
$$\int_0^{1} \sum_{n=1}^{\infty}\frac{(-1)^ny}{(n^2r^2 + y^2)^{\frac32}}\, dy =  \sum_{n=1}^{\infty}\int_0^{1}\frac{(-1)^ny}{(n^2r^2 + y^2)^{\frac32}}\, dy,$$
Thus,
$$\int_0^{\infty} \sum_{n=1}^{\infty}\frac{(-1)^ny}{(n^2r^2 + y^2)^{\frac32}}\, dy =  \sum_{n=1}^{\infty}\int_0^{1}\frac{(-1)^ny}{(n^2r^2 + y^2)^{\frac32}}\, dy + \sum_{n=1}^{\infty}\int_1^{\infty}\frac{(-1)^ny}{(n^2r^2 + y^2)^{\frac32}}\, dy \\ = \sum_{n=1}^{\infty}\int_0^{\infty}\frac{(-1)^ny}{(n^2r^2 + y^2)^{\frac32}}\, dy$$
