In analytic number theory, when deriving the integral formula for the completed zeta function, one makes use of the following line of equation: $$\pi^{-s}\Gamma(s)\zeta(2s)=\sum_{n=1}^{\infty}\int_0^{\infty}e^{-n^2\pi r}r^{s-1}\text{d}r=\int_0^{\infty}\sum_{n=1}^{\infty}e^{-n^2\pi r}r^{s-1}\text{d}r$$ where $\text{Re} \ s > 1$.
Question: Why is exchanging sum and integral in this situation allowed?
Some authors like Neukirch in his $Algebraic$ $Number$ $Theory$, Davenport in his $Multiplicative$ $Number$ $Theory$ or Montgomery&Vaughan in their $Multiplicative$ $Number$ $Theory$ $I$ seem to suggest that this follows by "absolute convergence" where they probably mean that $$\sum_{n=1}^{\infty}\int_0^{\infty}|e^{-n^2\pi r}r^{s-1}|\text{d}r< \infty.$$ I completely understand why this inequality holds but how does this form of absolute convergence implies that exchanging sum and integral is allowed? What type of convergence theorem is used here to see this?
Since the authors I mentioned do not give any further explanation, I guess there should be a quick way to see this. However, all my own attempts in showing this have resulted in long and not-so-clear arguments so far. For example, when using dominated convergence, I can show that $$\sum_{n=1}^{\infty}\int_0^{\infty}e^{-n^2\pi r}r^{s-1}\text{d}r=\lim_{k\to \infty} \sum_{n=1}^{\infty}\int^k_0e^{-n^2\pi r}r^{s-1} \text{d}r.$$ I could then try to treat the right-hand-side by using the functional equation for the $\theta$-function but since this way is somewhat involved and the authors do not discuss it, I don't think this is the intended way...