Weierstrass's M-test example for uniform convergence and switching Sum and Integral. How would I go about finding $M_n$ in 
\begin{equation}
\sum_{n=1}^{\infty} \int_{0}^\infty x^{\frac{s}{2}-1}e^{-\pi n^{2}x}dx
\end{equation}
to show that it is uniformly convergent?
UPDATE:
Just coming back to this and having trouble understanding this. Without relating this integral  to the zeta and gamma functions, how do I show the sum has uniform convergence ( Weierstrass)? For $Re(s) > 1$. And can therefore swap integral with summation by fubinis theorem?
 A: You may observe that, using a standard integral representation of the gamma function, we have
$$
\int _0^{\infty }x^{s/2-1}e^{-\pi n^2x}dx=\frac{1}{n^s}\pi^{-s/2}\Gamma(s/2)
$$ then your series rewrites
$$
\sum_{n=1}^{\infty} \int_{0}^\infty x^{\frac{s}{2}-1}e^{-\pi n^{2}x}dx=\pi^{-s/2}\Gamma(s/2)\sum_{n=1}^{\infty} \frac{1}{n^s}=\pi^{-s/2}\Gamma(s/2)\zeta(s)
$$ and the convergence of the series is uniform on each $[a,b] \ni s$, $1<a<b$, since
$$
\left|\frac{1}{n^s}\right|\leq\frac{1}{n^a}, \quad n\geq1.
$$
A: Lemma: For any fixed $a>0,$
$$\lim_{p\to \infty}\int_0^\infty x^pe^{-ax}\, dx = \infty.$$
Proof: The integral is greater than
$$\int_1^2 x^p e^{-ax}\  dx \ge e^{-2a}\int_1^2 x^p\ dx = e^{-2a}\cdot\frac{2^{p+1}-1}{(p+1)}.$$
The last expression $\to \infty$ as $p\to \infty.$
In your problem, let $f_n(s)= \int_0^\infty x^{s/2-1}e^{-\pi n^2x}\,dx.$ You wonder  if $\sum_n f_n(s)$ converges uniformly on $(1,\infty).$ It doesn't. Why? The lemma shows each $f_n(s) \to \infty$ as $s\to \infty.$ So each $f_n$ is unbounded on $(1,\infty).$ Elementary fact: No series of unbounded bounded functions on a set can converge uniformly on that set.
Now you also bring up the question of whether sum and integral can be interhanaged in this problem. They can. A simple way to see that is the monotone convergence theorem, a corollary of which says, to be brief: If each $f_n$ is nonnegative, then $\sum_n \int f_n = \int (\sum_n f_n).$
