Is the following function of time integrable? In [Da Prato, Giuseppe, and Jerzy Zabczyk. Stochastic equations in infinite dimensions. Cambridge university press, 2014.
APA] pag. 200 it is claimed that the function $$K^2(t)= \sum_{h=1}^{\infty}e^{-2 \pi^2h^2t}$$
is square integrable on $(0,T]$ with $T< \infty$ fixed.
How do you show it?
 A: 1. Let $f(t)$ by
$$ f(t) = \sum_{h=1}^{\infty} e^{-2\pi^2 h^2 t}. $$
By using the fact that $h \mapsto e^{-2\pi^2 h^2 t}$ is non-increasing, we get
$$ f(t) \leq \int_{0}^{\infty} e^{-2\pi h^2 t} \, \mathrm{d}h = \frac{1}{\sqrt{8\pi t}} $$
and
$$ f(t) = \sum_{h=0}^{\infty} e^{-2\pi^2 h^2 t} - 1 \geq \int_{0}^{\infty} e^{-2\pi h^2 t} \, \mathrm{d}h - 1 = \frac{1}{\sqrt{8\pi t}} - 1, $$
Therefore
$$ f(t)^2 \sim \frac{1}{8\pi t} \quad\text{as}\quad t \to 0^+ $$
and hence $\int_{0}^{T} f(t)^2 \, \mathrm{d}t = +\infty$ for any $T > 0$ by the limit comparison test.
2. Now why I am bothering to denote OP's function by $f(t)$ is that, in the reference, the actual formula is given by
$$ K^2(t) = \sum_{h=1}^{\infty} e^{-2\pi^2 h^2 t} $$
as we see from the following screenshot:



That is, the sum is already for $K(t)^2$ and not for $K(t)$. Then the above inequality shows that $K(t)^2$ is indeed square-integrable on $(0, T]$ for any $T > 0$.
A: Let $f_h(t) = e^{-2 \pi^2h^2t}$, then $\|f_h\|_{L^1} \le C h^{-2}$ with a constant $C$ that is independent of $T$.
Therefore $\sum_{h = 1}^\infty f_n$ converges in $L^1(0,T)$, for all $T \le \infty$.
