Integral inequality for a Cauchy exponential series product My goal is to get an inequality $\forall t>0$ for the following integral
$$
\int_0^t \left(\sum_{n=1}^\infty \exp(-n^2 t_0)\right)^2\,\mathrm{d}t_0
\le f(t).
$$
The goal is to at least lose the the square of the series. Initially I thought it was possible to find a reasonable $f(t)$ using the Mean value theorem. However, there is not an easy upper-bound for $\sum_{n=1}^\infty \exp(-n^2 t_0)$ since it explodes around $t_0 \approx 0$. Does anybody have a clue?
 A: Well, your integral is $\infty$ whenever $t>0$. We get for $u>0$:
$$ \left(\sum_{n=1}^{\infty}\exp(-n^2u)\right)^2 = \sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\exp(-(n^2+m^2)u)$$
By the Fubini-Tonelli theorem we can exchange the order of integration and summation for a nonnegative function.  Fix $t>0$ and define $\mathcal{A}_t=\{(n,m)\in \mathbb{N}^2: (n^2+m^2)t\geq 1\}$. Then:
\begin{align} 
&\int_0^t  \left(\sum_{n=1}^{\infty}\exp(-n^2u)\right)^2 du \\
&= \sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\int_0^t \exp(-(n^2+m^2)u)du \\
&= \sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\frac{1-\exp(-t(n^2+m^2))}{n^2+m^2}\\
&\geq \sum_{(n,m)\in\mathcal{A}_t} \frac{1-\exp(-t(n^2+m^2))}{n^2+m^2}\\
&\geq (1-e^{-1})\sum_{(n,m)\in\mathcal{A}_t} \frac{1}{n^2+m^2}\\
&=\infty
\end{align}
where the final equality can be shown because $\sum_{n=1}^{\infty}\sum_{m=1}^{\infty} \frac{1}{n^2+m^2}=\infty$, and $\sum_{(n,m)\in\mathcal{A}_t}$ only neglects a finite number of terms in comparison to $\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}$. 
