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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?

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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 $\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}$.

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  • $\begingroup$ I do not think you did the expansion correct, since the correct Cauchy product expansion should be: $\sum_{n=1}^\infty\sum_{m=1}^{n-1} \exp((-m^2-(n-m)^2)t_0)$. This would lead to the integral result $\int_0^{t} \sum_{n=1}^\infty\sum_{m=1}^{n-1} \exp((-m^2-(n-m)^2)t_0) = \sum_{n=1}^\infty\sum_{m=1}^{n-1} \frac{\exp((-m^2-(n-m)^2)t_0) - 1}{(-m^2-(n-m)^2}$. $\endgroup$ – Amp Mar 13 '16 at 19:17
  • $\begingroup$ @Amp : Why do you sum only from $m=1$ to $n-1$? $\endgroup$ – Michael Mar 13 '16 at 19:23
  • $\begingroup$ Excuse me, the integral should be $\int_0^{t} \sum_{n=1}^\infty\sum_{m=1}^{n-1} \exp((-m^2-(n-m)^2)t_0) dt_0= \sum_{n=1}^\infty\sum_{m=1}^{n-1} \frac{\exp((-m^2-(n-m)^2)t) - 1}{-m^2-(n-m)^2}$ $\endgroup$ – Amp Mar 13 '16 at 19:23
  • $\begingroup$ I have used the Cauchy product rule here, en.wikipedia.org/wiki/Cauchy_product $\endgroup$ – Amp Mar 13 '16 at 19:24
  • $\begingroup$ Before integrating, you need a correct value of the double sum. $\endgroup$ – Michael Mar 13 '16 at 19:24

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