3
$\begingroup$

Consider $f(x)$, a rapidly decreasing function, such that $\int_0^{\infty} f(x)=0$ and for $x$ near zero: $f(x)=O(x^a)$ (wit $a>0$). Can we interchange the sum and integral and write as below: $$\int_0^{\infty}\sum_{n=1}^{\infty} f(nx)= \sum_{n=1}^{\infty} \int_0^{\infty} f(nx) =0$$

Note that, as $\int_0^{\infty} f(x)=0$, the Poisson summation formula (thanks to the limit of $f(x)$ in zero) ensures that: $\sum_{n=1}^{\infty} f(nx)\sim O(x^a) \;\; (x \to 0)$ so the integral on the left in above expression is well defined.

We cannot apply directly the classical interchange theorems as: $\sum_{n=1}^{\infty} |f(nx)| \sim O(\frac{1}{x})$ and even if we have simple convergence of the sum, the partial sum $|\sum_{n=1}^{N} f(nx)|$ cannot be bounded near zero for all N (even if the complete sum is absolutely integrable). So is there way to show we can interchange or two sums are different, and if it is the case how can we show it ?

$\endgroup$
  • $\begingroup$ You are right but what I mean is that to apply the interchange therorem we need to find $g(x)$ integrable such that for all $N$ we have $|\sum_{n<N} f(nx)| < g(x)$ and this is not possible. I admit my comment is not completely clear I will clarify. $\endgroup$ – Bertrand Dec 2 '17 at 18:48
1
$\begingroup$

It is not possible to interchange integral and sum, but it seems possible to estimate the sum and define condition to have the integral null:

As by Poisson summation formula we have $\sum_{n=1}^{\infty} f(nx) \sim O(x^a)$ near zero we can split the integral in to:

$$\int_0^{\infty}\sum_{n=1}^{\infty} f(nx) dx= \int_0^{\epsilon}\sum_{n=1}^{\infty} f(nx) dx + \int_{\epsilon}^{\infty}\sum_{n=1}^{\infty} f(nx) dx$$

We can chose $\epsilon$ small to have first integral as small as we want, on the second one we can interchange sum and integral and if we note $F(x)$ the primitive of $f(x)$ such that $F(0)=0$ we have (as $\lim_{x \to 0} F(x)=0$ due to the condition imposedon $f(x)$: $\int_0^{\infty} f(x) dx=0$):

$$\int_{\epsilon}^{\infty}\sum_{n=1}^{\infty} f(nx) dx = -\sum_{n=1}^{\infty} \frac{1}{n} F(n \epsilon)$$

Now we can apply the Poisson summation formula (assuming $\lim_{x \to 0}\frac{F(x)}{x}=0$) to obtain (we note $\mathcal{F}$ the Fourier tansform):

$$ \sum_{n=1}^{\infty} \frac{1}{n \epsilon} F(n \epsilon) = \frac{1}{\epsilon} \sum_{n=1}^{\infty} \mathcal{F}(\frac{F(|x|)}{|x|})(\frac{n}{\epsilon})+\frac{1}{2 \epsilon} \mathcal{F}(\frac{F(|x|)}{|x|})(0)$$

And we see that, as the terms $\mathcal{F}(\frac{F(|x|)}{|x|})(\frac{n}{\epsilon})$ can be as small as we want for $\epsilon$ small, only one term remains and taking the limit with $\epsilon \to 0$:

$$\int_{0}^{\infty}\sum_{n=1}^{\infty} f(nx) dx = -\frac{1}{2} \mathcal{F}(\frac{F(|x|)}{|x|})(0) $$

So finally it seems the initial integral is zero only if:

$$\int_{0}^{\infty} \frac{F(x)}{x} dx=0$$

Comments are welcome, any reference for a similar treatment of a sum using Poisson summation formula ?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.