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Given the formula $$F(x)= \sum_{n=-\infty}^{\infty}f(x+n) $$ We know that is invariant under translations of the form $y=x+n$ for any integer $n$. However can we find a similar formula for dilations ??

I believe that $ H(x)= \sum_{Q}f(qx) $ is invariant under dilations of the form $ y=nx$ for integer $n$, where $\mathbb{ Q}$ is the set of rational numbers.

In this last case , what would be the Mellin transform of $ H(x) $? I believe that it will be $ F(s)\zeta(s) \zeta (-s) $, but I am not sure , here $ F(s)= \int_{0}^{\infty}dtH(t)t^{s-1} $

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Can the sum converge? Continuity of $f$ at $x=a$ and $f(a)\neq 0$ seems to imply that the sum diverges at that point. – Sangchul Lee Mar 30 '12 at 9:00
oK, let us suppose the SUm is ALWAYS convergent and also $f(0)=0$ and $ \int_{0}^{\infty}f(t)dt =0 $ – Jose Garcia Mar 30 '12 at 9:47
I think that there's little hope that the sum $\sum_{\Bbb Q}$ converges unless $\Bbb Q$ is discrete. This is realized, for instance, in the embedding of $\Bbb Q$ in its ring of adeles $\Bbb A=\Bbb Q\Bbb R\hat{\Bbb Z}$. – Andrea Mori Mar 30 '12 at 9:50
for example , if we use the fundamental theorem of arithmetic i guess thati could write $ H(x)= \sum_{m=-\infty}^{\infty}\sum_{p}f(p^{m}) $ so it involves a sum over primes and prime powers – Jose Garcia Mar 31 '12 at 12:43

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