Logarithmic weights on number theoretic sums

Suppose we are interested in the sum

$\sum _{n\leq x}a_n.$

The study of the sum

$\sum _{n\leq x}a_n\log (x/n)$

may be easier.

What can one say about the first sum from knowing the behaviour of the second?

In the case I have in mind, I have

$\sum _{n\leq x}a_n\log (x/n)=x+\mathcal O\left (x^{1/2}\right )$

and would like a similar result for the sum without weights.

How feasible is this? I'm quite sure an asymptotic formula holds for the sum without weights, but I don't know if I should expect to lose a power saving. (Logarithms and epsilons in the error are not of concern.)

I'm not sure which properties exactly of the $a_n$ are important. It may be useful that they are all non-negative.

Any pointers as to what one should expect in such a situation would be very much appreciated.

If it seems that one can't really say much without knowing at least something more about the $a_n$ I can give some more details. But I think my problem has more to do with the fact that I'm lacking some general principles in dealing with weights, so I'll leave the $a_n$ arbitrary for now.

• The standard technique is partial summation. More precisely, if $S(x)=\sum_{n\leq x}a_n\log n$, then the first sum is $\int_{2-}^x\frac{dS(t)}{\log t}$, which can be manipulated by integration by parts (see en.wikipedia.org/wiki/…). Have you tried this? – GH from MO Sep 25 '15 at 17:49
• Another idea is to compare the Mellin transforms of the two summatory functions: the transforms are very close to each other. I don't have time to elaborate on this now. – GH from MO Sep 25 '15 at 17:58
• I tried the "amateur version" of summation by parts (not thinking in terms of Riemann-Stieltjes integrals), but dismissed it almost immediately since the term $\sum _{n\leq t}a_n\log (x/n)$ comes up, which we don't understand. But I don't know if this is a correct objection, since I don't know "philosophically" why it doesn't work. (In your example, after applying integration by parts, the sum $\sum _{n\leq t}a_n\log n)$ comes up, which we don't understand either.) – Tomos Parry Sep 25 '15 at 20:04
• Mellin transforms: I'm sort of going the other way. I tried to study the first sum through its Mellin transform, but wasn't successful. However, I could do the second sum with the Mellin transform approach. (My thinking was: "well, if I can do it with logs then the whole sum shouldn't really be any bigger so it should be fine" - perhaps I oversimplified and didn't understand that the problem without the logs is significantly harder, even though it's not much bigger - is this correct?) – Tomos Parry Sep 25 '15 at 20:16
• Hey so it looks like you're trying to use a truncated Perron formula --- have you checked the standard 'Perron formula with error term' that would e.g. appear in Montgomery-Vaughan? As for the weights, $\sum_{n\leq x+x^{3/4}} a_n\log((x+x^{3/4})/n) - \sum_{n\leq x} a_n\log(x/n) = (\sum_{n\leq x} a_n)\log(1+x^{-1/4}) + \sum_{x < n\leq x+x^{3/4}} a_n\log((x+x^{3/4})/n)$. The latter sum, if $|a_n|\ll 1$, is $\ll x^{1/2}$. The left-hand side, by hypothesis, was $x^{3/4} + O(x^{1/2})$. Thus (rearranging) $\sum_{n\leq x} a_n = x + O(x^{3/4})$. Hope I haven't made a mistake in the above! – alpoge Sep 26 '15 at 6:59

Put $A(x) =\sum_{n\le x} a_n$, and $B(x) =\sum_{n\le x} a_n\log x/n$. Then $$B(x) = \int_1^x A(t)\frac{dt}{t}.$$ So information about $A(x)$ readily translates to information about $B(x)$ and there is no loss since integration (which makes things smoother) is involved. But to pass from $B(x)$ to $A(x)$ we need to differentiate, and based on the situation this could be either impossible, or could involve some loss.
Suppose first that the $a_n$ are bounded. Then $$B(x+h) - B(x) = \int_{x}^{x+h} A(t) \frac{dt}{t} = (A(x)+O(h)) \log\frac{x+h}{x},$$ and choosing $h=x^{3/4}$ we obtain $A(x)=x+O(x^{\frac 34})$. This was noted in Alpoge's comment above, and also holds if the $a_n$ are given to be non-negative rather than bounded. However there is a loss in this argument and the $x^{3/4}$ error term cannot be improved, even for bounded non-negative $a_n$. Here is an example: Take $a_n=2$ if $n \in [m^4,(m^4+(m+1)^4)/2)$ for some integer $m$, and $a_n=0$ if $n\in ((m^4+(m+1)^4)/2,(m+1)^4)$. Then you can check that $B(x) = x +O(x^{\frac 12})$ holds, but no estimate sharper than $A(x) = x +O(x^{\frac 34})$ can hold.
Finally without some restrictions on $a_n$, one can say nothing at all. For example, take $a_n=1+ (-1)^n n$. Then you can check that $A(n)$ alternates between about $n/2$ and about $3n/2$, whereas $B(n)$ is very close to $n$.