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Sometimes we meet with situation when we don't know asymptotics of a given but we know its average behaviour on $[1,n]$. For example, we can easily get that

$\sum_{i=1}^n \tau(n) = n \log (n) (1 + o(1))$

but we have only upper bound on $\tau(n)$ itself.

So my question is the next: is there some research (maybe for another NT fucntion) where authors found out minimal number of consecutive arguments (or its growth), such that average value of a given function on such arguments has clear asymptotics, i. e. a function $g(n)$ such that:

$\frac{1}{g(n)}\sum_{i=n+1}^{n+g(n)} \tau (i) = S(n) (1+o(1))$

for some function $S(n)$, which is a combination of polynomials, logarithms and exponents?

Edit: Thanks, Fedor and Gerry, I just try to summarize the best unconditional results on this moment. Really, this problem is closely related with a problem of error tern estimating for an average. From Huxley work "Exponential sums and lattice points III" (2003) it follows that $\Delta (x) << x^{131/416+\epsilon}$, where

$ \Delta (x) = \sum_{n\leq x} \tau (n) - x(\log x + 2\gamma -1)$,

so one can easily get using a method desribed by Fedor, that

$\frac{1}{h(x)}\sum_{x<n\leq x} \tau (n) = \log (x)(1 + o(1))$

for $h(x) = x^{\theta}$ with $\theta > 131/146$.

Also Shiu in his work "A Brun-Titchmarsh theorem for multiplicative functions" (1980) proved that

$\sum_{x<n\leq x + h(x)} \tau (n) << h(x) \log (x)$

for $x^\epsilon \leq h(x) \leq x$ with no additional restictions on positive $\epsilon$.

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    $\begingroup$ If you have reasobable bounds for the remainder in the asymptotic formula for the sum $T(n)$ (for $i$ from 1 to $n$), you may subtract $T(n+g(n))-T(n)$ and still get an asymptotics, if $g(n)$ is not too small. $\endgroup$
    – Fedor Petrov
    Commented Jun 6, 2017 at 19:26
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    $\begingroup$ $O(1)$ means any bounded function. So it makes little sense to write $1+O(1)$, because it is just another way of writing $O(1)$. You probably meant $1+o(1)$ throghout. ($o(1)$ means any function tending to zero.) $\endgroup$
    – GH from MO
    Commented Jun 7, 2017 at 15:46

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It is known that $\sum_{x<n\le x+h(x)}\tau(n)$ is asymptotic to $h(x)\log x$ for $h(x)=x^{\theta}$ with $\theta>131/416$, and it is conjectured that the asymptotic holds for $h(x)\gg x^{\epsilon}$. See, e.g., Danilo Bazzanella, On the divisor function in short intervals, Arch Math 97 (2011) 453-458.

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