# Recovering $\sum_{n \leq x} a(n)$ from $\sum_{n \leq x} a(n)e^{-n/x}$

In the theory of automorphic forms and multiple Dirichlet series, we often take inverse Mellin transforms of Dirichlet series to come up with Tauberian theorems, like the Ikehara Tauberian method. In particular, from

$\displaystyle \sum_n \frac{a(n)}{n^s}$, we might expect to use Mellin to get asymptotics for $\displaystyle \sum_{n \leq x} a(n)$.

Sometimes, the series doesn't behave well enough to integrate. A common way around that seems to be to multiply by the gamma function to get better decay. But this has the side effect of 'smoothing' the sum to get asymptotics for sums that look like $\displaystyle \sum_{n \leq x} a(n) e^{-n/x}$.

For a while, I just accepted that sometimes we need to accept smoothed sums. But I began to wonder how hard it would be to recover the regular, un-smoothed sum. So I'm prompted to ask:

Suppose we can calculate $F(x) := \displaystyle \sum_{n \leq x} a(n) e^{-n/x}$ for any given $x > 1$. For a fixed $N$, can one recover $\displaystyle \sigma_N := \sum_{n \leq N} a(n)$ from $F(x)$? In particular, it is ever possible to recover $\sigma_N$ by calculating $F(x_i)$ for a suitably chosen, but finite set of $x_i \in \mathbb{R}$?

• Note $F(1) = a(1) e^{-1/x}$, $F(2) = F(1) + a(2) e^{-2/x}$, etc. so given these integer values of $F(x)$ you can recover the $a(n)$'s and hence $\sigma_N$. Of course the number of samples needs to grow with $x$. And by the way, I wouldn't call the sum $\sum_{n \leq x} a(n) e^{-n/x}$ smooth, because you are still truncating the sum at $x$. An example of a smoothed sum would be $\sum_{n=1}^{\infty} a(n) e^{-n/x}$. – Matt Young Mar 25 '13 at 18:19

A standard approach to think about is partial summation. Suppose that $A(x)=\sum_{n\leq x} a(n)$ and $S(x)=\sum_{n\leq x} a(n)e^{-n/x}$. Then we can relate these two sums in the following way:
$$\sum_{n\leq x}a(n)e^{-n/x}=\int_{1}^{x}e^{-t/x}d\left(A(t)\right)=\frac{1}{e}A(x)+\frac{1}{x}\int_{1}^{x}A(t)e^{-t/x}dx.\ \ \ \ \ \ (1)$$
$$\sum_{n\leq x}a(n)e^{-n/x}e^{n/x}=\int_{1}^{x}e^{t/x}d\left(S(x)\right)=eS(x)-\frac{1}{x}\int_{1}^{x}e^{t/x}S(t)dt.\ \ \ \ \ \ (2)$$
Equation $(1)$ shows that we can write $S(x)$ in terms of $A(x)$, and if we know an asymptotic for $A(x)$ we can recover one for $S(x)$. At first glance, $(2)$ does the same thing, but things could potentially go wrong with the negative sign, that is there could be cancellation and we would need a better expansion for $S(t)$. Since $e^{-n/x}$ is a nice factor between $1$ and $1/e$, I don't think there should be any problems with using $(2)$ for most sequences $a(n)$ that we would be interested in.
In many applications, you can use the fact that $$e^{-n/x}=1 + O\Big(\frac{n}{x}\Big)$$ uniformly for $n\le x$ and therefore $$\sum_{n\le x} a_n e^{-n/x} = \sum_{n\le x} a_n+ O\left( \frac{1}{x} \sum_{n\le x}n \ a_n\right).$$
This works well for sequences $a_n$ generated by $L$-functions near the $1$-line. For instance, if $$\sum_{n\le x} a_n \asymp (\log x)^k$$ for some $k$.