If $f : t \to e^{-xt}$ with $x \geqslant 1$, and $d_n$ is the number of positive integers that divide $n$, I can show that $$ \lim_{\epsilon \to 0^+} \sum_{n\geqslant 1}\frac{\epsilon^2 d_n f(\epsilon\ln n)}{n} = \frac{1}{x^2} = \int_{0}^{+\infty} tf(t)dt$$

I would like to generalize the above, at least for a function $f : \mathbb{R}+ mathbb{R}_+ \to \mathbb{R}+$, mathbb{R}_+$, such that $f(t)=O(e^{-t})$ near $+\infty$.

I thought it would be easy using the following classical formula $$\sum_{n\geqslant 1} a_ne^{-s\ln \lambda_n} = \frac{1}{\Gamma(s)} \int_{0}^{+\infty} t^{s-1}\sum_{n\geqslant 1} a_ne^{-x \lambda_n}dx$$

but either it is not, or I'm missing a trick here.

If $f : t \to e^{-xt}$ with $x \geqslant 1$, and $d_n$ is the number of positive integers that divide $n$, I can show that $$ \lim_{\epsilon \to 0^+} \sum_{n\geqslant 1}\frac{\epsilon^2 d_n f(\epsilon\ln n)}{n} = \frac{1}{x^2} = \int_{0}^{+\infty} tf(t)dt$$

I would like to generalize the above, at least for a function $f : \mathbb{R}+ \to \mathbb{R}+$, such that $f(t)=O(e^{-t})$ near $+\infty$.

I thought it would be easy using the following classical formula $$\sum_{n\geqslant 1} a_ne^{-s\ln \lambda_n} = \frac{1}{\Gamma(s)} \int_{0}^{+\infty} t^{s-1}\sum_{n\geqslant 1} a_ne^{-x \lambda_n}dx$$

but either it is not, or I'm missing a trick here.