Dirichlet series and Laplace transform

Question

Let $\displaystyle\sum_{n=1}^\infty \dfrac{a_n}{n^s}$ be a Dirichlet series. It can be represented as a Riemann-Stieltjes integral as follows:

$$\displaystyle\sum_{n=1}^\infty \dfrac{a_n}{n^s}=\int_1^\infty\dfrac{1}{x^s}d(A(x))=\lim_{x\to\infty}\dfrac{A(x)}{x^s}+s\int_1^\infty \dfrac{A(x)}{x^{s+1}}dx$$

Now, observing that the integral resembles a Laplace transform with $x=e^t$, we have:

$$\displaystyle\sum_{n=1}^\infty \dfrac{a_n}{n^s}=\lim_{t\to\infty}\dfrac{A(e^t)}{e^{st}}+s\int_0^\infty A(e^t) e^{-st}dt$$

However, some texts simplify this expression to:

$$\displaystyle\sum_{n=1}^\infty \dfrac{a_n}{n^s}=s\int_0^\infty A(e^t) e^{-st}dt$$

The question arises regarding why the limit term $\lim_{t\to\infty}\dfrac{A(e^t)}{e^{st}}$ vanishes. While for certain functions like the Chebyshev function this holds true due to $A(x)=O(x)$, What about other Dirichlet series?

Answer 1

Today I found a theorem that answers my question. This can always be guaranteed when the Dirichlet series has a half-plane of convergence with $\sigma > 0$, $s=\sigma + it$. This result is known as Kronecker's lemma and can be checked in Hildebrand's notes on analytic number theory, Page 59.