We proved in a course this version of Ikehara's tauberian theorem when the Dirichlet serie had a simple pole at s=1. First, I was wondering how we can use this result to get an asymptotic asimtation when f extends analytically to $Re(s)>=\alpha$ with a simple pole at $\alpha$ with residue equal to 1. I would be thinking of $\sum_{n \leq x} a_{n} \sim x^{1/\alpha}$.
I would be glad to have references for any generalisation of these results when the Dirichlet series has a different form (higher multiplicity poles for example?). Thank you in advance
Thanks to Reuns I found out how to conclude. First, as written above, i needed to prove that $g(t) = o(t^{\beta})$ for any $\beta > \alpha$. I will consider another $\beta'$ lying in $]\alpha,\beta[$. I will use the fact that the convergence of $f$ at $s=\beta'$ tells us after partial summation that $\int_{1}^{\infty} \frac{g(t)}{t^{\beta' + 1}}$ converges and that the Cauchy tranches will be close to 0 when we go the infinity. If $g(t)$ isn't $o(t^{\beta})$, one can choose $\epsilon >0$ and a monotic sequence $(x_{n})$ going to infinity such that $g(x_{i}) \geq \epsilon x_{i}^{\beta}$. We will now work on this integral: $$ |\int_{x_{n}}^{x_{n}^{\frac{\beta}{\beta'}}} \frac{g(t)}{t^{\beta' + 1}} dt| \geq \beta \epsilon x_{n}^{\beta} \int_{x_{n}}^{x_{n}^{\frac{\beta}{\beta'}}} \frac{1}{t^{\beta}}dt = \epsilon (x_{n}^{\beta - \beta'} - 1)dt $$ The right side of the inequality goes to $\infty$ as n goes to $\infty$. It is thus absurd.