This is the text of the exercise:
Let $\mathbb{C+}$=$\{a+ib\in\mathbb{C}|a>0\}$ and $\mu_{_F}(a)$ the Lindelof function of $F$.
Let $F(s)=\sum_{n\geq1}\frac{a_n}{n^s}$=$\frac{4}{s-1}+\frac{1}{(s-4/5)^2}-\frac{2}{s-4/5}+h(s)$
Where $h(s)$ is an holomorphic function over $\mathbb{C}+$ and $a_n<<(log(n))^3$ $\forall n \in \mathbb{N}$
Suppose that $\mu_{_F}(a) = \begin{cases} 0, & \text{if $a>1$} \\ 1/2-a/2, & \text{if $0<a\leq1$ is odd} \end{cases}$
Find an asymptotic formula for $\sum_{n\leq x}a_n$ with an error $O(x^{1/3+\epsilon})$
$\textit{I have problem in optimizing the rest. So i explane what is my idea and where is the problem}$.
To get the formula i use the Riemann's method, so:
1) $\sum_{n\leq x}a_n=\frac{1}{2 \pi i}\int^{c+iT}_{c-iT} \frac{F(s)x^s}{s}ds$+$O(\frac{x^{1+\epsilon}}{T})$ $\forall \epsilon \geq0$ (Perron's formula with rest).
2)Using the Cauchy residue formula i move the integration line using the hypotesis on Lindelof function in order to get an error term depending on $a$:
$\sum_{n\leq x}a_n=\frac{1}{2 \pi i}\int^{c+iT}_{c-iT} \frac{F(s)x^s}{s}ds$+$O(\frac{x^{1+\epsilon}}{T})=$
$=\frac{1}{2 \pi i}\int^{a-iT}_{a+iT} \frac{F(s)x^s}{s}ds+\frac{1}{2 \pi i}\int^{a+iT}_{c+iT} \frac{F(s)x^s}{s}ds+\frac{1}{2 \pi i}\int^{c-iT}_{a-iT} \frac{F(s)x^s}{s}ds+\sum_{Poles} res(\frac{F(s)x^s}{s})+O(\frac{x^{1+\epsilon}}{T})$.
The modulus of the integrals parallel to the $\mathbb{R}$ axis is irrelevant (it comes from the convexity of the integrating function).
So i get: $\sum_{n\leq x}a_n=\sum_{Poles} res(\frac{F(s)x^s}{s})+O(\int_{-T}^T|t|^{\mu_{_{F}}(a)-1+\delta}x^a dt)+O(\frac{x^{1+\epsilon}}{T})$.
The error term is than $O(|T|^{\mu_{_{F}}(a)+\delta}x^a+\frac{x^{1+\epsilon}}{T})$ (the exponent of $T$ is $\leq1$ and positive).
The last formula is valid for every positive $\epsilon$ and $\delta$ but i don't know how to minimize the function.
I mean the equation should be $T^{\mu_{_{F}}(a)+\delta}x^a=\frac{x^{1+\epsilon}}{T}$ and so $T=x^\frac{1+\epsilon -a}{\mu_{_{F}}(a)+\delta +1}$.
Than $1-\frac{1+\epsilon -a}{\mu_{_{F}}(a)+\delta +1}=\frac{1}{3}$
What i do here is ignore $\epsilon$ and $\delta$ explicit the Lindelof function and get the $a=0$.
I don't understand formally why i can ignore $\delta$ and $\epsilon$ and if $a=0$ is lecit (this is becouse of the first problem, i mean if i can understand the expression with $\epsilon$ and $\delta$ i put $a=\epsilon ^{'}$ and than using the definition of limit i think i can get my result).
Any advice please?