Ikehara's Tauberian theorem for Dirichlet series states that if $$F(s)=\sum_{1}^{\infty}\frac{f(n)}{n^s}$$ with $f(n)\geq 0$ is such that $$F(s)=\frac{G(s)}{s-1}+H(s)$$ for $\sigma>1$ with $G,H$ analytic and $G(1)\neq 0$, then the continuity of $F(1+it)$, $t\neq 0$, implies that $$\frac{1}{x}\sum_{n\leq x}f(n)\sim G(1).$$

I am wondering if there is a converse to this statement, i.e. Do we know that if $F(1+it)$ has other discontinuities at say $t=\pm a_i$, $i=1,2,...$, then $$\frac{1}{x}\sum_{n\leq x}f(n)\not\sim G(1)?$$

If not, I can imagine that there is a converse under stricter assumptions on $F$, say if we assume $F(s)$ is analytic in a neighbourhood of every point of the line $\sigma=1$ except at $t=0$ and $t=\pm a_i$, or under stricter assumptions on the $a_i's$, then $$\frac{1}{x}\sum_{n\leq x}f(n)\not\sim G(1).$$ Do such theorems exist?


We apply the method outlined (Theorem 15.3) in Montgomery & Vaughan 'Multiplicative Number Theory', volume 1, chapter 15.

We assume much stronger assumptions for $F(s)$ on the line $\sigma=1$ that it has simple poles at $s=1$, $s=1+it_0$ and $s=1-it_0$ with $t_0>0$. Also, assume that the Dirichlet series $F(s)=\sum a_n n^{-s}$ has $\sigma_c=1$. Then we have for $A(x)=\sum_{n\le x} a_n$, $$ B(s):=\frac{F(s)}s-\frac{c}{s-1}=\int_1^{\infty} \left(A(x) - c x\right)x^{-s-1} dx $$ holds for $\sigma>1$.

If $A(x)\geq cx$ for $x>X_0$, then for $\sigma>1$, $$ B(s)+\frac12 e^{i\phi} B(s+it_0)+\frac12 e^{-i\phi} B(s-it_0)=\int_1^{\infty} (A(x)-cx) (1+\cos(\phi-t_0\log x)) x^{-s-1}dx \ \ \ \ (*) $$ The LHS of (*) has a simple pole at $s=1$ with residue $$ G(1)-c+\frac12 e^{i\phi} \beta + \frac12 e^{-i\phi} \overline{\beta} $$ where $\beta = \mathrm{Res}_{s=1+it_0} F(s)/s\neq 0$. On the other hand, the RHS does not tend to $-\infty$ as $s\rightarrow 1+$, because the integrand is nonnegative for $x\geq X_0$. We now take $\phi$ so that $$ \frac12 e^{i\phi} \beta + \frac12 e^{-i\phi} \overline{\beta} = -|\beta|. $$ Then $$ G(1)-c-|\beta|\geq 0 $$ yiending $$ c\leq G(1)-|\beta|. $$ Thus, $\liminf\limits_{x\rightarrow\infty} A(x)/x \leq G(1)-|\beta|$. Similarly, we can prove that $\limsup\limits_{x\rightarrow\infty} A(x)/x \geq G(1)+|\beta|$. Then we have the desired result $A(x)/x \not\sim G(1)$.


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