# Doubt in proof of theorem 8.20 of Apostol Modular functions and Dirichlet series in number theory

I am self studying number theory from Tom M Apostol Dirichlet Series and Modular functions in number theory and I have a doubt in theorem 8.20 of the book. I am attaching images of relevant results.

The Theorem--

My doubt is in last paragraph of proof how by integral analog of Landau theorem function on left is analytic in half plane $$\sigma$$ > 1/2 .

Landau theorem -

What I could deduce using Landau theorem--> The function on right is analytic. So, the function on left converges for $$\sigma$$ >1 , which is not similar to result deduced by Apostol.

• Do you know Landau's Theorem? – Conrad Dec 16 '19 at 21:45
• @Conrad it's statement is given in question . Is Landau theorem something else? – Tim Dec 17 '19 at 4:58
• Landau's theorem is the statement that a Dirichlet series/integral with coefficients (integrating function) of constant sign must have a singularity at the real value on the abscissa of convergence; it is usually used in the counterpositive (no singularity at a real value and constant sign coefficients implies abscissa is strictly to the left) or to show omega results (we know the abscisssa and no singularity, hence coefficients/integrating function must change sign indefinitely, hence something cannot be always greater than something else, but they must switch order indefinitely) – Conrad Dec 17 '19 at 12:50

• The first part is to assume that for some $$\Re(s) > 1$$, $$L_n(s)=\sum_{n\le x}\lambda(n)n^{-s}= 0$$, then take a sequence $$t_k$$ such that for $$p\le x$$, $$p^{-s-it_k}\to -p^{-s}$$ which implies that $$n^{-s-it_k}\to\lambda(n)n^{-s}$$ and hence $$\zeta_n(z+s+it_k)\to L_n(z+s)$$ uniformly around $$z=0$$ which means that for $$k$$ large enough $$\zeta_n(s+it_k+z)$$ has a zero near $$z=0$$.

• The last part is that $$\forall x > X,\sum_{n\le x} \lambda(n)n^{-1}\ge 0$$ implies $$F(s)=\int_X^\infty (\sum_{n\le x}\lambda(n)n^{-1})x^{-s}dx$$ has a singularity at its abscissa of convergence $$\sigma$$.

Proof : if $$F$$ is analytic then $$F(\sigma-\epsilon) = \sum_{k\ge 0}\frac{\epsilon^k}{k!} \int_X^\infty (\sum_{n\le x}\lambda(n)n^{-1})x^{-\sigma} (\log x)^k dx$$ if it is finite then we can invert $$\sum,\int$$ obtaining that $$\int_X^\infty (\sum_{n\le x}\lambda(n)n^{-1})x^{-\sigma+\epsilon}dx$$ is finite.

The analyticity of the LHS shows $$\sigma=1/2$$.

I don't know about the converse, assuming the RH what can we say about the zeros of $$\zeta_n$$ and the sign of $$\sum_{n\le x}\lambda(n)n^{-1}$$ ?

• my doubt was how to deduce that function on left is analytic everywhere in half plane $\sigma$ >1/2 using integral analog of Landau 's theorem . But you are perhaps explaining something else. – Tim Dec 17 '19 at 5:15
• The $\ge 0$ condition implies the RHS has a singularity on the real axis at its abscissa of convergence $\sigma$. The LHS is analytic on $(1/2,\infty)$ thus it must be $\sigma=1/2$. Once the RHS converges for $\Re(s) > 1/2$ then the LHS is analytic for $\Re(s) > 1/2$ (ie. the RH is true). – reuns Dec 17 '19 at 5:17
• I understand why LHS is analytic on (1/2, $\infty$) . But why condition that C(x) is nonnegative implies that RHS has a singularity on real axis at its abcisaa of convergence? – Tim Dec 17 '19 at 5:27
• Expand $F(\sigma-\epsilon)$ in a Taylor series in $\epsilon$, every term is non-negative and their formal sum is $\int_{n_0}^\infty C(x)x^{-\sigma+\epsilon}dx$ which has to be finite – reuns Dec 17 '19 at 5:30
• It is obvious that the RHS is complex differentiable, holomorphic, analytic on its half-plane of convergence. The equality between LHS and RHS is true everywhere both sides are analytic (this is called analytic continuation) – reuns Dec 17 '19 at 5:32