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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-- enter image description here

enter image description here

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 - enter image description here

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.

Can somebody please help to make deduction given in the book .

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  • $\begingroup$ Do you know Landau's Theorem? $\endgroup$ – Conrad Dec 16 '19 at 21:45
  • $\begingroup$ @Conrad it's statement is given in question . Is Landau theorem something else? $\endgroup$ – Tim Dec 17 '19 at 4:58
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    $\begingroup$ 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) $\endgroup$ – Conrad Dec 17 '19 at 12:50
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  • 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 $\sum_{n\le x}\lambda(n)n^{-1}$ ?

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  • $\begingroup$ 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. $\endgroup$ – Tim Dec 17 '19 at 5:15
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    $\begingroup$ 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). $\endgroup$ – reuns Dec 17 '19 at 5:17
  • $\begingroup$ 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? $\endgroup$ – Tim Dec 17 '19 at 5:27
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    $\begingroup$ 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 $\endgroup$ – reuns Dec 17 '19 at 5:30
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    $\begingroup$ 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) $\endgroup$ – reuns Dec 17 '19 at 5:32

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