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Given the totient summatory function

$$\Phi(x)=\sum\limits_{n=1}^x\varphi(n)\tag{1}$$

and the related Dirichlet series

$$\frac{\zeta(s-1)}{\zeta(s)}=\underset{N\to\infty}{\text{lim}}\left(\sum\limits_{n=1}^N\frac{\phi(n)}{n^s}\right)\,,\quad\Re(s)>2\tag{2}$$

the explicit formula for

$$\Phi_o(x)=\underset{\epsilon\to 0}{\text{lim}}\left(\frac{\Phi(x-\epsilon)+\Phi(x+\epsilon)}{2}\right)\tag{3}$$

derived from the residues at the poles of

$$\frac{\zeta(s-1)}{\zeta(s)}\,\frac{x^s}{s}\tag{4}$$

is

$$\Phi_o(x)=\frac{3\,x^2}{\pi^2}+\underset{T\to\infty}{\text{lim}}\left(\sum\limits_{|\Im(\rho)|\le T}\frac{x^\rho\,\zeta(\rho-1)}{\rho\,\zeta'(\rho)}\right)+\frac{1}{6}+\underset{N\to\infty}{\text{lim}}\left(\sum\limits_{n=1}^N\frac{x^{-2\,n}\,\zeta(-2\,n-1)}{(-2\,n)\,\zeta'(-2\,n)}\right)\tag{5}$$

where $\rho$ is a non-trivial zero of $\zeta(s)$.


I've been told the explicit formula for $\Phi_o(x)$ defined in formula (5) above doesn't converge.


Peter Humphries explains the reason for this in a comment on my my related question.

The explicit formula for $\sum_{n \leq x} \varphi(x)$ is wrong; when you shift the contour, the shifted contour integral is not small. One can use this to show that the error term for this sum is at least as large as a constant multiple of $x\sqrt{\log \log x}$ infinitely often.

Peter provides further clarification in another comment.

What I mean is that $\sum_{n \leq x} \varphi(x)$ has a main term, coming from the pole of $\zeta(s - 1)/\zeta(s)$ at $s = 2$, and an error term of size at least $x\sqrt{\log \log x}$, which does not come from the poles of $\zeta(s - 1)/\zeta(s)$ at the zeroes of $\zeta(s)$.


Note that in the two comments quoted above $\varphi(x)$ should have been $\varphi(n)$.


Question (1):

Given the summatory function

$$f(x)=\sum\limits_{n=1}^x a(n)\tag{6}$$

and assuming the related Dirichlet series

$$F(s)=\underset{N\to\infty}{\text{lim}}\left(\sum\limits_{n=1}^N\frac{a(n)}{n^s}\right)\,,\quad\Re(s)>\alpha\ge 2\tag{7}$$

doesn't converge at $\Re(s)=2$, is it true in general that the explicit formula for

$$f_o(x)=\underset{\epsilon\to 0}{\text{lim}}\left(\frac{f(x-\epsilon)+f(x+\epsilon)}{2}\right)\tag{8}$$

derived from the residues at the poles of

$$F(s)\,\frac{x^s}{s}\tag{9}$$

doesn't converge?


Question (2):

Assuming the answer to Question (1) above is yes, is $2$ a special number or can $2$ in question (1) above be replaced with some $\lambda$ where $1<\lambda<2$ where the answer to the modified question would still be yes?


Question (3):

Assuming the answer to Question (2) above is no:

Question (3a): What is an example of an explicit formula that converges associated with the conditions stated in Question (1) above?

Question (3b): Can $2$ in question (1) above be replaced with some $\lambda$ where $\lambda>2$ where the answer to the modified question would be yes?

Steven Clark
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  • There is nothing special about $2$. You can rescale without problem. I think $\zeta(s/2)$ is an example of what you're looking for in your third question, but I find your phrasing there a bit ambiguous so I could be wrong. But I think the "real" answer to your question is that you should examine Perron's formula (or other Tauberian theorems) to see how one extracts those summatory results. Or look at a proof of various explicit formulas. I'll note that it's not obvious to me that your claimed explicit formula doesn't converge, but it is wrong (for the reason that Peter mentioned). – davidlowryduda Mar 21 '22 at 20:04
  • @davidlowryduda With respect to convergence, I meant it doesn't converge to the definition $\Phi_o(x)=\underset{\epsilon\to 0}{\text{lim}}\left(\frac{\Phi(x-\epsilon)+\Phi(x+\epsilon)}{2}\right)$. Since the non-convergence to the definition of $\Phi_o(x)$ seems to be related to shifting the contour, I wondered is there's a general statement that can be made with respect to shifting contours. I don't really understand your example $\zeta(s/2)$. What is the definition of the coefficient $a(n)$ associated with the Dirichlet series $\zeta(s/2)=\sum\limits_{n=1}^\infty\frac{a(n)}{n^s}$? – Steven Clark Mar 21 '22 at 21:59
  • @davidlowryduda Also, my interest is in explicit formulas where $F(s)$ has an infinite number of poles related to the zeros of $\zeta(s)$, and your example $\zeta(s/2)$ doesn't meet this condition. – Steven Clark Mar 21 '22 at 21:59

1 Answers1

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For $$F(s)=\frac{\zeta(s-1)}{\zeta(s)}$$ you should look first at the residue theorem for $$\int_1^x\int_1^t \Phi(y)dydt=\frac1{2i\pi}\int_{(3)}\frac{F(s)}{s(s+1)(s+2)} x^{s+2}ds,\qquad x > 1$$ because $$\lim_{n\to -\infty} \int_{(2n+1)} |\frac{F(s)}{s(s+1)(s+2)} x^{s+2}|d|s|=0$$ $$ \lim_{T\to \infty} \int_{-2+iT}^{-\infty+iT} |\frac{F(s)}{s(s+1)(s+2)} x^{s+2}|d|s|=0$$ So to close the contour all we need is to "traverse the critical strip" that is to check if there is a sequence $T_n$ such that $$\lim_{n\to \infty}\int_{-2+iT_n}^{3+iT_n} |\frac{F(s)}{s(s+1)(s+2)}x^{s+2}|d|s| = 0$$ (this is the hardest step, needing Jensen, Borel-Catheodory,Phragmen Lindelof, Hadamard 3 circles, the density of zeros and so on)

in which case we get the explicit formula $$\int_1^x\int_1^t \Phi(y)dydt=\lim_{n\to \infty} \sum_{|\Im(z)|<T_n} Res(\frac{F(s)}{s(s+1)(s+2)}x^{s+2},z)$$

$F(s)\asymp s$ as $\Re(s)\to -\infty$ (away from the even integers) which causes many problems when trying to do the same with $\Phi(x)=\frac1{2i\pi}\int_{(3)}\frac{F(s)}{s} x^sds$.

As you can see the analytic/meromorphic continuation to $\Re(s) < -c$ (from the functional equation), its poles and growth, and the growth and poles in the crticial strip are some essential parts of the explicit formula.

reuns
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