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I am self studying a research paper in analytic number theory (Ball and Rivoal - Irrationalité d’une infinité de valeurs de la fonction zêta aux entiers impairs) and I am unable to think about an argument in 1 of it's lemma.

The paper is in French but I am adding images of formulas references only. Rest of text which could be useful I will write. (For more details see 7th page, logical p. 199, of research paper.)

equation (8) of linked paper

auxiliary results

Assume $\tilde S(z)$ to be RHS of equation 8.

Assume that it has been proved $\tilde S(z)$ is continuous on $E$ and it has also been proved that $S_n(z)$ is continuous on $E$ where $E =\{z \in \mathbb C : \lvert z\rvert \geq1\}$. Also, $a$, $b$, $n$ are integers and $1\leq b \leq a$ and $1 \leq r < a/2$.

I am not able to think why they must be equal on the given domain. On one side is an infinite series and on other side an integral with some other terms.

Can someone please help, I have no clue regarding this and no further explanation has been given in the research paper.

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  • $\begingroup$ I did some light editing. You mention supposing that $S_n'(z)$ is equal to the right-hand side of (8), but (8) is an equation involving $S_n(z)$, not $S_n'(z)$. Is that really what you meant? $\endgroup$
    – LSpice
    Commented Apr 7, 2020 at 12:56
  • $\begingroup$ @Lspice I wrote Tilde (~) over S(z) as S'(z) as I didn't knew how to write ~ over S(z) in math Jax. $\endgroup$
    – Arnold
    Commented Apr 16, 2020 at 8:34
  • $\begingroup$ Same as in TeX: $\tilde S(z)$ \tilde S(z). $\endgroup$
    – LSpice
    Commented Apr 16, 2020 at 10:02
  • $\begingroup$ @Gaussian has already answered it. $\endgroup$
    – LSpice
    Commented Apr 16, 2020 at 12:55
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    $\begingroup$ This question already has an accepted answer; why have you bumped it with an edit? $\endgroup$
    – Yemon Choi
    Commented Jan 6, 2022 at 0:44

1 Answer 1

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Since I can't make comment, let me write this as an answer. You know that the LHS and the RHS are continuous on $|z|\geq1$, so to prove that they are equal on $E$, it is enough to show that they are equal on $|z|>1$.

Actually, the condition $|z|>1$ helps you writing the integrand of $I_n(z)$ as a power series, the variable of which is $z$: then, you will have to check that you can switch the integral and the sum... and you'll have to check that this is equal to the power series defining $S_n(z)$.

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  • $\begingroup$ can you please write a detailed answer on how it will be actually done!! $\endgroup$
    – Arnold
    Commented Apr 16, 2020 at 12:19
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    $\begingroup$ @Dxdxdade: I don't think MO is meant for people to do your work. $\endgroup$
    – DamienC
    Commented Apr 16, 2020 at 13:52
  • $\begingroup$ @DamienC , it ok I was already thinking of trying it myself once again and then if I have any questions I will ask to Gaussian. $\endgroup$
    – Arnold
    Commented Apr 16, 2020 at 13:54
  • $\begingroup$ @Gaussian , why condition |z|>1 allows to write integrand of $I_n(z) $ as power series? $\endgroup$
    – Arnold
    Commented Apr 16, 2020 at 17:01

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