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I would like to know if is it possible to combine in a nice way the infinite product that satisifies the Möbius function and the formula for "general integration by parts".

Motivation. Since one can define for each integer $k\geq 1$ and $ \left| x \right| <1$ $$f_k(x)= \left( 1-x^k \right)^{\frac{\mu(k)}{k}} ,$$ where $\mu(n)$ is the Möbius function and we know from this MSE, Really advanced techniques of integration the shape of a "formula for general integration by parts". I think that could be possible/feasible to combine this formula with the infnite product $(16)$ of previous MathWorld's article, that is the infinite product that satisfies the Möbius function to get a limit $\lim_{n\to\infty}$ from the formula of integration by parts, that provide us a statement.

Question. What could be a nice combination of previous statements? Can you provide me details and hints, or hints and your final statements? I believe that could be a nice exercise. Many thanks.

I know also that, if there are no mistakes in this calculation that $$f'_k(x)= \mu(k)\left( 1-x^k \right)^{\frac{\mu(k)}{k}}\frac{x^{k-1}}{x^k-1}. $$

Final remark. I don't know if also is possible to combine with Lambert series for the Möbius function, the Prime Number Theorem... I don't know and I am saying this as motivation, thus only is required to combine the infinite product with integration by parts.

Rerences:

Möbius, Über eine besondere Art von Umkehrung der Reihen, Journal für die reine und angewandte Mathematik (1832), Vol. 9, page 120.

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  • Thus what am I asking is how to exploit both statements to get a nice closed form. Is required to do the $$\lim_{n\to\infty}$$ to do feasible the combination of both statements. Many thanks all users. –  Feb 25 '17 at 20:25
  • I don't know what you mean by "combining the infinite product" with integration by parts. Do you have any examples at all to illustrate what you mean by combining an infinite product with integration? Also, you write about infinite products, but you don't actually give one, so it's not clear what infinite product you have in mind. – Gerry Myerson Feb 26 '17 at 00:38
  • I was disconnected, many thanks for your question @GerryMyerson . The problem is that I can not state all calculations and related justifications of the statements, that if there are no mistakes in my calculations that is the specialization of the integration by parts with the $f_k(x)'s$, my attempt, but I would like to know if it's possible to do more calculations in RHS:

    $$\lim_{n\to\infty}\int_0^x-\prod_{k=2}^n f_k(y)dy=e^{-y}-\lim_{n\to\infty}\sum_{k=2}^n\int_0^x \mu(m)(1-y^m)^{\frac{\mu(m)}{m}}\frac{y^{m-1}}{y^m-1}\prod_{\substack{k=1\k\neq m}}^nf_k(y)dy.$$

     –  Feb 26 '17 at 08:33

1 Answers1

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Please stop with those kind of questions (which doesn't mean anything, except that you didn't understand some formula and that you want to apply it anyway)

$$-\sum_{n=1}^\infty \frac{\mu(n)}{n}\log (1-x^n) = x \quad \implies \quad \prod_{n=1}^\infty (1-x^n)^{\mu(n)/n} = e^{-x}$$ is just the definition of $\mu(n)$ together with $-\log (1-x^n) = \sum_{k=1}^\infty \frac{x^{nk}}{k}$

What to say more ? You want to play with this and integration by parts ? Ok have fun, but this isn't the way you'll learn some maths.

reuns
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  • Many thanks I have no a strategy to apply the formula in other identity, only was an exercise to participate in this site. I believe that is interesting try to combine the infnite product with the integration by parts that I've refered in the link. Feel free to update your answer if you want. Any case many thanks a good week. –  Feb 25 '17 at 23:03