5
$\begingroup$

Do we know the sum, $$\sum_p \frac{\chi\left(p\right)}{p}\textrm{?}$$ for $\chi\left(n\right)$ being $1$ if $n=1\textrm{ mod }4$, $-1$ if $n=3\textrm{ mod }4$ and zero otherwise.

I'm looking for an exact value, if there is one, and a derivation.

$\endgroup$
  • 2
    $\begingroup$ I strongly doubt that this sum has a nice closed form. $\endgroup$ – Peter Sep 19 '19 at 12:51
  • $\begingroup$ You should know from the Euler product for $L(s,\chi)$ that there is never a closed-form for $\sum_p \frac{\chi(p)}{p}$, only for $\log L(1,\chi)=\sum_{p^k} \frac{\chi(p^k)}{k p^k}$ when $\chi(-1) = -1$ and for $\log L(2,\chi)$ when $\chi(-1)=1$ $\endgroup$ – reuns Sep 19 '19 at 14:12
3
$\begingroup$

Since $|\chi(p)|\leq 1$, we may approximate $\frac{\chi(p)}{p}$ with $-\log\left(1-\frac{\chi(p)}{p}\right)$ and deduce that $$ \sum_{p}\frac{\chi(p)}{p} \approx \log\prod_{p}\left(1-\frac{\chi(p)}{p}\right)^{-1}=\log L(1)=\log\frac{\pi}{4} $$ where $L(s)$ stands for the Dirichlet $L$-function $L(s)=\sum_{n\geq 1}\frac{\chi(n)}{n^s}$. With the same mechanism leading to a series representation for the prime $\zeta$ function (i.e. Moebius' inversion formula) we may state that

$$ \sum_{p}\frac{\chi(p)}{p} = \sum_{n\geq 1}\mu(n)\frac{\log L(n)}{n} $$ where the series on the RHS has a reasonable convergence speed, due to $L(n)\approx 1-\frac{1}{3^n}\Rightarrow\log L(n)\approx -\frac{1}{3^n}$. Numerically we have $$ \sum_{p}\frac{\chi(p)}{p}\approx -0.18654766883298535284. $$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Then, it would seem that from a certain perspective, there are "slightly" more primes of the $4n + 3$ form than those of the $4n + 1$ form. Beautiful mystery! $\endgroup$ – Piquito Sep 19 '19 at 13:29
  • 2
    $\begingroup$ @Piquito: the phenomenon has been deeply investigated: en.wikipedia.org/wiki/Chebyshev%27s_bias $\endgroup$ – Jack D'Aurizio Sep 19 '19 at 13:31
  • 2
    $\begingroup$ Every time in mathematics the expression "it has been deeply investigated" is used, that means that a large part of the human population is not invited to the party. $\endgroup$ – Piquito Sep 19 '19 at 13:39
  • 2
    $\begingroup$ Well, I guess it is safe to say that a large part of the human population does not give a damn about Number Theory, so that is not utterly surprising. $\endgroup$ – Jack D'Aurizio Sep 19 '19 at 13:43
  • $\begingroup$ However, I wanted to refer to the mathematical population of the world (my English is poor). Do not respond to this last comment. Regards $\endgroup$ – Piquito Sep 19 '19 at 13:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.