- #1

zetafunction

- 391

- 0

[tex] \pi (x) = \sum _{n=0}^{\infty}a_n log(x) [/tex]

and we have the expression for the logarithmic integral

[tex] Li (x) = \sum _{n=0}^{\infty}b_n log(x) [/tex]

where the numbers a(n) and b(n) are known , then my question is , what could one expect about the difference expansion

[tex] \pi (x) - Li(x) = \sum _{n=0}^{\infty}(a_n - b_n) log(x) [/tex] ??