eljose
Mar20-06, 04:30 AM
I assume you will know the Polylogarithm.. won,t you?..:redface: :redface: well my question is about the convergence of the integral:
\int_0^{\infty}dxe^{-x}Li_{-a}(x) (1)
where LI is the Polylogarithm (don,t confuse it with Logarithmic integral)..my question is if the integral above converges for a>0.
This question is deeply related with Borel Resummation of the divergent series:
1+2^{a}+3^{a}+............... you can check that multiplying
and dividing each term by n! and taking the definition of n! as an integral you check te integral (1).
Or else you can take the next realtionship:
(1-2^{a}+3^{a}-...........)=(1-2^{1+a})(1+2^{a}+3^{a}+...)
where again using Borel resummation techniques we have:
\sum_{n=0}^{\infty})(-1)^{n}n^{a}=\int_0^{\infty}dxe^{-x}Li_{-a}(-x) (2)
so using (1) or (2) we can define a "sum" for the divergen series:
\sum_{n=0}^{\infty}n^{a}
\int_0^{\infty}dxe^{-x}Li_{-a}(x) (1)
where LI is the Polylogarithm (don,t confuse it with Logarithmic integral)..my question is if the integral above converges for a>0.
This question is deeply related with Borel Resummation of the divergent series:
1+2^{a}+3^{a}+............... you can check that multiplying
and dividing each term by n! and taking the definition of n! as an integral you check te integral (1).
Or else you can take the next realtionship:
(1-2^{a}+3^{a}-...........)=(1-2^{1+a})(1+2^{a}+3^{a}+...)
where again using Borel resummation techniques we have:
\sum_{n=0}^{\infty})(-1)^{n}n^{a}=\int_0^{\infty}dxe^{-x}Li_{-a}(-x) (2)
so using (1) or (2) we can define a "sum" for the divergen series:
\sum_{n=0}^{\infty}n^{a}