What is the function of proviron

using the notation of umbral calculus where each power ζ r {\displaystyle \zeta ^{r}} is to be replaced by ζ ( r ) {\displaystyle \zeta (r)} , so . for k = 2 {\displaystyle k=2} we have ∫ 0 ∞ x n e x ( e x − 1 ) 2 d x = n ! ζ ( n ) , {\displaystyle \int _{0}^{\infty }{\frac {x^{n}e^{x}}{(e^{x}-1)^{2}}}\mathrm {d} x={n!}\zeta (n),} while for k = 4 {\displaystyle k=4} this becomes