Finding definite integral in interval (positive infinity ~ e) ∫ 1 / X (LNX) ^ P DX

∫[e,+∞]1/[x(lnx)^p] dx
= ∫[e,+∞](lnx)^(-p) dlnx
= 1/(lnx)^(p-1) * 1 / (-p+1)
= 0 - 1/(lne)^(p-1) * 1/(1-p)
= -1/(1-p)
= 1/(p-1)

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