From the first post we know that the integral series for log(x) is f(x) = s1(x) + s2(x). And:

s2 has the pattern of an exponential generating function because part. The harmonic series is then the series that is generating our result. If we expand s2 by each power of x we can see what the harmonic series looks like in a generating function.

A nice part of exponential generating functions is that taking the gf’s derivative just slides down the coefficient to one lower power.

Each term of s2′ has one more reciprocal being summed than the same term in s2 itself. s2 for x^1 has 1/1, but s2′ has 1/1 + 1/2. For each term, the difference between s2′ – s2 is

So, replacing s2 with f to make differential equations easier, we have:

This is a pretty interesting and not immediately obvious sum. But:

Which is just f’ – f with k shifted forward. Since we can’t use k = 0 anyway, we can make this shift to make the sum actually valid starting from 0. Which means:

This integral does not have an elementary solution, but according to Wolfram Alpha it is Ei(2x) – Ei(x) where Ei is the Exponential Integral special function.

So the final answer is not clean or neat but there it is.