What is the infinite sum of logarithm lnK divided by k(K+1)?

Question

I was trying to calculate the following integral: $\displaystyle\int_1^\infty\frac{dx}{x \lfloor x \rfloor}=? $
which I found to be equivalent to $\displaystyle\sum_{k=2}^\infty\frac{\ln(k)}{k(k+1)} $. There is a close relative of this sum which is $\displaystyle\sum_{k=2}^\infty\frac{\ln(k)}{k^2}=-\zeta^\prime(2)$ and its value is known in terms of Glashier-Kinkelin constant (A):
$\displaystyle\sum_{k=2}^\infty\frac{\ln(k)}{k^2}= \frac{\pi^2}{6}[12\ln(A)-\gamma-\ln(2\pi)]$
My idea for solving this was to assume a function as $\xi(s)=\displaystyle\sum_{k=1}^\infty\frac{1}{k^s+k^{s-1}}=\zeta(s-1)-\xi(s-1)$ with properties like $\xi(1)=\zeta(1)-1$, $\xi(2)=1$, $\xi(3)=\frac{\pi^2}{6}-1$, and ...
the answer would be $\displaystyle\int_1^\infty\frac{dx}{x \lfloor x \rfloor}=-\xi^\prime(2) $ but I don't know how to calculate this.

Answer 1

The logarithmic series is related to the Alladi-Grinstead constant, and its decimal expansion can be

found on OEIS. It has no known closed form, but it can also be expressed as $~\displaystyle\sum_{n=1}^\infty\frac{\zeta(n+1)-1}n$.

However, your integral is not equivalent to it, but rather to the similar $~\displaystyle\sum_{k=2}^\infty\frac{\ln(k)}{k(k\color{red}-1)},~$ which is

also found on OEIS, being equivalent to $~-\displaystyle\sum_{n=2}^\infty\zeta'(n),~$ which does not have a known closed form either.