The harmonic number is well known to be defined as the finite sum of inverses of consecutive integers starting with 1:
$$H_n = \sum_{k=1}^n \frac{1}{k}\tag{1}$$
Playing with other possibilities I wondered happens if we replace $k$ by $\log(k)$ (skipping over the term with $k=1$), i.e. what can be said about the sum
$$h_n = \sum_{k=2}^n \frac{1}{\log(k)}\tag{2}$$
or, more generally,
$$h_{n}^{(p)} = \sum_{k=2}^n \frac{1}{\log(k)^p}\tag{3}$$
Let us first find an integral representation.
Letting for $k\ge 2$
$$ \frac{1}{\log(k)} = \int_{0}^{\infty} e^{-t \log(k)} \,dt=\int_{0}^{\infty} k^{-t}\tag{4}$$
and doing the sum under the integral
$$\sum_{k=2}^{n} k^{-t} = H_n^{(t)}-1\tag{5}$$
where $H_n^{(t)}$ is the generalized harmonic number of order $t$ defined by $(5)$,
we find the interesting expression
$$i_{n} =\int_{0}^{\infty}( H_n^{(t)}-1)\,dt\tag{6}$$
which contains an integral over the whole of the generalized harmonic number with respect to the second index.
Question 1: can you find a closed expression for this integral $(6)$ or the sum $(2)$ as a function of $n$?
Question 2: what is the asymptotic behaviour of $h_n$ or $i_n$ for $n\to \infty$?
Similarly with
$$\frac{1}{\log(k)^p} = \frac{1}{\Gamma(p)} \int_0^{\infty } t^{p-1} \exp (t (-\log (k))) \, dt\tag{7}$$
we have
$$h_{n}^{(p)} =\frac{1}{\Gamma(p)}\int_{0}^{\infty}t^{p-1}( H_n^{(t)}-1)\,dt\tag{8}$$
and the same questions as above for integer $p\ge 2$.