We have Euler's well known result
$$ \sum_{r = 1}^n\frac{1}{r} = \log n + \gamma + O\Big(\frac{1}{n}\Big) $$
where $\gamma$ is the Euler-mascheroni constant. I experimentally observed a generalisation of the above. We have for $a > 1$,
$$ \sum_{r = 1}^{n-1}{1 \over r \sqrt[a]{\log \big(\frac{\log n}{\log r}\big)}} = \Gamma\Big(1 - \frac{1}{a}\Big)\log n + C_a + O\Big(\frac{1}{n}\Big) $$
where $C_a$ is a constant that depends only on $a$. If this is true than Euler's result will correspond to the special case $a \to \infty$.
Question: Can this be proved or disproved?
$$S_n=\sum_{r = 1}^{n-1}{1 \over r \sqrt[a]{\ln \big(\frac{\ln n}{\ln r}\big)}}=\sum_{r = 1}^{n-1}e^{-\ln r} \ln^{-\frac1a} \big(\frac{\ln n}{\ln r}\big)$$ Switching to integration and using $dr=d(e^{\ln r})=e^{\ln r}d(\ln r)$ $$S_n\sim\int_1^{n-1}e^{-\ln r} \ln^{-\frac1a} \big(\frac{\ln n}{\ln r}\big)dr=\int_1^{n-1}\ln^{-\frac1a} \big(\frac{\ln n}{\ln r}\big)d(\ln r)$$ $$\overset{x=\frac{\ln r}{\ln n}}{=}\ln n\int_0^\frac{\ln (n-1)}{\ln n}(-\ln x)^{-\frac1a} \,dx\overset{x=e^{-s}}{=}\ln n\int_{\ln(\frac{\ln n}{\ln(n-1)})}^\infty e^{-s}s^{-\frac1a}ds$$ $$=\ln n\int_0^\infty e^{-s}s^{-\frac1a}ds-\ln n\int_0^\frac{\ln (n-1)}{\ln n} e^{-s}s^{-\frac1a}ds$$ $$=\ln n\,\Gamma\left(1-\frac1a\right)-\frac{\ln^\frac1a n}{(1-\frac1a)\,n^{1-\frac1a}}+O\left(\frac{\ln^\frac1a n}{n^{2-\frac1a}}\right)$$ A numeric evaluation shows that there is also a constant term (for example, for $a=2\,\,C_{a=2}\approx -\,0.1982...\,)$, though I'm not sure whether we can get it in a closed form.
The Euler–Maclaurin summation formula suggests that $$S_n= \ln n\,\Gamma\left(1-\frac1a\right)+C_a+O\left(\frac{\ln^\frac1a n}{n^{1-\frac1a}}\right);\quad a> 1$$