If $$ H(x)=\sum_{n \leq x} \frac{1}{n} $$ what is its Mellin transform? I was able to find the Mellin transform of $\log(x+1)$ and of $\frac{1}{x+1}$, but I'm quite a bit inexperienced so I haven't been able to deduce the result. I've also seen results of more general functions, but I'm having trouble interpreting all the notation.
Note: this is not a homework or test problem, but merely a curiousity.
Ok I'm going to put an attempt at a solution based on @robjohn 's suggestions. Using the identity he provided, we have
$$ \int_0^{\infty} H(x) x^{s-1} dx=\int_0^{\infty} \sum_{k=1}^{\infty} \frac{x^s}{k(k+x)} dx=\sum_{k=1}^{\infty} \int_0^{\infty} \frac{x^s}{k(k+x)} dx= $$
$$ \sum_{k=1}^{\infty} k^{s-1}\int_0^{\infty} \frac{y^s}{1+y} dy=\sum_{k=1}^{\infty} k^{s-1}\pi \csc(\pi(s+1))=-\frac{\pi}{\sin(\pi s)}\zeta(1-s) $$
Does this seem correct? I feel like this is wrong.