Is the approximation in terms of series for the logarithm
$$\log(z)= \sum_{n=0}^{\infty}\frac{2}{2n+1}\Bigl(\frac{z-1}{z+1}\Bigr)^{2n+1} $$
a good approximation if I replace this series inside the integral
$$ \int_{a}^{\infty}f(z)\log(z)\,\mathrm dz $$
at least if $a$ is a big number, let us say $ a >100 $?
If you're taking a power series of $log(x)$, it will only be accurate up to some order right, but when you try to integrate this to infinity that order, no matter how large will give you "infinite" error.
It Doesn't depend on a at all.
The first term doesn't converge under integration at all, no mater what a you pick
