The prime counting function $ \pi(x) $ satisfies the integral equation
$$ \log\zeta (s)= s\int_{0}^{\infty}dx \frac{ \pi (e^{t})}{e^{st}-1} \tag{0}$$
and it has the solution in terms of Gram's series $$ \pi (x) \sim 1+ \sum_{n=1}^{\infty} \frac{\log^{n}(x)}{\zeta (n+1)\Gamma (n+1)}.$$
My question here is, if this method can be extended to integral equations of the form
$$ \frac{g(s)}{s}=\int_{0}^{\infty}f(x)K(st)dt \tag{1} $$
by expanding the function $$ \frac{g(s)}{s}=a_{0}+ \sum_{n=1}^{\infty}a_{n}s^{-n},\tag{2}$$
in this case I believe that the solution inside (1) can be written as
$$ f(x)= \sum_{n=1}^{\infty}a_{n}\frac{x^{n}}{M(n+1)}$$
with $$ M(s)= \int_{0}^{\infty}K(t)t^{s-1} .$$
Am I right?