# Gram's series for integral equation

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?

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(i) What does $f(x)$ in $(1)$ correspond to in $(0)$, $\pi(e^t)$? (ii) Is $K(st)=1/(e^{st}-1)$? (iii) Why $g(s)/s$? Looks like you imply a logarithmic derivative somewhere... –  draks ... Jan 3 at 22:18
Shouldn't be $dt$ in $(0)$? Does or how does $(0)$ relate to one of my questions: math.stackexchange.com/q/280984/19341 I would be glad to read your opinion... –  draks ... Jan 21 at 9:42