Weird solution to an integral equation So according to this wiki link
https://en.m.wikipedia.org/wiki/Integral_equation
The solution to the Fredholm equation of the first type of the form
$$g(s)=s\int_{0}^{\infty}dtK(st)f(t)$$
For a given function $g$ and kernel $K$ is
$$f(t)=\sum_{n=0}^{\infty}{\frac{a_n}{M(n+1)}t^n}$$
Where the coefficients can be found in the expansion of $g$
$$g(s)=\sum_{n=0}^{\infty}a_n s^{-n}$$
and
$$M(n+1)=\int_{0}^{\infty}dtK(t)t^n$$
is the Mellin transform of $K$ if it exists.
Now the identity for $g(s) $ is a little weird. Why is the sum over the negative powers instead?
Also if it is correct (or corrected) please direct me to a proof or something. I can't for the life of me figure out how to derive this.
 A: Here's a basic solution which I think shows the essential transformative steps; it can be made more rigorous, though perhaps with some restrictions on $f(t)$, $g(s)$, and $K(st)$, by carefully validating operations such as the interchanges of integration and summation and the range of $s$ (e.g., we should perhaps stipulate $s > 0$); nevertheless:
Given the integral equation
$g(s) = s\displaystyle \int_0^\infty K(st) f(t) \; dt, \tag 0$
we set
$u = st; \tag 1$
holding $s$ fixed, we have
$du = sdt, \tag 2$
$t = \dfrac{u}{s} = us^{-1}; \tag 3$
then
$s \displaystyle \int_0^\infty K(st)f(t) \; dt = \int_0^\infty K(st) f(t) s \; dt = \int_0^\infty K(u) f(us^{-1}) \; du; \tag 4$
we expand $f(us^{-1})$ into a power series
$f(us^{-1}) = f(t) = \displaystyle \sum_0^\infty f_n t^n = \sum_0^\infty f_n (us^{-1})^n = \sum_0^n f_n u^n s^{-n}; \tag 5$
we find
$g(s) = s \displaystyle \int_0^\infty K(st)f(t) \; dt$
$= \displaystyle \int_0^\infty K(u) \left ( \sum_0^\infty f_n u^n  s^{-n} \right )\; du = \sum_0^\infty f_n \left ( \int_0^\infty K(u) u^n \; du \right ) s^{-n}; \tag 6$
with 
$g(s) = \displaystyle \sum_0^\infty a_n s^{-n}, \tag 7$
and
$M(n + 1) = \displaystyle \int_0^\infty K(u) u^n \; du; \tag 8$
we re-assemble (6) via (7) and (8):
$\displaystyle \sum_0^\infty a_n s^{-n} = \sum_0^\infty f_n M(n + 1) s^{-n}; \tag 9$
term-by-term comparison of these series yields
$f_n = \dfrac{a_n}{M(n + 1)}, \tag{10}$
whence
$f(t) = \displaystyle \sum_0^\infty \dfrac{a_n}{M(n + 1)} t^n, \tag{11}$
as per request. $OE\Delta$
