Analytic Approximation of Fredholm equation of second kind Consider an integral equation of the type
\begin{equation}
f(x)=g(x)+\int_{-a}^{a} K(x-x')f(x')\mathrm dx'
\end{equation}
Say $a$ is very small. Is there a systematic way to approximate the solution $f(x)$ as a power series in $a$?
 A: The integral on the RHS of your equation may be expanded
$$\tag{1}
\int\limits_{-a}^a dx' \ K(x-x')f(x') \sim 2aK(x)f(0)+a^2\left[K(x)f'(0)-K'(x)f(0) \right] \ ,\quad a\to0
$$
I've expanded the integrand as a Taylor series around $x'=0$ then integrated term by term. You can continue to get more terms. Let
$$\tag{2}
f(x)=\sum\limits_{n=0}^\infty a^nf_n(x)
$$
Substitute (1) and (2) into your equation then match coefficients of $a$. The first few are
$$\tag{3}
f_0(x)=g(x) \\
f_1(x)=2K(x)f_0(0)\\
f_2(x)=2K(x)f_1(0)+K(x)f_0'(0)-K'(x)f_0(0)
$$
There is a pattern (which may sound more confusing than it is). Consider $\int\limits_{-a}^a dx' \ K(x-x')f(x')$, and we ask what is the coefficient of $a^k$ in this expression. There are two contributions:

*

*A factor of $2 a^m/m!$ from the $m$th term in the integrated Taylor series, which sits next to the $m$th derivative of $(Kf)|_{x'=0}$. This can be simplified using the binomial theorem, if you like.


*A factor of $a^n$ coming from the expansion $f=\sum\limits_n a^nf_n$.
The overall coefficient of $a^k$ coming from this expression is the sum of all the coefficients such that $k=m+n$.
Actually, all of this amounts to forming the expression
$$\tag{4}
\sum\limits_{n=0}^\infty a^n f_n(x)=g(x)+\sum\limits_{n=0}^\infty a^n \int\limits_{-a}^a dx' K(x-x')f_n(x')
$$
Then differentiating (4) $k$ times wrt $a$ and setting $a=0$, then solving for the $f_n$ recursively. To differentiate the integral bounds you use Leibniz integral rule.
