I'm trying to solve the following problem: find
$$ \frac{\delta J[f](x)}{\delta f(x')} $$ where
$$ J[f](x) = \int_a^x \frac{1}{f(x')} \left( \int_b^{x'} f(x'')dx'' \right) dx' $$
I can't seem to find any resources that discuss how to deal with the nested structure here. For example, if one simply tries to vary f, i.e. perform
$$ f(x) + \epsilon \eta(x) := f(x) + \delta f(x)$$
in order to compute
$$ \delta J := J[f] - J[f+\delta f] $$
one finds
$$ \delta J = -\int_a^x \frac{1}{f^2(x')} \left( \int_b^{x'} f(x'')dx'' \right) \delta f(x') dx' + \int_a^x \frac{1}{f(x')} \left( \int_b^{x'} \delta f(x'') dx'' \right) dx' $$
The first term is already in the form that allows me to deduce the functional derivative, whereas the second is not. It would require me to somehow shift the $\delta f(x'')$ outside of the nested integral (without the limit depending on $ x' $, one would just switch variables $ x'' \leftrightarrow x' $ and to find the functional derivative).
Do you have any thoughts on this?
Thanks. Nick
This just amounts to writing the integral carefully. We can pull the $\frac{1}{f(x')}$ term into the inner integral to obtain $$\int_a^x \frac{1}{f(x')} \left( \int_b^{x'} \delta f(x'') dx'' \right) dx' = \int_a^x\int_b^{x'} \frac{1}{f(x')} \delta f(x'')\; dx'' dx'$$ This is simply a double integral over the region $R = \{(x', x'') : a \leq x' \leq x, b \leq x'' \leq x'\}$. You can check that this region can also be defined by the inequalities $b \leq x'' \leq x, \min(x', a) \leq x' \leq x$. Thus, the integral can be re-written as $$\int_a^x\int_b^{x'} \frac{1}{f(x')} \delta f(x'')\; dx'' dx' = \int_b^x\int_{\min(a, x'')}^{x} \frac{1}{f(x')} \delta f(x'')\; dx'dx'' $$ $$ = \int_b^x \delta f(x'')\left(\int_{\min(a, x'')}^{x} \frac{1}{f(x')}\; dx'\right)\; dx''$$ This should allow you to find the first variation (but be careful: the limits on the integral we just obtained are $[b, x]$, not $[a,x]$!)