# Calculus of Variations: Contains an integral of my goal function

OK, using the calculus of variations, I want to find a function $f$ that maximizes:

$$J = \int_0^n L(x,f(x)) \text{d}x$$.

But $L$ has multiple integrals in it (for example, $\displaystyle \int_0^n y f(y) \text{d}y$). Can I still use the Euler-Lagrange equation to find f?

Here's the full equation: I want to maximize

$$J = \frac{\int_0^n x f(x) \text{d}x}{\sqrt{\int_0^n \left( y - \int_0^n z f(z) \text{dz} \right)^2 f(y) \text{d}y}} = \int_0^n \frac{x f(x)}{\sqrt{\int_0^n \left( y - \int_0^n z f(z) \text{d}z \right)^2 f(y) dy}} \text{dx}$$

Edit: Actually, I should say, I don't care whether I use the calculus of variations or not, I just want to find $f$. I assumed calculus of variations is the way to do that.
What are the constraints on $f$? If $f$ is unsigned, the denominator could possibly be 0... –  Willie Wong Jun 24 '11 at 23:00
If I understand correctly, you're just dividing the mean by the standard deviation. You can make the standard deviation arbitrarily small by making $f$ arbitrarily peaked; as $f$ tends towards a delta distribution around any value $x_0\neq0$, $J$ grows without limits. –  joriki Jun 25 '11 at 5:07