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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

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

Thanks in advance!

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.

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What are the constraints on $f$? If $f$ is unsigned, the denominator could possibly be 0... –  Willie Wong Jun 24 '11 at 23:00
f is a probability distribution, so it is between zero and one. (I'm not sure how to enforce that, however....) –  usul Jun 24 '11 at 23:40
chapter 7 of books.google.com/… might help –  yoyo Jun 25 '11 at 0:57
Thanks. Actually I'm thinking now that the denominator can be zero. Does that mean I'm screwed? (I'd still like to be able to find function that produces the "max" of infinity if possible....) –  usul Jun 25 '11 at 4:33
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
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