1. ## Optimization

Hello everyone, while I was at a leadership conference my mind wandered and I found myself pondering the optimization of functions of the form

$f(\theta)=\int_a^{b}f(x,\theta)~dx\quad\text{where }a\ne{b}$

So I basically treated it as a function of one variable, and proceeding as such I arrived at a question.

$f'(\theta)=\frac{d}{d\theta}\int_a^{b}f(x,\theta)~ dx$

Now by Leibniz's rule the above equation is equivalent to

$f'(\theta)=\int_a^{b}\frac{\partial{f(x,\theta)}}{ \partial\theta}~dx$

Now I next looked for critical points.

If we let $F(\theta)=\int_a^{b}\frac{\partial{f(x,\theta)}}{\ partial\theta}~dx$

Then the critical points occur at four points, namely :

When $F(\theta)=0$
Where $F(\theta)$ is undefined
Less obviously, where $\frac{\partial{f(x,\theta)}}{\partial\theta}=0$. This obviously arises from the fact that $\int_a^{b}0~dx=0$
And also where $\frac{\partial{f(x,\theta)}}{d\theta}$ is undefined.

Now here is where I ran into an anomaly, I found multiple cases where $c\in\mathbb{R}$, satisfies two conatradicting critical point criteria, namely.

$F(c)$ is undefined, but $\frac{\partial{f(x,c)}}{\partial\theta}=0$.

So this implies that this is a multi-valued function, so what does one do from here? Does one just sweep this doubly critical point under the carpet and continue? I hope not.

Any guidance would be appreciated.

Alex

2. Just off the top of my head, Leibniz's Rule does not state equivalence willy-nilly. There are requirements before the differential operator can be moved under the integral. Have you met those requirements? If you do meet them. does this vaporize your anomaly?

3. Originally Posted by TKHunny
Just off the top of my head, Leibniz's Rule does not state equivalence willy-nilly. There are requirements before the differential operator can be moved under the integral. Have you met those requirements? If you do meet them. does this vaporize your anomaly?
Thank you for your response, yes I looked at the requirements, but it does not get rid of all of them. Once again, thank you.