Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How do we solve $\frac{d}{dx_t}[\int_0^T y(x_s) ds]$ where T>t and $y(x_t)$ is a function of $x_t$ and $x_t$ is a function of $t$

share|cite|improve this question
I don't understand your notation. What does $\frac{d}{dx_t}$ mean? – copper.hat Jul 25 '12 at 7:37

You can't take the derivative with respect to a single function value of the function $x_t$; the value of the integral wouldn't change if you changed just that one function value.

The usual approach to varying functions would go like this: Consider the functional

$$I[f]=\int_0^Ty(f(s))\,\mathrm ds\;,$$

where $f(s)$ corresponds to your $x_s$. If $y$ is differentiable, the (first) variation of $I$ is

$$\delta I=\frac{\partial I[f+\epsilon\delta f]}{\partial\epsilon}=\frac{\partial}{\partial\epsilon}\int_0^Ty(f(s)+\epsilon\delta f(s))\,\mathrm ds=\int_0^Ty'(f(s))\delta f(s)\,\mathrm ds\;.$$

Now you can ask things like when is $I$ stationary, i.e. for which $f$ does the first variation vanish for all $\delta f$. You can also take the limit $\delta f(s)=\delta(s-t)$ to get something similar to what you asked for, a measure of the change in $I$ if you change $f$ only in an arbitrarily small neighbourhood of $t$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.