I have an integral which is of the form:
$$ I = \int_0^\infty g(t, x(t)) \, dt. $$
I'm trying to demonstrate how incremental changes in $x(t)$, for any given $t$, affect $I$. Informally/intuitively, I'm fairly sure the answer I'm looking for is $g_2(t,x(t))$, where $g_2$ is the partial derivative with respect to the second argument. But I'm not sure what I'm technically mathematically doing here.
I can think of three possibilities:
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Maybe I'm trying to take some kind of functional derivative? I don't know much about functional derivatives, so I'm not sure whether they're applicable here.
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Maybe this involves some version of Leibniz's rule, though the situation appears different because the second argument $x(t)$ is a function of the first – not an independent parameter – and I'm considering the effect of a change for a given value of $t$.
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In that case that $g(t, x(t)) = f(t)x(t)$, where $x(t)$ is a density function. Then $f$ can be viewed as the Radon-Nikodym derivative of a probability measure for $t$, $\mathbb{P}$, with respect to the Lebesgue measure. Then $I$ can be written
$$ I = \int_{\mathbb{R}_+} f(t)x(t) \, dt = \int_{\mathbb{R}_+} f \, d\mathbb{P} $$
so taking the derivative with respect to $x(t)$ for some $t$ involves differentiating the integral with respect to the measure (if that makes any sense)?
I would greatly appreciate if anyone could point me in the right direction for this problem and/or any of these three possibilities. Maybe some of these are different ways of approaching the same problem? Or maybe they are all off track.
Thanks!
Best Answer
This is essentially the calculus of variations, which measures the change in functionals as we vary the input functions. We first should note that $I$ is a function of a function $x$, which we will then write
$$I(x)=\int_0^\infty g(t,x(t))dt$$
Let's suppose that we perturb $x$ by some function $\eta$, but examine what happens as we shrink this perturbation, so we consider the new function $x+\epsilon\eta$ and consider what happens when $\epsilon$ is very small. We then have
$$I(x+\epsilon\eta)=\int_0^\infty g(t,(x+\epsilon\eta)(t))dt$$
We can then define an auxiliary function $\Phi_\eta(\epsilon)$ as $\Phi_\eta(\epsilon)=I(x+\epsilon\eta)$, which is purely a scalar function of $\epsilon$. We differentiate this and get
$$\Phi_\eta'(\epsilon)=\int_0^\infty\eta(t)g_2(t,(x+\epsilon\eta)(t))dt$$
If we let $\epsilon\to 0$, this represents how $I$ changes when we perturb $x$ by the function $\eta$. This gives
$$\Phi_\eta'(0)=\int_0^\infty\eta(t)g_2(t,x(t))dt$$
In other words, this integral measures how an incremental change in $x$ in the form of $\eta$ changes the value of $I$.