Can some one help me out this derivative:
$$
\frac{\partial\int_{-\infty}^\infty f(x)g(x)\,dx}{\partial g}
$$
Appreciate any explanation!
Many thanks to those who answered or commented on my question!
Maybe I am not very clear about my question or I might have misunderstood what I need. I was reading a Quantum Field Theory textbook and trying to verify the following equation:
$$
\langle 0\mid 0\rangle_{f, h} = \int Dp\,Dq \exp \left[i\int_{-\infty}^\infty dt(p\dot q-H_0(p, q)-H_1(p, q)+fq+hp)\right]
$$
$$
=\exp \left[ -i\int_{-\infty}^\infty dt \, H_1 \left( \frac 1 i \frac \delta {\delta h(t)}, \frac 1 i \frac{\delta}{\delta f(t)} \right) \right]\times \int Dp\,Dq\, \exp \left[ i\int_{-\infty}^\infty dt\,(p\dot q-H_0(p, q)+fq+hp) \right]
$$
I expanded the first exponential of the second line to a Taylor series. Thus every term is a derivative with respect to either $h(t)$ or $g(t)$. And then I got lost.
I guess this is related to functional derivatives and I tried to search some examples but I failed. Or maybe it's just I didn't realize I have seen the same thing because of my poor math.
Best Answer
The integral $\displaystyle\int_{-\infty}^\infty f(x)g(x)\,dx$ is just a number, not a function of $x$. But maybe what you want to differentiate is $\displaystyle g\mapsto\int_{-\infty}^\infty f(x)g(x)\,dx$, i.e. the derivative with respect to $g$ rather than with respect to $g(x)$. If so, I'd omit the "$x$" from the notation; there isn't any $x$.
Now recall from calculus that $\displaystyle g\mapsto\int_{-\infty}^\infty f(x)g(x)\,dx$ is linear in $g$, i.e. if $g=g_1+g_2$ then the output will be the sum of the output when the two inputs are $g_1$ and $g_2$ and if $g=cg_1$ where $c$ is constant, then the output will be $c$ times whatever it is when the input is $g_1$.