If we can interpret the line integral of a function over a path as a functional. What would be its functional derivative?
For instance, in this example, let $\gamma$ be a path in $\mathbb{R}^3$. And consider $\boldsymbol{A}(\boldsymbol{r})$ a smooth function from $\mathbb{R}^3$ to $\mathbb{R}^3$. Then we can interpret the line integral over path $\gamma$ as a functional, since, for a fixed path, the integral sends functions to real numbers. $$ F[\boldsymbol{A}(\boldsymbol{r})] = \int_\gamma \boldsymbol{A}(\boldsymbol{r}) \cdot d\boldsymbol{r}. $$
Using the definition of Wikipedia (https://en.wikipedia.org/wiki/Functional_derivative), the variation is given by $$ \delta F[\boldsymbol{A},\boldsymbol{\phi}] = \lim_{\epsilon \to 0} \frac{F[\boldsymbol{A}+\epsilon \boldsymbol{\phi}]-F[\boldsymbol{A}]}{\epsilon}, $$ where $\boldsymbol{\phi}$ is the test function.
After substitution, I get $$ \delta F[\boldsymbol{A},\boldsymbol{\phi}] = \int_\gamma \boldsymbol{\phi}(\boldsymbol{r})\cdot d\boldsymbol{r}. $$ To obtain the functional derivative we need an expression of the form $$ \delta F[\boldsymbol{A},\boldsymbol{\phi}] = \int_{\mathbb{R}^3} \sum_j \frac{\delta F}{\delta \phi_j}(\boldsymbol{r}) \phi_j(\boldsymbol{r}') dV', $$ where $\phi_j$ is the j-component of $\boldsymbol{\phi}$. However, I fail to identify $$ \int_{\mathbb{R}^3} \sum_j \frac{\delta F}{\delta \phi_j}(\boldsymbol{r}') \phi_j(\boldsymbol{r}') dV' \stackrel{?}{=} \int_\gamma \phi(\boldsymbol{r})\cdot d\boldsymbol{r}. $$ I would guess that the functional derivative will be some kind of delta distribution in the directions perpendicular to the curve $\gamma$ but I am not sure how to proceed.
Thanks!