Vector valued line integral I am confronted by a line integral of the form
$$ \pmb{A}_C(\pmb{x}) = \int_{\pmb{x_1}}^{\pmb{x_2}} \delta^3(\pmb{x}-\pmb{z}) \; d\pmb{z}, $$
where $C$ is a path between two points $\pmb{x_1}$ and $\pmb{x_2}$, and the $\pmb{x}$-dependence of the result is ensured by the 3D Dirac delta distribution.
So this object to me is quite confusing, I can't find either a nice intuitive way nor a methodological way to do anything with this integral. Logically I tried to examine this object with taking $C$ to be the straight line, meaning that the path is given by
$$\pmb{z}(t) = \pmb{x_1} + t (\pmb{x_2}-\pmb{x_1}), $$
but I am a bit puzzled on how to actually use this path parametrization in the integral itself. The  way I've learnt to do line integrals is $\int_c \pmb{A}(\pmb{r}) \; d\pmb{z} = \int_{t_0}^{t_1} \pmb{A}(\pmb{r}(t)) \cdot \pmb{z}'(t) \; dt $, but I am confused how to actually do this with a Dirac delta distribution. To remove this difficulty I've also tried to do Fourier transformation to get
$$ \pmb{A}_C(\pmb{x}) = \int_{\pmb{x_1}}^{\pmb{x_2}} \delta^3(\pmb{x}-\pmb{z}) \; d\pmb{z}, = \int_{\pmb{x_1}}^{\pmb{x_2}} e^{i \pmb{k} \pmb{z}} \; d\pmb{z},$$
but I still can't really see how could I adapt my path here. I guess the thing that's confusing me is that usually I take the path integral of a vector field and get a scalar, but here I essentially take the path integral of a scalar field to get a vector field. I also understand that the result should still be a distribution, as in the only thing that will be a 'normal' function is $\int \pmb{A}_C(\pmb{x}) f(x) \; dx$, but I still want to have some insight into the mathematical meaning of the distribution formula as well as how exactly to calculate with it.
 A: After reading a paper where the formula appears, I now understand better. I find it easier to construct the integral than to interpret it, so let me take it that way.
Let $C$ be a smooth directed curve in $\mathbb{R}^3.$ In the papers they want to construct a field $\mathbf A_C$ that lives only on $C$ and has the direction of $C$. I will do this by splitting the curve into line segments. So let $\mathbf x_0, \mathbf x_1, \ldots, \mathbf x_n$ be ordered points on the curve, with $\mathbf x_0$ being the start point of $C$ and $\mathbf x_n$ being the end point.
Below I will show that the field living on the line segment from $\mathbf x_k$ to $\mathbf x_{k+1}$ can be expressed as
$$
\mathbf A_{\mathbf x_k, \mathbf x_{k+1}}(\mathbf x)
= -\int_{\mathbf x_k}^{\mathbf x_{k+1}} \delta^3(\mathbf x-\mathbf x') \, d\mathbf x'
.
$$
The field along the polygonal chain $\mathbf x_0,\ldots,\mathbf x_n$ then is
$$
\mathbf A_{\mathbf x_0,\ldots,\mathbf x_n}(\mathbf x)
= \sum_{k=0}^{n-1} \mathbf A_{\mathbf x_k, \mathbf x_{k+1}}(\mathbf x)
= -\sum_{k=0}^{n-1} \int_{\mathbf x_k}^{\mathbf x_{k+1}} \delta^3(\mathbf x-\mathbf x') \, d\mathbf x'
.
$$
Taking limits by refining the segmentation of $C$ we arrive at
$$
\mathbf A_C(\mathbf x) = -\int_C \delta^3(\mathbf x-\mathbf x') \, d\mathbf x'
.
$$
Now to the field along a line segment. First we consider a line segment parallel to the $x$-axis, from point $\mathbf x_0=(x_0,y_0,z_0)$ to point $\mathbf x_1=(x_1,y_0,z_0).$ The field on this can be expressed as
$$
\mathbf A_{\mathbf x_0, \mathbf x_1}(x,y,z)
= 1_{[x_0,x_1]}(x) \delta(y-y_0) \delta(z-z_0) \, \mathbf e_x,
$$
where $\mathbf e_x$ is the unit vector in the $x$ direction and $1_{[x_0,x_1]}$ is the indicator function on the interval $[x_0,x_1]$ (in one dimension).
Now we can write $1_{[x_0,x_1]}(x)$ as an integral:
$$
1_{[x_0,x_1]}(x)
= H(x-x_0) - H(x-x_1)
= -\int_{x_0}^{x_1} \delta(x-x') \, dx'
.
$$
Inserting this into the expression for $\mathbf A_{\mathbf x_0,\mathbf x_1}(\mathbf x)$ gives
$$\begin{align}
\mathbf A_{(x_0,y_0,z_0), (x_1,y_0,z_0)}(x,y,z)
&= \left(-\int_{x_0}^{x_1} \delta(x-x') \, dx'\right) \, \delta(y-y_0) \delta(z-z_0) \, \mathbf e_x \\
&= -\int_{x_0}^{x_1} \delta(x-x') \delta(y-y_0) \delta(z-z_0) \, \mathbf e_x \, dx' \\
&= -\int_{(x_0,y_0,z_0)}^{(x_1,y_0,z_0)} \delta(x-x') \delta(y-y') \delta(z-z') \, (\mathbf e_x\,dx'+\mathbf e_y\,dy'+\mathbf e_z\,dz') \\
&= -\int_{\mathbf x_0}^{\mathbf x_1} \delta^3(\mathbf x - \mathbf x') \, d\mathbf x', \\
\end{align}$$
where it has been used that $y=y_0$, $z=z_0$ and $dy=dz=0$ along the line segment.
Since the last expression is inherently independent of direction of the line segment, we conclude that the field living on the line segment from $\mathbf x_k$ to $\mathbf x_{k+1}$ can be expressed as
$$
\mathbf A_{\mathbf x_k, \mathbf x_{k+1}}(\mathbf x)
= -\int_{\mathbf x_k}^{\mathbf x_{k+1}} \delta^3(\mathbf x-\mathbf x') \, d\mathbf x'
.
$$

My first interpretation
This is my interpretation:
$$
\mathbf{A}_C(\mathbf{x}) = \int_{\mathbf{x_1}}^{\mathbf{x_2}} \delta^3(\mathbf{x}-\mathbf{z}) \, d\mathbf{z}
= \int_{\mathbb{R}^3} \mathbf{1}_C(\mathbf z)\,\delta^3(\mathbf{x}-\mathbf{z})\,d\mathbf{z}
= \mathbf{1}_C(\mathbf x)
= \begin{cases}
1 & \text{if $\mathbf x\in C$} \\
0 & \text{if $\mathbf x\not\in C$} \\
\end{cases}
$$
Here, $\mathbf{1}_C$ is the indicator function of $C$.
