Integral Represation of Four Current with proper time I have a question about the four current in covariant representation.
the four current is defined as
$$
J^{\alpha} = \binom{c\rho}{\vec{j}}
$$
and i'm having a point charge,
$$\rho(\vec{x},t)=e\delta(\vec{x}-\vec{r}(t)),$$
$$\vec{j}(\vec{x},t)=e\frac{d\vec{r}}{dt}\delta(\vec{x}-\vec{r}(t)).$$
Now, the four-vector $r=(r^0, \vec{r})$ is the space-time-point on the trajectory of my point charge and $x=(x^0, \vec{x})$ is the observation point.
With $r^{\alpha}(\tau)$ as function of proper time and the four-velocity $V^{\alpha}$ the four current can be written like
$$J^{\alpha} = e \int{d\tau V^{\alpha}\delta^{(4)}(x-r(\tau))}.$$
My Question: why and how? I just can't verify this expression, particulary i don't understand where the integral is coming from, and i'm a little bit confused about how $r^{\alpha}(\tau)$ looks like. I mean i know
$$r^{\alpha} = \binom{ct}{\vec{r}(t)},$$ but how to handle this with proper time and also what's about $V^{\alpha}(\tau)$ and $V^{\alpha}(t)$?
 A: Note: $c=1$ in the following.
Every time-like worldline can be parametrized by its proper time. If you are given $\vec r(t)$, then the proper time at $t_0$ is given by 
$$\tau(t_0) = \int_0^{t}\sqrt{1-\left(\frac{\mathrm{d}\vec r}{\mathrm{d}t}\right)^2}\mathrm{d}t $$ 
and inverting this expression to get $t(\tau)$ gives you the worldline $r^\mu = (t(\tau),\vec r(t(\tau))$ parametrized by its proper time.
Now, insert $1 =\frac{\mathrm{d}t}{\mathrm{d}t}$ into your expression for $\rho$. Then, your four-current becomes
$$ j^\mu = e \frac{\mathrm{d}r^\mu}{\mathrm{d}t}\delta(\vec x-\vec r(t))$$
and using the chain rule
$$ j^\mu(\vec x,t) = e \frac{\mathrm{d}r^\mu}{\mathrm{d}\tau}\frac{\mathrm{d}\tau}{\mathrm{d}t}\delta(\vec x - \vec r(t)) = e u^\mu \frac{\mathrm{d}\tau}{\mathrm{d}t}\delta(\vec x - \vec r(t))$$
Now, we use
$$ j^\mu(\vec x,t')= = \int j^\mu(\vec x,t)\delta(t'-t)\mathrm{d}t=\int e u^\mu \frac{\mathrm{d}\tau}{\mathrm{d}t}\delta(\vec x - \vec r(t))\delta(t'-t)\mathrm{d}t$$
and again by the chain rule
$$ j^\mu(\vec x ,t') =  \int e u^\mu \delta(\vec x - \vec r(t))\delta(t'-t)\mathrm{d}\tau$$
Finally, $t= r^0$, $t' = x^0$ and $\delta(x - r) = \delta(\vec x - \vec r)\delta(x^0 - r^0)$ give the formula you asked about.
