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I had an exercise (Carroll Chapter 4, exercise 3) where I was asked to prove that writing the energy-momentum tensor for a single particle as a dust fluid, meaning $$ T^{\mu \nu}\left(y^{\sigma}\right)=m \int_{M}\left[\frac{\delta^{(4)}\left(y^{\sigma}-x^{\sigma}(\tau)\right)}{\sqrt{-g}}\right] \frac{d x^{\mu}}{d \tau} \frac{d x^{\nu}}{d \tau} d \tau $$ and the conservation of $T^{\mu \nu}$ I can get the geodesic equations. I got to the correct result also shown in this answer here:

\begin{aligned} \nabla_{\mu}^{(y)} T^{\mu \nu}(y)= \frac{m}{\sqrt{-g(y)}} \int_{\tau_{i}}^{\tau_{f}} \mathrm{~d} \tau \underbrace{\left\{\ddot{x}^{\nu}+\Gamma_{\mu \lambda}^{\nu}(x(\tau)) \dot{x}^{\mu} \dot{x}^{\lambda}\right\}}_{\text {geodesic eq. }} \delta^{4}(y-x(\tau)) \\ -\frac{m}{\sqrt{-g(y)}}\left[\dot{x}^{\nu} \delta^{4}(y-x(\tau))\right]_{\tau=\tau_{i}}^{\tau=\tau_{f}} \\ \stackrel{\text { geodesic eq. }}{=}-\frac{m}{\sqrt{-g(y)}} \underbrace{\left[\dot{x}^{\nu} \delta^{4}(y-x(\tau))\right]_{\tau=\tau_{i}}^{\tau=\tau_{f}}}_{\text {source terms }} . \end{aligned} I am interested in understanding the meaning of the delta function here. I see how to argue the result are the geodesic equations but what should I do if I actually wanted to solve that integral with the delta function?

How should I procede since $\tau$ is the parameter of $x$. Can I relate this to an integral in $d^4x$? Where I would know how to use the Dirac delta.

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