My question is related to Derivation of the geodesic equation from the continuity equation for the energy momentum tensor
I need to understand one step in derivation.
Let's consider the Energy-momentum tensor of point particle: \begin{equation}\label{1} T^{\mu\nu}(x) = \frac{m}{\sqrt{-g(x)}}\int d\tau \frac{dX^{\mu}}{d\tau}\frac{dX^{\nu}}{d\tau}\delta^{(4)}(x - X(\tau)) \end{equation}
We want to find a covariant derivative of $T^{\mu\nu}$. For arbitrary symmetric tensor, the covariant derivative is: \begin{equation}\label{2} \nabla_{\mu} T^{\mu\nu} = \frac{1}{\sqrt{-g}} \frac{\partial \left( \sqrt{-g} T^{\mu\nu}\right) }{\partial x^{\mu}} + \Gamma^{\nu}_{\mu\lambda}T^{\mu\lambda} \end{equation}
And for our case, let's consider derivative $\frac{1}{\sqrt{-g(x)}} \frac{\partial \left( \sqrt{-g(x)} T^{\mu\nu}\right) }{\partial x^{\mu}}$:
\begin{multline} \frac{1}{\sqrt{-g(x)}} \frac{\partial \left( \sqrt{-g(x)} T^{\mu\nu}\right) }{\partial x^{\mu}} = \\ = \frac{1}{\sqrt{-g(x)}} m \int d\tau \frac{dX^{\mu}}{d\tau}\frac{dX^{\nu}}{d\tau}\frac{\partial }{\partial x^{\mu}} \left[ \delta^{(4)}(x - X(\tau))\right] = \\ = - \frac{1}{\sqrt{-g(x)}} m \int d\tau \frac{dX^{\mu}}{d\tau}\frac{dX^{\nu}}{d\tau}\frac{\partial}{\partial X^{\mu}} \left[ \delta^{(4)}(x - X(\tau)) \right] = \\ = - \frac{1}{\sqrt{-g(x)}} m \int d\tau \frac{dX^{\nu}}{d\tau}\frac{d}{d\tau}\left[ \delta^{(4)}(x - X(\tau))\right] = ?\\ \end{multline}
What correct property of $\delta$-function should I use for the next step? Intergrating by parth I think is no completely correct.
