The integral of delta function on the whole real line is well-defined, but one idea came to mind for indefinite integral. But it seems not straightforward and I cannot proceed the following \begin{align} \int e^{-at}\delta \left(t-T\right)dt&=\int e^{-at}dH(t-T)=e^{-at}H(t-T)+a\int H(t-T)e^{-at}dt \\ &=e^{-aT}H(t-T)+\ldots \end{align} where $H(t)$ is Heaviside function.
Now let the lower limit to be zero and integrate the above to $\infty$. \begin{align} \int_0^\infty e^{-at}\delta \left(t-T\right)dt=e^{-at}H(t-T)|_0^\infty+\ldots \end{align}
We have $e^{-at}\delta(t-T) = e^{-aT}\delta(t-T)$ so the primitive distribution (indefinite integral) of $e^{-at}\delta(t-T)$ is $e^{-aT}H(t-T) + C,$ where $C$ is a constant.