what is the result of $\delta(t-t_0)*\delta(t+t_0):$
$\int_{-\infty}^\infty \delta(t+t_0-x)\delta(x-t_0)dx$
Im not sure how to proceed with that question. Normally when we use sifting property, the function is defined at at least where delta function lies. But these are distinct deltas where they are defined at distinct points. Any help will be appreciated
The convolution of two distributions $S, T \in \def\D{\mathcal D}\D'(\def \R{\mathbf R}\R)$, one of which has compact support, is the unique $S*T \in \D'(\R)$ such that $$ (S * T) * \varphi = S*(T*\varphi), \qquad \phi \in \D(\R) $$ Now, for $\varphi \in \D(\R)$, we have \begin{align*} \delta(\cdot + t_0) * \varphi &= \varphi(\cdot + t_0)\\ \delta(\cdot - t_0) * \bigl(\delta(\cdot + t_0) * \varphi\bigr) &= \delta(\cdot - t_0) * \varphi(\cdot + t_0)\\ &= \varphi \end{align*} Hence, $\delta(\cdot + t_0) * \delta(\cdot - t_0)$ is the unit of the convolution $\D(\R) \to \D(\R)$, that is, $\delta$. Hence $$ \delta(\cdot + t_0) * \delta(\cdot -t_0) = \delta $$
In other words: As you wrote in your initial post, the result of the convolution of $\delta(\cdot + t_0)$ and $\delta(\cdot - t_0)$ cannot be computed by standard means as a function. So, we will try to see how it acts unter integration, it's like $\delta$ is defined by the property $$ \int_{\R} \delta(t)\phi(t)\, dt = \phi(0) $$ for smooth functions $\phi$. So let $\phi$ be a smooth function, we have \begin{align*} \int_{\R} \varphi(t)\bigl(\delta(t + t_0) * \delta(t-t_0)\bigr)\, dt &= \int_{\R} \varphi(t)\int_\R \delta(t+ t_0 - x)\delta(x - t_0)\,dx\, dt\\ &=\int_\R\int_\R \varphi(t)\delta(t+t_0-x)\, dt \,\delta(x-t_0) \,dx\\ &= \int_\R \varphi(x-t_0)\delta(x-t_0)\, dx\\ &= \varphi(0) \end{align*} So, for all smooth $\varphi$, we have $$ \int_{\R} \varphi(t)\bigl(\delta(t + t_0) * \delta(t-t_0)\bigr)\, dt = \varphi(0) $$ that is $\delta(\cdot + t_0) * \delta(\cdot - t_0)$ has the same properties under integration as $\delta$. As this is the defining property $$ \delta(\cdot + t_0) * \delta(\cdot - t_0) = \delta $$