This is going to be a dumb question, but I can't figure it out, so here goes
$ f(t)\quad \bigotimes \quad \delta \quad (t\quad -\quad { t }_{ o }) $ = $\int { f(\tau )\delta (t\quad -\quad { t }_{ o }\quad -\quad \tau ) } \quad =\quad f(t\quad-\quad { t }_{ o })$
but when I try to do the convolution integral, I get
$ \int { f(\tau) \delta (t\quad -\quad (\tau \quad -\quad { t }_{ o }) } \quad d\tau $
which is not quite the same thing. What's my mistake?
Typically a convolution is of the form:
$$ (f * g)(t) = \int f(\tau)g(t - \tau) d\tau $$
In your case, the function $g(t) = \delta (t - t_0)$. We then get
$$ (f * g)(t) = \int f(\tau) \delta((t - \tau) - t_0) d\tau = \int f(\tau) \delta(t - t_0 - \tau) d \tau$$
Using the fact that $g(t - \tau) = \delta((t - \tau) - t_0)$
Of course, the right hand term using properties of the $\delta$ function can be evaluated to $f(t - t_0)$.