How to simplify this double integral with a dirac delta function in it?

Question

I am having trouble simplifying the following integral $$\int_{-\infty}^t\int_{-\infty}^s f(t,\tau)f(s,\xi) \delta (\tau-\xi)d\xi d\tau,$$ given that $t>s$ and where $\delta$ is the dirac-delta function.

I don't know what the exact requirements for $f$ are so that the following works $$\int_{-\infty}^tf(t,\tau)\delta(t-\tau)d\tau=\frac{f(t,t)}{2}$$ but I think we can assume that these requirements hold true in this case. My guess at this problem is then that $$\int_{-\infty}^t\int_{-\infty}^s f(t,\tau)f(s,\xi) \delta (\tau-\xi)d\xi d\tau = \frac{1}{2}\int_{-\infty}^t f(t,\tau)f(s,s) d\tau$$ but I am unsure whether this is true. It is just my intuition given that $t>s$ and $\tau$ is integrated up to $t$. I cannot, however, provide a proper argument for this. Any hints on how to simplify this integral are appreciated.

Answer 1

As a preliminary, mathematically speaking expressions like $$ \int_0^\infty \delta(x) f(x)dx = f(0)/2 $$ don't really make much sense. I'm guessing this probably comes from considering something like $ \lim_{n} \int_0^\infty f(x) \delta_n(x) dx $ where all $\delta_n$ are even. But you might as well consider some $\delta_n$ that aren't even, then you could make $ \int_0^\infty \delta(x) dx $ any number between $0$ and $1$ and still have the determining property $$ \int_{-\infty}^\infty f(x) \delta(x) dx = f(0). $$

In order to avoid confusion like that I will simply consider

$$ \int_A f(x) d\mu(x) = \chi_A(0) f(0), $$ so the integral is zero if $0 \not \in A$ and $f(0)$ otherwise. In other words $\mu$ is simply the dirac measure where $\mu(A)$ is $1$ if $0 \in A$ and $0$ if $0 \not \in A$. Given this we can work out that for fixed $s,t, \tau$ $$ h(\tau) = \int_{(-\infty, s)} f(t, \tau) f(s, \tau+\xi) d\mu(\xi)= \left \{ \begin{matrix} f(t,\tau) f(s, \tau) & \tau \in (-\infty, s) \\ 0 & \tau \not \in (-\infty,s) \end{matrix} \right. $$ At last we then have $$ \int_{(-\infty, t)} h(\tau) d\tau = \int_{(-\infty, s)} f(t,\tau) f(s,\tau) d\tau, $$ because $t>s$ and where $d\tau$ refers to the Lebesgue measure.