So I have this
$$
x=0.5\delta(t-1)+0.5\delta(t+2)\\
y=0.3\delta(t+1)+0.7\delta(t-2)
$$
and I have to find the convolution. They give me the integral of $\delta(a-t)\delta(t-b)=\delta(a-b)$
I have searched on google and in the book but I have not found a similar problem anywhere else.
I tried to do the convolution but I must be missing a property or something because it doesn't get to the type I want and need.
$$(x*y)(t)=\int_{-\infty}^{+\infty}{x(\tau)y(t-\tau)d\tau}$$
$$(x*y)(t)=\int_{-\infty}^{+\infty}{[0.5\delta(\tau-1)+0.5\delta(\tau+2)][0.3\delta(t-\tau+1)+0.7\delta(t-\tau-2)]d\tau}$$
$$(x*y)(t)=\int_{-\infty}^{+\infty}{0.15\delta(\tau-1)\delta((t+1)-\tau)\,d\tau} + \int_{-\infty}^{+\infty}{0.35\delta(\tau-1)\delta((t-2)-\tau)\,d\tau} + \int_{-\infty}^{+\infty}{0.15\delta(\tau-(-2))\delta((t+1)-\tau)\,d\tau} + \int_{-\infty}^{+\infty}{0.35\delta(\tau-(-2))\delta((t-2)-\tau)\,d\tau}$$
Knowing that
$$ \int_{-\infty}^{+\infty}{\delta(a-x)\delta(x-b)\,dx} = \delta(b-a) $$
your convolution becomes
$$(x*y)(t)= 0.15\delta(1-(t+1)) + 0.35\delta(1-(t-2)) + 0.15\delta(-2-(t+1)) + 0.35\delta(-2-(t-2))$$
$$(x*y)(t)= 0.15\delta(-t) + 0.35\delta(3-t) + 0.15\delta(-3-t) + 0.35\delta(t)$$
$$(x*y)(t)= 0.5\delta(t) + 0.35\delta(t-3) + 0.15\delta(t+3)$$