Find a reduction for $(H(t+3)-H(t-5))\cdot(\delta(t+2)+\delta(t-3)+\delta(t-9))$. This exercise is from a Complex Analysis course, more explicitly inside the "Laplce Transform" chapter:

Find a reduction for $$(H(t+3)-H(t-5))\cdot(\delta(t+2)+\delta(t-3)+\delta(t-9)),$$ where $H(t)$ is the Heaviside function: $$H(t-a)=\begin{cases}1,&t>a,\\0,&t<a,\end{cases}$$ and $\delta(t)$ is the Dirac function: $$\delta(t)=\lim_{\tau\to0}F_{\delta}(t),\quad F_{\delta}(t)=\begin{cases}1/\tau,&0\leq t<\tau\\0,&t>\tau.\end{cases}$$


I plotted the functions:
$H(t+3)-H(t-5)$:

$\delta(t+2)+\delta(t-3)+\delta(t-9)$:

Finally, the answer would be $(H(t+3)-H(t-5))\cdot(\delta(t+2)+\delta(t-3)+\delta(t-9))$:

As an expression:
$$
(H(t+3)-H(t-5))\cdot(\delta(t+2)+\delta(t-3)+\delta(t-9))=
\begin{cases}
0,&x<-2,\\
1/2,&-2\leq x<0,\\
4/9,&0\leq x<3,\\
1/9,&3\leq x<5,\\
0,&x>5.
\end{cases}
$$
Is this right? What other method would you use to find the product?
 A: Using sifting property of delta function:
$$\int_{-\infty}^{+\infty}(H(t+3)-H(t-5))(\delta(t+2)+\delta(t-3)+\delta(t-9))dt$$
$$ = \int_{-\infty}^{+\infty} H(1)\delta(t+2)dt - \int_{-\infty}^{+\infty}H(-7)\delta(t+2)dt +$$ $$\int_{-\infty}^{+\infty}H(6)\delta(t-3)dt - \int_{-\infty}^{+\infty}H(-2)\delta(t-3)dt + $$ $$\int_{-\infty}^{+\infty}H(12)\delta(t-9)dt - \int_{-\infty}^{+\infty}H(4)\delta(t-9)dt  = 2$$
A: The singularities of the step functions that appear in the problem are at $t=-3$ and $t=+5$. The singularities of the delta functions are at $t=-2$, $t=+3$ and $t=+9$, which fortunately are different than the previous ones. Therefore we can multiply these distributions without problem. For example $H(t+3)\delta(t+2)=\delta(t+2)$ because at $t=-2$ we have that $H(t+3)$ equals one. Similarly, $H(t-5)\delta(t-3)=0$ because at $t=3$ we have that $H(t-5)$ equals zero. So simplifying we have:
$$ \begin{align}
(H(t+3)−H(t−5))⋅(\delta(t+2)+\delta(t−3)+\delta(t−9))&=\\
\delta(t+2)+\delta(t−3)+\delta(t−9)-\delta(t−9)&=\\
\delta(t+2)+\delta(t−3)
\end{align}$$
