How to calculate $$\int_{-t}^{t} \delta (\tau) d\tau$$
My attempt is $$\int_{-t}^{t} \delta (\tau) d\tau = \int_{-\infty}^{t} \delta (\tau) d\tau + \int_{-t}^{\infty} \delta (\tau) d\tau - \int_{-\infty}^{\infty} \delta (\tau) d\tau= u(t) + u(t) - 1$$
Am I correct? Not quite sure if I use the correct properties of dirac function. Thanks!
The Dirac delta is defined by its integrals, not by values, as it is not a function. $$\int_a^b \delta(t)dt = \begin{cases} 1 & a \leq 0 \leq b \\ 0 & a>0 \text{ or } b < 0\end{cases}$$
Since you are integrating from $-t$ to $t$, your interval includes $0$ so the answer is $1$. The three integrals you broke it up into evaluate to $1,1,1$ so the sum is $1$, as it should be. Technically some of those do integrate to the step function, but since $t$ is a given constant, I wasn't really thinking of the step function at all. However, that is not the usual way to break up an integral's interval.
I think the breakdown you wrote is valid, but I've never seen it done that way. In effect you integrated over $-t$ to $t$ by doing it over all real numbers but including the interval of interest twice and then subtracting the integral over all real numbers. But again, all of this is unnecessary because there is nothing to prove, the claim is true by definition of the Dirac Delta.