How to solve this integral with dirac delta functions?

Question

$$F(t)=\displaystyle\int_{-\infty}^{\infty}\left[e^{-2t^2}\left\{\dot\delta(t-2)\right\}+\delta(t^2-16)\right]\,dt$$ How to get rid of the derivative ? and for the second function i wrote it as $\delta\left\{(t-4)(t+4)\right\}$ and then substitute $t-4=u$ but after that i'm again getting a squared function inside, i don't know how to get rid of it either.

Answer 1

i believe These two will get the job done

$$\delta(t^{2}-a^{2})= \dfrac{1}{2|a|} \left \{\delta(t-a)+\delta(t+a ) \right \}\tag{1}$$

$$\int_{-\infty}^{\infty}x(t).\delta^{n}(t-t_0)dt=(-1)^{n} \left \{ \dfrac{d^{n}}{dt^{n}}x(t)\Big{|}_{t=t_0} \right \}\tag{2}$$

Answer 2

Hints: for any test function $f(t)$,

$$ \int_{-\infty}^\infty f(t)\dot{\delta}(t-\tau)dt = -\dot{f}(\tau) $$ This follows from "integration by parts". Then,

$$ \delta(g(t)) = \sum_{\tau:g(\tau) = 0}\frac{\delta(t-\tau)}{\vert \dot{g}(\tau)\vert} $$