Show that $\int_{t_1}^{t_2} x(t)\delta^{(n)}(t-t_0)dt=(-1)^n x^{(n)}(t_0)$.

Question

The unit impulse function $\delta(t)$ is defined in terms of the integral $$\int_{-\infty}^\infty x(t)\delta(t)dt=x(0)$$ where $x(t)$ is any test function that is continuous at $t=0$. Show that $$\int_{t_1}^{t_2} x(t)\delta^{(n)}(t-t_0)dt=(-1)^n x^{(n)}(t_0),\quad t_1<t_0<t_2,$$ where the superscript $n$ denotes the $n$th derivative; $x(t)$ and its first $n$ derivatives are assumed continuous at $t=t_0$.

Answer 1

HINT:

For $n=1$ you get via integration by parts

$$\int_{t_1}^{t_2}x(t)\delta'(t-t_0)dt=x(t)\delta(t-t_0)\Big|_{t_1}^{t_2}-\int_{t_1}^{t_2}x'(t)\delta(t-t_0)dt=\\= -\int_{t_1}^{t_2}x'(t)\delta(t-t_0)dt=-x'(t_0)\tag{1}$$

Repeated application of (1) gives the desired formula.

Answer 2

My hint would be:

Do not use integration by parts, as this is not something that is defined for distributions, but use the definition of the derivative of a distribution.

But this really depends on how your course has introduced distributions and how they will be handled. In a mathematically sound way, one introduced distributions as functionals and then says that locally integrable functions induce some distributions by integration, then derives that this motivates a nice notion of derivatives of a distribution and than your formula for the derivation of $\delta$ is basically this definition for this case (using the fundamental theorem of calculus).