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 $t=t_0$.
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Use integration by parts. Be aware however, that these calculations are purely formal as there is no such thing as an "impulse function". The correct formal setting is the theory of distributions (generalized functions). – J.R. Sep 29 '14 at 07:42
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@YourAdHere or measures – ShakesBeer Sep 29 '14 at 07:43
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1(Strong) Induction may be useful too (to simplify the working of IBP) – ShakesBeer Sep 29 '14 at 07:44
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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.
Matt L.
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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).
Dirk
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