We know that using properties of unit impulse function it can be shown that

$$\int_{t_1}^{t_2} x(t) \delta^{(n)}(t-t_0) dt=(-1)^nx^{(n)}(t_0),\quad t_1<t_0<t_2$$ (source: Continuous and Discrete Signals and Systems- Soliman)

But what would be the case for the problems like these $$\int_{-1}^{2}(3t^3+9)\delta''(t+1) dt$$

I've tried in this way:

$$\int_{-1}^{2}x(t)\delta''(t+1) dt=\\ \big[x(-1)\delta'(t+1)-x'(-1)\delta(t+1)\big]_{-1}^{2}-\int_{-1}^{2}x'(t)\delta'(t+1)dt$$

can't figure out the first part inside the third bracket.

  • $\begingroup$ What do you mean by $\delta^{''}$ ? Second derivative of the delta function ? $\endgroup$
    – Hilmar
    Aug 20, 2019 at 17:18
  • $\begingroup$ yes $\delta''$ means second derivative $\endgroup$
    – John
    Aug 20, 2019 at 18:06
  • $\begingroup$ Is this an actual problem from a book or a test, or did you come up with it yourself? I'm asking because I think evaluating an integral with a Dirac impulse (or its derivative) exactly at a limit of the integral is ill-defined. $\endgroup$
    – Matt L.
    Aug 20, 2019 at 18:12

2 Answers 2


Integrals of the form


with $\delta^{(n)}(t)$ being the $n^{th}$ generalized derivative of the Dirac delta impulse, are undefined. What is well-defined are integrals of the form




In $(2)$, the Dirac impulse (or its derivative) is outside the integration interval and, consequently, the integral evaluates to zero, whereas in $(3)$ $\delta^{(n)}(t-a)$ is inside the integration interval and the value of the integral is given by the first formula of the question.


Just using your initial formula, you get

$$Y = (-1)^2 \cdot \frac{\partial^2 }{\partial t^2}\left [ 3t^3+9 \right ]_{t=-1} =$$

$$\left [ 18 \cdot t \right ]_{t=-1}=-18$$

EDIT: per Matt's answer: this is only correct if the integration interval explicitly includes $-1$. If it's explicitly excluded, the answer would be $0$, it it's undefined, the answer is undefined.

  • $\begingroup$ but the initial formula can't be applied at $t=-1$ $\endgroup$
    – John
    Aug 20, 2019 at 18:06
  • $\begingroup$ Correct. See Matt L's answer about the exact definition of the integration intervals. $\endgroup$
    – Hilmar
    Aug 20, 2019 at 18:26

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