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Came across a question that requires evaluation of $\int_{-3}^{+1} \left(x^{3}-3x^{2}+2x-1\right)\delta\left(x+2\right)dx$

Here's my attempt:

Recall that:

$\int_{-\infty}^{\infty} f\left(x\right)\delta\left(x-a\right)dx=f\left(a\right)\int_{-\infty}^{+\infty} \delta\left(x-a\right)dx=f\left(a\right)$

Note that a=-2 relative to the question.

Then,$$\int_{-2-1}^{-2+3} \left((-2)^{3}-3\left(-2\right)^{2}-2\left(-2\right)-1\right)\delta\left(x+2\right)dx= (-2)^{3}-3\left(-2\right)^{2}-2\left(-2\right)-1 $$

I'm still fairly uncomfortable dealing with dirac delta function due to my sparse exposure to them. My guess is that I'm doing this question wrongly and that the domain of integration requires shifting so that the domain of integration is symmetric about the point x=0. Any help is appreciated

Mathematicing
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  • $f(x) = x^3 - 3 x^2 + 2x - 1$ but the result is $f(-2)$ which isn't what you wrote. remember : $\delta(x+2) \ne 0$ only in a neighborhood of $x+2 = 0$ – reuns Jan 03 '16 at 04:20
  • @user1952009 let me correct that! It is a typo! – Mathematicing Jan 03 '16 at 04:21
  • and if you are unconfortable, go back to the definition : $$\delta(x) = \frac12 \lim_{a \to \infty} a e^{-|a x|}$$

    (or any function which has its peak more and more concentrated at the origin)

     – reuns Jan 03 '16 at 04:22
  • @user1952009 I think it helps that I went back to the definition and it says that the dirac delta function has an area of 1 at the point a and integrating this domain produces 1 at x=a while being zero everywhere else. – Mathematicing Jan 03 '16 at 04:45

2 Answers2

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In THIS ANSWER and THIS ONE, I provided primers on the Dirac Delta.

Here, using the Unit Step Function $u(x)$ defined by

$$ u(x)= \begin{cases}1&,x>0\\\\ 0&,x<0\\\\ 1/2&,x=0 \end{cases} $$

We interpret the notation $\int_a^b f(x)\delta(x-x')\,dx$ using the unit step function and write

$$\begin{align} \mathscr{D_{x';a,b}}\{f\}&=\int_a^b f(x)\delta(x-x')\,dx\\\\ &=\int_{-\infty}^{\infty}f(x)\left(u(x-a)-u(x-b)\right)\delta(x-x')\,dx\\\\ &=f(x')\left(u(x'-a)-u(x'-b)\right) \end{align}$$

Now, depending on $x'$ relative to $a$ and $b$, we have

$$\begin{align} \mathscr{D_{x';a,b}}\{f\}&= \begin{cases} f(x')&,a<x'<b\\\\ \frac12 f(x')&,x'=a\,\,\text{or}\,\,x'=b\\\\ 0&, \text{otherwise} \end{cases} \end{align}$$

Community
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Mark Viola
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  • But the question is not about the more general case else about a Dirac delta inside of the interval of integration. – cosmoscalibur Jan 12 '16 at 20:10
  • @cosmoscalibur The OP stated "I'm still fairly uncomfortable dealing with dirac delta function due to my sparse exposure to them. My guess is that I'm doing this question wrongly and that the domain of integration requires shifting so that the domain of integration is symmetric about the point x=0. "

    I posted a direct reply to the issue regarding how to interpret the notation $\int_a^b f(x)\delta(x-x'),dx$ for the Dirac Delta.

     – Mark Viola Jan 12 '16 at 20:20
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In general, in a definite domain
$ \int_b^c f(x)\delta(x-a)dx = f(a)$ if $b < a < c$,
then $\int_{-3}^{1} (x^3 - 3x^2 + 2x -1) \delta(x+2) dx = (-2)^3 - 3(-2)^2 + 2(-2) - 1$ because $-3 < -2 < 1$.

cosmoscalibur
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  • I have corrected the typos in the OP. Should I have been correct then? – Mathematicing Jan 03 '16 at 04:23
  • Integration the dirac delta not require a symmetric domain. Symmetric or asymmetric only is important if the center of delta in inside of the domain. Correction of typo from 2 to -2 is correct. – cosmoscalibur Jan 03 '16 at 04:27
  • Edit:cancel edit – Mathematicing Jan 03 '16 at 04:34
  • If you check that center of the dirac delta is inside of the domain, only evaluete the function in that value, otherwise is zero. – cosmoscalibur Jan 03 '16 at 04:37
  • Very clear! thanks! – Mathematicing Jan 03 '16 at 04:38
  • a last question. what happens if the center of the dirac delta x=a is not in the domain of integration? What do we have to say then? – Mathematicing Jan 03 '16 at 04:48
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    As I mentioned you in the last comment, if the center is zero. – cosmoscalibur Jan 03 '16 at 04:49
  • But if the integrand $f(x)\delta(x-2)$ with f(x) being an even function and the domain runs from -1 to 1, can I not change the domain of integration? The dirac delta seems to be an even function too. – Mathematicing Jan 03 '16 at 04:54
  • If you change domain you change the problem. Domain is changed when using variable change as technique of integration but new and old integral are equivalents. In this case, you want an arbitrary change. If you change the domain, function and center of dirac delta also change. – cosmoscalibur Jan 03 '16 at 05:01