# Integrating a dirac delta 'function' on a definite domain

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

• $f(x) = x^3 - 3 x^2 + 2x - 1$ 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 3 '16 at 4:20
• @user1952009 let me correct that! It is a typo! – Mathematicing Jan 3 '16 at 4:21
• 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 3 '16 at 4:22
• @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 3 '16 at 4:45

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 $-3 < -2 < 1$.

• I have corrected the typos in the OP. Should I have been correct then? – Mathematicing Jan 3 '16 at 4:23
• 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 3 '16 at 4:27
• Edit:cancel edit – Mathematicing Jan 3 '16 at 4:34
• 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 3 '16 at 4:37
• As I mentioned you in the last comment, if the center is zero. – cosmoscalibur Jan 3 '16 at 4:49

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}

• 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
• Math Magician. Please let me know how I can improve my answer. I really want to give you the best answer I can. If you don't find it useful, then I am happy to delete it. So, if you would, please let me know either way. - Mark – Mark Viola Jan 17 '16 at 18:37
• Math Magician. Shall I delete my answer then? - Mark – Mark Viola Jan 25 '16 at 3:25