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I'm learning about Dirac Delta functions, and I have a question about when one of the bounds is 0.

An example I'm working on is $\int_{-3}^{0}\,dx\,\delta(x-1)$.

How would I evaluate this at 0?

Would it just be undefined?

Thanks!

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    $\begingroup$ Whatever the definition, the result is $0$ because $1$ is not into the interval of integration $[-3,0]$. $\endgroup$ – Jean Marie Oct 12 '16 at 22:16
  • $\begingroup$ Yes of course. If you meant $\int_{-3}^0 \delta(x) dx$ then It depends on your definition of $\delta$. So what is your definition ? A common one is $\int_a^b \delta(x) f(x) dx = \lim_{\epsilon \to 0} \int_a^b \frac{1_{|x| < \epsilon}}{2 \epsilon} f(x) dx = \lim_{\epsilon \to 0} \frac{1}{2 \epsilon}\int_{[a,b] \cap [-\epsilon,\epsilon]} f(x)dx$. $\endgroup$ – reuns Oct 12 '16 at 22:16
  • $\begingroup$ @Bye_World and this definition is consistent only for $\phi(x)$ continuous at $x= 0$ and $A$ an open set $\endgroup$ – reuns Oct 12 '16 at 22:17
  • $\begingroup$ I think JeanMarie's answer makes sense. Thanks for the help! $\endgroup$ – Spuds Oct 12 '16 at 22:19
  • $\begingroup$ If you really meant $\int_{-3}^0 \delta(x-1)dx$ then it is $0$ under your definition, since $\int_{-3}^0 \delta(x-1)dx = \int_{(-4,-1)} \delta(x)\phi(x)dx$ with $\phi(x) = 1$ continuous at $x=0$ $\endgroup$ – reuns Oct 12 '16 at 22:23
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In THIS ANSWER and THIS ONE, I provided primers on the Dirac Delta.


The notation $\int_a^b f(x)\delta(x-c)\,dx$ is interpreted to mean the functional $\langle fp_{ab},\delta_c\rangle$.

Here, $p_{ab}$ is the "rectangular pulse" function, $p_{ab}(x)=u(x-a)-u(x-b)$, and $u$ is the unit step (or Heaviside Function) where

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

Note that there are various conventions for the value $u(0)$.

Therefore, we have

$$\begin{align} \int_a^b f(x)\delta(x-c)\,dx&=\langle fp_{ab},\delta_c\rangle\\\\ &=\begin{cases}f(c)&,c\in(a,b)\\\\0&,c\notin [a,b]\end{cases} \end{align}$$

In the case at hand, we have $a=-3$, $b=0$, and $c=1$. Therefore, we have

$$\int_{-3}^0 f(x)\delta(x-1)=0$$

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  • $\begingroup$ Please let me know how I can improve my answer. I really want to give you the best answer I can. And Happy Holidays! -Mark $\endgroup$ – Mark Viola Dec 6 '16 at 19:39

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