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I know some basic properties of the Dirac delta 'function', and I'm familiar with the sifting property. In any case, I'm stuck trying to evaluate this double integral $$ \int\limits_{-r}^{0}\int\limits_{0}^{\theta}\sin(2\omega\xi)\cos(\omega\theta)\delta(\theta+r)d\xi d\theta, $$ where $r,\omega\in\mathbb{R}^{+}$. Here's my initial attempt. $$ \int\limits_{-r}^{0}\int\limits_{0}^{\theta}\sin(2\omega\xi)\cos(\omega\theta)\delta(\theta+r)d\xi d\theta =\left(\int\limits_{-r}^{0}\left(\int\limits_{0}^{\theta}\sin(2\omega\xi)d\xi\right)\cos(\omega\theta)\delta(\theta+r)d\theta\right)= \left\{ \begin{array}{ccl} -\frac{1}{2\omega}\sin(\omega r)\sin(2\omega r) &,&~~\text{if}~r\in(-r,0)\\ 0 &,&~r\notin [-r,0] \end{array} \right. $$ This does not make any sense to me since $r\in\mathbb{R}^{+}$. My guess is that the value of this integral is zero. All suggestions are appreciated.

user775349
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2 Answers2

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First integrate with respect to $\xi$: $$ \int_{-r}^0 \int_0^{\theta}\sin(2\omega\xi)\cos(\omega\theta)\delta(\theta+r)\,d\xi\,d\theta =\int_{-r}^0\frac{(1-\cos(2\omega\theta))}{2\omega}\cos(\omega\theta)\delta(\theta+r)\,d\theta. \tag{1} $$ As discussed here and here, there is some ambiguity in the definition of an integral containing a Dirac delta when the latter is centered at the boundary. In the context of Fourier series, it's usual to define $\int_0^{\infty}f(x)\delta(x)\,dx=\frac{1}{2}f(0^+)$, in which case the result of $(1)$ would be $$ \frac{1}{4\omega}(1-\cos(2\omega r))\cos(\omega r). \tag{2} $$ In the context of the Laplace transform, however, it's usual to define $\int_0^{\infty}f(x)\delta(x)\,dx=f(0^+)$, in which case the result of $(1)$ would be twice the expression in $(2)$.

Gonçalo
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  • Please note that $-\frac{1}{2\omega}sin(\omega r)\sin(2\omega r)=\frac{1}{2\omega}\left(1- \cos(2\omega r)\right)\cos(\omega r)$. In other words, we both arrive at the same result if $r\in (-r,0)$. But how does this even make any sense since $r>0$? – user775349 Jul 27 '24 at 22:58
  • Why do you write $r\in(-r,0)$? The variable that must be in that interval is $\theta$. – Gonçalo Jul 27 '24 at 23:06
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    https://math.stackexchange.com/questions/4371653/double-integral-with-dirac-delta – user775349 Jul 27 '24 at 23:13
  • That post is about a slightly different problem. Suppose you have an integral of the form $\int_{-R}^0f(\theta)\delta(\theta+r),dr$ (note that $r$ and $R$ are different parameters). Then the result is $f(-r)$ if $-r\in(-R,0)$, and $0$ if $-r\notin(-R,0)$. Your integral corresponds to the case $-r=-R$. In this case, the result is not well defined and depends on the context, as discussed in the posts that I mentioned in my answer. – Gonçalo Jul 27 '24 at 23:36
  • $\int\limits_{-R}^{0}f(\theta)\delta(\theta+r)d\theta=?$ – user775349 Jul 28 '24 at 01:00
  • I answered this question in my previous comment. – Gonçalo Jul 28 '24 at 01:12
  • $dr\neq d\theta$. – user775349 Jul 28 '24 at 01:15
  • Sorry, you are right. That was a typo. The integral in that comment was supposed to be $\int_{-R}^0f(\theta)\delta(\theta+r),d\theta$. – Gonçalo Jul 28 '24 at 01:18
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OP's double integral is of the form

$$\begin{align} J~=~&\int_{\mathbb{R}}\!\mathrm{d}\theta f(\theta)\delta(\theta+r)I(\theta)\cr ~=~&f(-r)I(-r)\cr ~\stackrel{(2)+(3)}{=}~&\color{red}{\frac{1}{2}}\cos(\omega r)\frac{1-\cos(2\omega r)}{2\omega},\end{align}\tag{1}$$ where $$f(\theta)~:=~1_{[-r,0]}(\theta)\cos(\omega\theta)\tag{2}$$ and $$ I(\theta)~:=~ \int_0^{\theta}\!\mathrm{d}\xi \sin(2\omega\xi) ~=~\frac{1-\cos(2\omega\theta)}{2\omega},\tag{3}$$ cf. e.g. this Phys.SE post.

Qmechanic
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