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For a function $f(t)$ of a single variable, I know the following property of the Dirac delta function:

$\int_{-\infty}^\infty f(t) \delta(t-a) \, dt = f(a).$

But, what if we have a function of two or more variables? Say, we have a function $f(x, t)\in L^1(\mathbb{R^2})$. Then can we have the property:

$\int_{-\infty}^\infty \int_{-\infty}^\infty f(x, t) \delta(x-a) \delta(t-b) \, dx \, dt = f(a, b) ?$

Or, do we have something like a "two-dimensional Dirac delta function,”

$\int_{-\infty}^\infty \int_{-\infty}^\infty f(x, t) \delta(x-a, t-b) \, dx \, dt = f(a, b) ?$

General Mathematics
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    Note that your first double integral can be written $$\int \delta(t-b) \left( \int f(x,t) \delta(x-a) ,\mathrm{d}x \right) ,\mathrm{d}t \text{,}$$ which you should recognize does exactly what you want. – Eric Towers Dec 22 '24 at 16:07
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    Both of these are already valid statements. Where things get more involved is in the context of different coordinate systems, e.g., the Dirac delta function in polar coordinates as discussed here. – Semiclassical Dec 22 '24 at 16:08
  • @EricTowers sir, in that case: $\int_{-\infty}^\infty \delta(t-b) ( \int_{-\infty}^\infty f(x, t) \delta(x-a) , dx) , dt$ hence $ =\int_{-\infty}^\infty \delta(t-b) f(a, t) dt = f(a, b)$ am I correct sir? – General Mathematics Dec 22 '24 at 16:13
  • @Semiclassical sir, so you mean there is two dimensional dirac delta function and the property what I write in the post for $\delta(x-a, t-b)$ holds? Moreover, is $\delta(x-a, t-b)=\delta(x-a)\delta(t-b)$? – General Mathematics Dec 22 '24 at 16:17
  • @EricTowers sir, still I am confused about my last comment. Is it correct? Moreover, how that inner integral reduce to $f(a, t)$? I mean $\int_{-\infty}^\infty f(x, t) \delta(x-a), dx = f(a, t) ?$ how? Isn't here $f$ is function of two variables and $\delta$ is function of one variable? – General Mathematics Dec 22 '24 at 16:25
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    That is correct. More generally, the ordinary n-dimensional Dirac delta function is $$\delta(\vec{x}-\vec{a})=\delta(x_1-a_1)\delta(x_2-a_2)\cdots \delta(x_n-a_n)$$ – Semiclassical Dec 22 '24 at 16:26
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    Isn't "shifting" instead of "sifting" ? – Jean Marie Dec 22 '24 at 16:46
  • @JeanMarie yes, that's a typo. I edited it. – General Mathematics Dec 22 '24 at 17:12
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    Sifting was correct: see for instance https://mathworld.wolfram.com/SiftingProperty.html. (Sifting means selecting for some particular object. In this case, integrating $\delta(x-a)$ against some function $f(x)$ sifts for the value $f(a)$ and ignores the behavior of $f$ everywhere else.) – Semiclassical Dec 22 '24 at 17:36
  • Change of variable? – psl2Z Dec 22 '24 at 17:57
  • @Semiclassical Thanks for the precision. – Jean Marie Dec 22 '24 at 18:30
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    The Dirac Delta is NOT a function, and those objects--written as integrals--are NOT integrals. The Dirac Delta is distribution (a linear functional) that maps functions into numbers. All of the hand-waving formality with integrals seems to work, but is not rigorous. If $\phi\in \mathbb{R}^n$ and $\phi\in C_C^\infty$, then $\langle \delta_{\vec a},\phi\rangle = \phi(\vec a)$, where $\vec a\in\mathbb{R}^n$. – Mark Viola Dec 24 '24 at 21:09

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This is, indeed, a well-defined operation. Consider the 2 Dirac deltas $\delta_a$ and $\delta_b$ where $$ \int_\mathbf{R} g(t) \delta_a(t) \ dt = \int_\mathbf{R} g(t) \delta(t-a) \ dt= g(a), $$ and $$ \int_\mathbf{R} g(t) \delta_b(t) \ dt = g(b), $$ for any smooth $g$. Then, the operation $$ \int_\mathbf{R} \delta_a \bigl( \int_\mathbf{R} f(t,s) \delta_b(t) \ dt \bigr) \ ds $$ is well-defined and equals to $f(a,b)$ for any smooth $f$ on $\mathbf{R^2}$. The "function" $\delta_a \otimes \delta_b$ is called the tensor-product of $\delta_a$ and $\delta_b$. In other words, the tensor-product operation $(g_1 \otimes g_2) (x,t) = g_1(t)g_2(x)$ for two functions $g_1$ and $g_2$ can also be performed for the dirac delta function.

Saleh
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