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Consider two n-dimensional vectors $\mathbf{x, x_0}$ and the expression $$ \int_0^\infty f(t) \delta(\mathbf x - t\mathbf x_0) dt $$ where $\delta$ is the n-dimensional Dirac delta function. Intuitively this means that $\mathbf x$ is collinear with $\mathbf x_0$ and that there is a weight $f(t)$ where $t$ is the "slope".

Now $\mathbf x = t \mathbf x_0, t > 0$ iff $\mathbf{x\cdot x_0} = \mathbf{|x||x_0|}$ with $t = \frac{\mathbf{x\cdot x_0}}{|\mathbf x_0|^2}$

So I am naively wondering if I may write $$ \int_0^\infty f(t) \delta(\mathbf x - t\mathbf x_0) dt = f\left( \frac{\mathbf{x\cdot x_0}}{|\mathbf x_0|^2} \right) \delta(\mathbf{x\cdot x_0 - |x||x_0|}) = f\left( \frac{|\mathbf x|}{|\mathbf x_0|} \right) \delta(\mathbf{x\cdot x_0 - |x||x_0|}) $$ and also whether there might even be a simpler expression?

Many thanks in advance! p

phaedo
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    You can drop the delta function since $$ \int_0^\infty f(t)\delta(g(x,t))dt = f(t_0) $$ where $t_0$ is obtained from finding $$ g(x,t_0) = 0 $$ which you have found. The other condition is that $$ t_0 \in (0, \infty) $$ – Chinny84 Sep 30 '19 at 22:39
  • except that the LHS depends on x whereas your RHS doesn't... ultimately the result "..." will be inserted in an expression that depends on x: $\int_{\mathbb R^n} g(\mathbf x) \cdots d\mathbf x$ – phaedo Sep 30 '19 at 23:36
  • The RHS does depend on $x$ implicitly via the zeros of $g(x,t)$. If there are many zeros then you will have a sum on the RHS. This is a property of Dirac delta functions. – Chinny84 Oct 01 '19 at 06:17
  • OK but I think your line of reasoning only works for 1-dimensional Dirac delta. As a counter-example take f(t) = 1, n =2, X = (x,y), X0 = (a,b), then $$\int \delta(x - at)\delta(y - bt)dt = a^{-1} \int \delta(x/a - t)\delta(y - bt)dt = a^{-1}\delta(y - b/a x) = \delta(ay - bx) $$ – phaedo Oct 01 '19 at 14:43

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