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So there is a certain proof for $\delta(x^2-a^2)$ property in my book which has a part that states enter image description here

So i would like a proof to this general statemement which i could not find anwhere $$\int_{0}^{\infty}f(g(x))\delta{(x-a)}dx=f(g(a))$$ I know that the definition of delta function states $$\int_{0}^{\infty}f(x)\delta{(x-a)}dx=f(a)$$ but this is completely different

  • Isn't it one definition of the delta function? – Fakemistake Mar 17 '20 at 11:07
  • @Fakemistake No –  Mar 17 '20 at 11:08
  • Hint: Write $f(g(x))$ as $(f\circ g)(x)$. It turns out that the value of the integral is the value of $f\circ g$ in $a$, which is $f(g(a))$. – Fakemistake Mar 17 '20 at 11:09
  • Why is this "completely different"? – Ben Grossmann Mar 17 '20 at 11:10
  • @Fakemistake regarding your curt "no", the answer is actually "yes". That is, we can completely define $\delta$ by saying that this applies for $a = 0$, and the rest is a consequence of shifting the integral with substitutions. See, for instance, the measure definition of the Dirac delta function. – Ben Grossmann Mar 17 '20 at 11:14
  • @Omnomnomnom I know the definition of the dirac delta function. The question in my first comment was not precise enough. It would have been better, when I had asked ,,Does the result not follow directly from the defintion applied to the function $f\circ g$?'' – Fakemistake Mar 17 '20 at 13:35
  • @Fakemistake Sorry, I directed the comment towards you, but that should have been directed to Ruvik instead – Ben Grossmann Mar 17 '20 at 16:34

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The "sifting property" of the Dirac delta function tells you that for any function $h(x)$, we have $$ \int_{-\infty}^\infty h(x) \delta(x - a)\,dx = h(a). $$ With that, if $a > 0$ and we set $$ h(x) = \begin{cases} 0 & x < 0\\ f(g(x)) & x \geq 0, \end{cases} $$ then we find that $$ \int_0^\infty f(g(x)) \delta(x-a)\,dx = \int_{-\infty}^\infty h(x) \delta(x - a)\,dx = h(a) = f(g(a)). $$

Ben Grossmann
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  • Thank you, I thought it has some sort of a rigorous proof as a physicist I shouldn't have even asked this question in the first place but thank you. –  Mar 17 '20 at 11:31
  • @Ruvik what is non-rigorous about this proof? – Ben Grossmann Mar 17 '20 at 11:32
  • IDK, maybe I am making it overcomplicated, I have spent 2 days trying to find a proof for this :D –  Mar 17 '20 at 11:37
  • @Ruvik I think it would be a useful exercise for you to figure out what kind of difficult step you thought was necessary for the proof, and then figure out why it turned out not to be necessary. Part of being good at finding proofs is being able to recognize when you've started looking in the wrong direction. – Ben Grossmann Mar 17 '20 at 11:42
  • also if you do not mind to explain, how does the book changes the integral boundaries from $-\infty,0$ to $0,\infty$, is that a property of Dirac function? –  Mar 18 '20 at 06:55