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I came across a really different integral that relates the product of a function and the delta Dirac written as follows

$\displaystyle\int \delta(y-f(x,m))\frac{1}{f(x,m)}dx$

Based on your broad experience, I ask for some suggestions for integration techniques that allow solving this particular integral.

Thanks in advance.

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    does this answer your question? https://math.stackexchange.com/questions/276583/dirac-delta-function-of-a-function – Simon Jun 02 '21 at 19:36
  • Hi @Simon. I hope you are doing well. I thank you for the post, it helps, but does not solve it. I have asked this question, because I have no ideia how to solve this particular type of integral. –  Jun 02 '21 at 20:07
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    Then you have to tell us something about the function $f$. For a general function, the linked answer is all I can tell you. You will have to find the values of $x$ such that $y-f(x,m)=0$. And at those values find the derivative (w.r.t. $x$) of $y-f(x,m)$. – Simon Jun 02 '21 at 20:11
  • @Simon. Well, the function $f(x,m)$, in most of the applications that I am interested in, can be written as polynomials that relate both $x$ and $m$. For instance, $f(x,m)=m-x^{2}$. –  Jun 02 '21 at 20:17
  • First, the Dirac Delta is NOT a function; it is a distribution. Second, this is NOT an integral; it is a linear functional. – Mark Viola Jun 02 '21 at 20:57

1 Answers1

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For the example $f(x,m)=m-x^2$, you can use the formula as in Dirac Delta Function of a Function .

Define $g(x)=y-f(x,m)$.

  • If $m>y$, the function $g$ has two zeros, namely $x_{1,2}=\pm\sqrt{m-y}$. Therefore the integral is: \begin{align} \int \delta(y-f(x,m))\frac{1}{f(x,m)}dx &= \frac{1}{f(x_1,m)}\frac{1}{|g'(x_1)|}+\frac{1}{f(x_2,m)}\frac{1} {|g'(x_2)|} \end{align}
  • if $m<y$, the function $g$ has no zeros, so the integral is zero.
  • if $m=y$, the function has a double zero at $x=0$, and things get somewhat complicated. Im not going into details here, but maybe look at Dirac delta of a function with zero derivative for some ideas.

This method is correct for any function $f$.

Important note: you do need to determine how many roots the function $g$ has, but you may not need to compute them explicitly. For example you can easily see that \begin{align} f(x_1,m)=y-g(x_1)=y \end{align} without knowing what the precise value of $x_1$ is. Similar tricks might be possible for the $g'(x_i)$ term. This of course depends on the function $f$, but with polynomials, your chances are somewhat good that you can simplify it somewhat.

Simon
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