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Is there any way to solve the integration below? or make it simple to eliminate the Dirac-delta function?

$$\int_{-\infty}^\infty m(x)\delta(G(x)-g_c)f_X(x)dx $$

where $f_X(x)$ is a probability density function (PDF) of random variable x.

It will be very helpful for any reference or clue to solve it.

Thank you.

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    $\begingroup$ If $G$ is continuously differentiable you can essentially change variable to $y=G(x)-g_c$. In general $\delta(f(x)) = \sum_j \delta( x-x_j)/| f'(x_j)|$ where $x_j$ are the roots of $f(x)$. $\endgroup$
    – lcv
    Commented Mar 4, 2019 at 11:15
  • $\begingroup$ You need to include some assumptions about $G$ and a motivation: why would you want to compute such integrals, what is the context that led you tom this question. $\endgroup$
    – Liviu Nicolaescu
    Commented Mar 5, 2019 at 8:52

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The usual way to solve problems such as this is to Fourier transform: Call your function $f(g_c)$, then its Fourier transform $$F(\xi)=\int_{-\infty}^\infty f(g_c)e^{i\xi g_c}dg_c=\int_{-\infty}^\infty m(x)f_X(x) e^{i\xi G(x)}\,dx$$ no longer contains the Dirac delta function. You can then recover $f(g_c)$ by an inverse Fourier transform, $$f(g_c)=\frac{1}{2\pi}\int_{-\infty}^\infty F(\xi)e^{-i\xi g_c}\,d\xi.$$ Whether or not this is doable in some closed form will of course depend on your choice of the functions $m$, $f_X$, and $G$.

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    $\begingroup$ @user136540 You have to be ca Dirac function is not a function, it is a generalized function. The next best thing you can do is approximate $\delta$ by a Gaussian with a a tiny, tiny variance. $\endgroup$
    – Liviu Nicolaescu
    Commented Mar 4, 2019 at 11:00
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    $\begingroup$ Carlo Beenakker, @LiviuNicolaescu I believe the answer is simpler than that :). $\endgroup$
    – lcv
    Commented Mar 4, 2019 at 11:17
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    $\begingroup$ @CarloBeenaker You have to know that I, as a pure mathematician, am a bit envious on theoretical physicists that integrate objects like $\delta(g(x))$ to get spectacular conclusions. To me $\delta(g(x))$ is the pullback of a distribution which is a rather ill defined operation, unless $0$ is a regular value of $g$. Suddenly I get worried. For my sanity I replace $\delta$ by a Gaussian with tiny variation and my worries disappear. $\endgroup$
    – Liviu Nicolaescu
    Commented Mar 4, 2019 at 14:39
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    $\begingroup$ @LiviuNicolescu The pullback of a distribution is not a rather ill defined object—-it was defined and studied in some detail in the 50‘s for the case where $g$ is a diffeomorphism by, amongst others, Heinz König and J. Sebastião e Silva. Using a standard localisation process („recollement des morceaux“), it can be extend to much more general situations. Thus $\delta \circ \sin (x)$ is well-defined since we can find a covering of the real line by open intervals on each of which either the sine function is a diffeomorphism or it takes its values in an open interval on which $\delta$ is smooth. $\endgroup$
    – user131781
    Commented Mar 7, 2019 at 10:33
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    $\begingroup$ @user131781. You gotta make up your mind. You said below that $G$b is not a nice function? What do you mean by that. Without any specific assumptions tthe best answer you can expect goes along tthe lines of another less known Heisenberg principle "You can say everything about nothing and nothing about everything." This is a math site. A bit of rigor and precision is expected of you. $\endgroup$
    – Liviu Nicolaescu
    Commented Mar 7, 2019 at 11:17
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To summarize Icv's comment: $\int_{-\infty}^{+\infty}\phi(x)\delta(G(x)-g_c)\ dx=\sum_{x\in G^{-1}(g_c)}\phi(x)/|G'(x)|$ whenever $G$ is nice. Then apply to $\phi=mf_X$.

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  • $\begingroup$ Thank you for answering. Unfortunately, G is not a nice function, so that I should calculate it in a numerical way, random samplings for random variable X. I was able to obtain analytical sets satisfying above condition (roots of $G(x)$-$g_c$) $\endgroup$
    – user136540
    Commented Mar 5, 2019 at 2:29
  • $\begingroup$ @user136540 If your problem is how to compute that "integral" numerically, there are certainly better answers. However note that (as Liviu Nicolaescu pointed out) if $G$ is not a "nice function" then your integral is not well defined. This has the practical meaning that if you try to compute that object with said methods you will get answers that will depend on the algorithm used, including errors. $\endgroup$
    – lcv
    Commented Mar 9, 2019 at 22:14
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Having been disturbed by the nature of some of the comments here I have reluctantly decided to give an answer to this query. Suppose that $G$ is a continuous function on the line and the equation $G(x)=g_c$ has at most a countable family $x_i$ of soltions (I am assuming that $g_c$ is a constant). Assume further that for each $i$, there is an open interval containing $x_i$ on which $G$ is a diffeomorphism. Then the integral above exists and is equal to $$\sum m(x_i) \frac 1{|G‘(x_¡)|}f_X(x_i).$$ This for continuous $m$ and $f_X$ for which the sum converges absolutely. This result is perfectly rigorous and uses a precise definition for the composition of a distribution (in this case, the $\delta$-distribution) with a function. I might add that this mathematics has been available for sixty years.

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