I would like to take care on my answer on the following question
Fourier transform involving a dirac delta function
I have tried to answer this question,of course did not know exact answer,just if it would be rejected or there will be some constraints,i would delete my answer, I don't want OP knows incorrect answer,so please help us to solve his problem correctly in case of my answer is wrong and help us to learn new things,it is just hypothetical question and also hypothetical answer from me,so we need you help,please help us
The delta distribution may be composed with a smooth function $g(x)$ in such a way that the familiar change of variables formula holds, that $$ \int_{\Bbb{R}} \delta\bigl(g(x)\bigr) f\bigl(g(x)\bigr) |g'(x)|\,dx = \int_{g(\Bbb{R})} \delta(u)f(u)\, du $$ provided that $g$ is a continuously differentiable function with $g'$ nowhere zero. That is, there is a unique way to assign meaning to the distribution $\delta\circ g$ so that this identity holds for all compactly supported test functions $f$. This distribution satisfies $δ(g(x)) = 0$ if $g$ is nowhere zero, and otherwise if $g$ has a real root at $x_0$, then $$ \delta(g(x)) = \frac{\delta(x-x_0)}{|g'(x_0)|}.$$
It is natural therefore to define the composition $δ(g(x))$ for continuously differentiable functions $g$ by $$ \delta(g(x)) = \sum_i \frac{\delta(x-x_i)}{|g'(x_i)|} $$ where the sum extends over all roots of $g(x)$, which are assumed to be simple.
Thus, for example $$ \delta\left(x^2-\alpha^2\right) = \frac{1}{2|\alpha|}\Big[\delta\left(x+\alpha\right)+\delta\left(x-\alpha\right)\Big]. $$ In the integral form the generalized scaling property may be written as $$ \int_{-\infty}^\infty f(x) \, \delta(g(x)) \, dx = \sum_{i}\frac{f(x_i)}{|g'(x_i)|}. $$