1

Even though integrating numerically gives $0$, we use the Dirac delta $\delta(x)$sifting property to potentially invert functions:

$$\int_{x-c}^{x+c}f(t)\delta(t-x)dt=\int_{-c}^cf(t+x)\delta(t)dt=f(x)\mathop\implies^{t\to f(t)} f^{-1}(x)= \int_{f^{-1}(x-c)}^{f^{-1}(x+c)}t\delta(f(t)-x)df(t),f^{-1}(x-c)<f^{-1}(x)<f^{-1}(x+c),c>0$$

Then rename the integral bounds:

$$f^{-1}(x)=\int_{a}^bt\delta(f(t)-x)df(t),a<f^{-1}(x)<b\tag 1$$

Confirmed here with the inverse error function. However, this integral reminds one of a floor function integral:

$$xe^x=1\implies x=\frac34-\int_0^\frac34\lfloor te^t \rfloor dt=\frac 34-\int_x^\frac34 dt$$

$\lfloor te^t \rfloor:$ enter image description here

which is trivial. A possible transformation is using a limit expansion of $\delta(x)$ which numerically works:

$$f^{-1}(x)=\lim_{c\to0}\frac c\pi\int_a^b \frac{t\,d f(t)}{|f(t)-x+ic|^2}$$

Does $(1)$ provide a trivial expression of the inverse function or how applicable is it to solving equations?

Тyma Gaidash
  • 12,081
  • 1
    Well for one thing it gives the sum of all the solutions to $f(t)=x$ in the interval. Besides that, analytically this is essentially trivial. It is a bit like those expressions for the nth prime that effectively implement an primality test and a loop inside a mathematical expression. Numerically, you could use an approximate identity to get some estimates perhaps. The problem is that if you think about doing this with something like adaptive quadrature, it is rather similar to just bisection. – Ian Jan 02 '23 at 03:51
  • About the Confirmed here link you provide, Did you check this? Looks like the valid interval is quite small for the inverse error function – Joako Jan 02 '23 at 03:54
  • 1
    @Joako The “catalan/eulergamma” constants were meant to show that any numbers $a>b$ could be plugged into the bounds. Try this example with $a=-10,b=10$. – Тyma Gaidash Jan 02 '23 at 12:16

0 Answers0