Inverse of a function's integral
The function $g$ is strictly positive. Let the function $f$ be defined as
$$f(x) = \int_0^x g(u) du$$
Is there a way to express $f^{-1}(x)$ in terms of $g$?
$\endgroup$ 21 Answer
$\begingroup$If $g$ takes on both negative and positive values, or is zero on some interval, then $f$ is not invertible, as mentioned in comments.
Assume $g$ is strictly positive (or strictly negative), hence $f^{-1}$ exists and is differentiable by inverse function theorem.
Then $f(f^{-1}(x))=x$, so by differentiating, we get that $f'(f^{-1}(x))(f^{-1})'(x) =1$, i.e, $g(f^{-1}(x))(f^{-1})'(x) = 1$.
Thus we see that $f^{-1}$ satisfies the differential equation $$y' = \frac{1}{g(y)}$$
For some functions $g$ this can be solved exactly (for example, $g(y) = e^y$ or $g(y) = y^2+1$), while for others it cannot be solved exactly (for example $g(y) = e^{y^2}$). Hence, you can get this differential formula but no explicit solution in general.
$\endgroup$ 4More in general
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