$\DeclareMathOperator \erf{erf}$ Wolfram Alpha gives the following $\erf^{-1}(z)$ series:
$$\sum_{n=1}^\infty\frac{z^n}{2\pi n}\int_0^{2\pi}e^{it}\erf(e^{it})^{-n}dt$$
which can be derived via Lagrange inversion and the inverse Z transform. $\erf$ is the error function:
$$\erf^{-1}(z)=\sum_{n=0}^\infty\frac{z^n}{n!}\left.\frac{d^{n-1}}{da^{n-1}}\left(\frac a{\erf(a)}\right)^n\right|_0$$
Combining it with $$\left.\frac{d^n}{dx^n}f(x)\right|_0=\frac{n!}{2\pi i}\oint_{|z|=1}f(z)z^{-n-1}dz$$
gives the blockquoted series. Additionally, we can evaluate the sum to get: $$ \begin{split} \erf^{-1}(z) &=-\frac1{2\pi i}\oint_{|w|=1}\ln\left(1-\frac z{\erf(w)}\right)dw\\ &=-\frac1{2\pi}\int_0^{2\pi}e^{it}\ln\left(1-\frac z{\erf(e^{it})}\right)dt \end{split} $$ shown here. Although it likely can be generalized, this method works for $f(z)$ monotonic near $z=0,f(0)=0$, and $f’(0)\ne0$. One concern with the method is that we find the formula: $$ \begin{split} g(f^{-1}(z)) &= g(0)-\frac1{2\pi i}\oint_{|w|=1}g’(w)\ln\left(1-\frac z{f(w)}\right)dw\\ &=g(0)-\frac1{2\pi}\int_0^{2\pi}e^{it}g’(e^{it})\ln\left(1-\frac z{f(e^{it})}\right)dt \end{split} $$ which seems like a tautology because the formula applies to many arbitrary $f(z)$ without seemingly simplifying. Additionally, the formula is reminiscent to the one derivable from this answer, a version of the Cauchy integral formula for $f^{-1}(z)$:
$$ \begin{split} f^{-1}(z) &=\frac1{2\pi i}\oint\frac{wf’(w)}{f(w)-z}dz\\ & =\frac1{2\pi}\int_0^{2\pi}\frac{e^{2it}f’(e^{it})}{f(e^{it})-z}dt\\ & =1-\frac1{2\pi i}\oint\ln(f(w)-z)dw\\ & =1-\frac1{2\pi}\int_0^{2\pi}e^{it}\ln(f(e^{it})-z)dt \end{split} $$
saying
But no formula of this type would be worth publishing in the 20th or 21st centuries
which implies the $g(f^{-1}(z))$ formula may not be as useful due to its similarity to the Cauchy integral formula.
How helpful is this $g(f^{-1}(z))$ formula/Lagrange-inversion-inverse-Z-transform method in deriving integral representations of $f^{-1}(z)$?
