# Inverse of a function defined by an integral

Hi, I have a function defined by an integral as follows. $$z=f(w) = \int_0^w \frac{(\zeta-a_1)^{\alpha_1}(\zeta-a_2)^{\alpha_2}...}{(\zeta-b_1)^{\beta_1}(\zeta-b_2)^{\beta_2}...}\ d\zeta$$ where $w$ is real, $a_i$ and $b_i$ are constants and $\alpha_i$ are integers. Thus the integrand is a rational function that can be nicely factored into linear terms.

Are there some theorems discussing, relevant properties about, or way of calculating the inverse $$w=f^{-1}(z)?$$ All suggestions are welcome.

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Apart from the usual "expand as a series, invert with Lagrange, use truncated version of inverse series as a starting point for Newton-Raphson or some other nonlinear equation solver"? – J. M. Sep 6 '10 at 9:17
Probably not useful to you, but have you looked at the Schwarz-Christoffel formula (about Riemann maps of discs onto polygons)? This is a very important special case. Also, I would suggest the "complex variables" tag, since this is the "natural" way to do this, at least locally. – Zen Harper Sep 6 '10 at 9:34
@zen harper: thanks, this problem comes indeed from the Schwarz-Christoffel formula. I now want to invert the map – Mermoz Sep 6 '10 at 10:19
@ J.M. could you give me some more info about "ivert with Lagrange"? – Mermoz Sep 6 '10 at 10:23
This is not going to be easy. The simplest cases already include sin, cos, and the <a href="mathworld.wolfram.com/JacobiEllipticFunctions.html">; functions of Jacobi</a> – Piero D'Ancona Sep 6 '10 at 13:57

You can always get a (non-linear) ordinary differential equation for $f^{-1}$. It is easy to see that $f$ satisfies a 2nd order linear ODE with polynomial coefficients with no order 0 term [the first order ODE has non-polynomial coefficients, so harder to work with]. From there, it is also mechanical to get an ODE for $f^{-1}$ by interchanging the roles of $f$ and $w$.

But since your original function is (in general, depending on the path of integration) a MeijerG function, very few of these have closed-form inverses. As Piero mentions in the comments, the trigonometric and Elliptic functions are some of the few cases where this 'works'.

It also depends on what you are trying to do 'next' with these functions. If you are looking at numerical evaluation, then there are closed-forms for the Lagrange inverse, some of which translate to closed-forms for the Hermite-Pade approximants, from which efficient approximations can be derived. [J.M.'s method works in general, here it turns out that this can be pushed in closed-form further than usual].

For any given case, there are useful tools in Maple (and Mathematica) to help you carry these computations out. Probably in other CASes as well, but I don't know.

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Question - how can I use the non-linear, second order ODE for $g = f^{−1}$ to get the first, say, 100 terms of the power series expansion for $g$ about zero? Just messing around with an example, it seems that the later coefficients depend, in a highly non-linear fashion, on the first non-zero coefficient... Is this standard? Is there a canonical reference? Best, – Sam Nead Aug 10 '12 at 5:21
The first order ODE is non-linear, but for your pattern the second order ODE is linear. Getting a power series from that is 'classical'. The easiest way is to use the recurrence relation for the coefficients (use any decent CAS to get that). The dependence on the first non-zero coefficient should be 'clear' from the recurrence. A lot of this information seems to be only found in (much) older texts on ODEs which focus on computation rather than on existence proofs; look at Forsyth's for example. – Jacques Carette Aug 12 '12 at 16:53
I don't see how to get a linear second order ode. For my example I get something like $p(g) g′′ = q(g) (g′)^2$, where $p,q$ are low-order polynomials and $g$ is the inverse of $f$. The dependence of the later coefficients of the power series of $g$ on the first non-zero coefficient is not linear. I've been trying to use SAGE, but they have not yet implemented power series solutions to ODEs (or if they have, I haven't found it). – Sam Nead Aug 29 '12 at 19:15
To be more precise - my exact question is about inverting the incomplete beta function (mathworld.wolfram.com/IncompleteBetaFunction.html) $B(z,a,b)$ for fixed (rational) $a$ and $b$. This is motivated by wanting to find the inverse to the Schwarz-Christoffel mapping from the upper-half plane to a triangle. – Sam Nead Aug 29 '12 at 19:27
Perhaps I was too hasty in abandoning the first order differential equation. It has some unpleasant exponents, but perhaps I can persuade it to give me a power series expansion for $g$... – Sam Nead Aug 29 '12 at 19:43