Resolution of the Final ODE

We use the gfun package to solve for the solution which corresponds to the integral to be computed. Due to its integral representation, this function is analytic at 0, hence admits a Taylor expansion at 0. We proceed to compute a closed form for by summation of this expansion. To this end, we determine a recurrence equation on the coefficients of the Taylor expansion using gfun[diffeqtorec] :

>

> rsol:=rsolve(ore,u(n));

After rearranging the terms in the sum, it is obvious that is non-zero for even only.

> collect(map(normal,rsol,expanded),u,factor);

We perform the corresponding change of variable, ,

> subs({n=2*p,(-1)^n=1,RootOf(_Z^2+1)^n=(-1)^p},");

so that Maple can sum the Taylor series:

> sum("*a^(2*p),p=0..infinity);

> h:=collect(value(expand(")),u);

It only remains to evaluate and . We first compute and find it is 0 by inversion of limits. Let

>

be the integrand. We have:

>

In the same way, each coefficient of the Taylor series for the integral is obtained by inversion of limits. In particular, , but Maple is not capable of integrating:

> kappa=int(coeff(series(normal(diff(f,a,a)),a=0),a,0)/2,x=0..infinity);

(This integral for cannot be computed by a call to int using the Release 4, but the next release will probably be able to integrate it.)

We obtain the following form for :

> -combine(normal(-subs({u(0)=0,u(2)=kappa},h)),ln,symbolic);

It only remains to be proved that . We do not do it, since computing this last integral which is a constant lies outside the scope of the theory of holonomy. With this example, we have reduced the problem of evaluating a parametrized integral to the evaluation of a non-parametrized integral. In case there were no closed form for , we could at least perform a simple numerical evaluation and return a result in terms of this numerical value and the series above for .