Integrating a functional of an InterpolatingFunction

Question

It is straightforward to Integrate an InterpolatingFunction. However, even for a simple functional of an InterpolatingFunction, Integrate returns unevaluated.

ifun1 =
  First[
    u /. 
      NDSolve[
        {u''[t] + u[t] == 0, u[0] == 0, u'[0] == 1},
        u,
        {t, 0, π}
      ]
  ]

Integrate[ifun1[t], {t, 0, π}]
Integrate[2*ifun1[t], {t, 0, π}]

The former Integrate returns a Real, the latter, the expression unevaluated.

How can the second integral be evaluated symbolically? (In my non-MWE, symbolically is important because the solution to my PDE has two arguments, over only one of which I would like to integrate, so NIntegrate will not work.) The solution should apply to more-complicated functionals, too (e.g. 2*t*ifun1[t]). If a solution applies only to functionals that are restricted to a certain family, such as "linear," it would still be helpful.

Answer 1

You should know that your first Integrate is numerically evaluated (so Mathematica calls NIntegrate on your interpolation function anyway). On the second I don't know why Mathematica doesn't call NIntegrate immediately but explicitly using NIntegrate yields a result.

In any case, this is building on your comment that you have a function with two arguments and you want to integrate over one and get a result as a function of the other. In my opinion b.gatessucks is spot on in that you need to use FunctionInterpolation.

Here's a random set of points:

points = MapThread[{#1[[1]], #1[[2]], #1[[3]] + #2} &, {Flatten[
     Array[{Sequence@##, Sin[Times@##]} &, {20, 20}], 1], 
    RandomReal[{-1, 1}, 400]}];

and here's an interpolation of the points:

f = Interpolation[points];

Now you can do whatever you want in the integrand over one of the arguments. FunctionInterpolation will evaluate the integral numerically at some points of your specified domain and create a function :

g = FunctionInterpolation[

  Integrate[2*f[t, y] - f[t, y]^2 + t Cos[f[t, y]], {t, 0, π}]
  ,
  {y, 1., 20.}
  ]

that behaves like any interpolating function. But note that "symbolically" and Interpolation are quite incompatible notions.