For iterative numerical application (for example NestList) I need to define a function, which depends on a function and outputs a function.

To describe the problem, here I give a simple example:

pint = Function[{ip }, Function[{t}, Integrate[(x + t x - 2 t) ip[x], {x, 0, t} ] ]];

This analytical function might be applied iteratively as expected

pint[Exp[-#]&][t] (*1 - t - E^-t (1 + t^2)*)
pint[pint[Exp[-#] &]][t] (*-9 - t + 1/2 (-3 + t) t^2 + E^-t (9 + 10 t + 7 t^2 + 3 t^3 + t^4)*)

Works fine, syntax seems to be ok.

Now my real problem, thereby I assume the integration can't be evaluated analytically:

Simply changing Integrate to NIntegrate gives

pintN = Function[{ip },Function[{t}, NIntegrate[(x + t x - 2 t) ip[x], {x, 0, t} ] ]] ;      

which evaluates as expected (same result as pint)

Plot[{pint[Exp[-#] &][u], pintN[Exp[-#] &][u]}, {u, 0, 5},PlotStyle -> {Thickness[.01], Automatic}]

enter image description here

Unfortunately the function pintN can't be applied iteratively! For example

Plot[pintN[pintN[Exp[-#] &]][u], {u, 0, 5}]

doesn't work!

How can I make the numerical version run? Thanks!

  • $\begingroup$ I think pintN = Function[{ip}, Function[{t}, Module[{x}, Quiet@NIntegrate[(x + t x - 2 t) ip[x], {x, 0, t}]]]] works to make the integration variable unique. NIntegrate splutters some messages (hence the Quiet), but I think the answer is correct. Note that pint also seems to suffer from localizations issues with x. $\endgroup$ Commented Jan 11, 2021 at 22:00
  • $\begingroup$ @SjoerdSmit Thanks, I tried your modified definition but Plot[Evaluate[pintN[pintN[Exp[-#] &]][t]] , {t, 0, Pi}] doesn't finish evaluation. $\endgroup$ Commented Jan 11, 2021 at 22:05
  • $\begingroup$ Ah, but that's because it makes no sense to evaluate a numerical function with at symbolic symbol t, right? You'll need some form of pattern matching to handle that. $\endgroup$ Commented Jan 11, 2021 at 22:11

1 Answer 1


With numerical functions, you have to be careful that NIntegrate will always get a valid integrand. The easiest way to do this is with pattern matching:

pintN[ip_][t_?NumericQ] := Module[{x},
  NIntegrate[(x + t x - 2 t) ip[x], {x, 0, t}]

This will now work, but it's very slow:

Plot[pintN[pintN[Exp[-#] &]][u], {u, 0, 5}, PlotPoints -> 20, MaxRecursion -> 0]

Note that sometimes you can cleverly use ParametricNDSolveValue to represent integrals. For example, to represent a function that numerically integrates another function you can use:

int[f_] := Module[{x, y, t},
   {y'[x] == f[x], y[0] == 0},
   {x, 0, t},
  • $\begingroup$ Thanks for your answer. Now I learned how to restrict t_?NumericQin the definition of pintN. It seems to be impossible to restrict t in my original defintion Function[{ip },Function[{t}… Thanks also for your hint concerning NDSolve . $\endgroup$ Commented Jan 12, 2021 at 8:37

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