Simplify integral with piecewise expression

Question

How can I simplify this integral?

Integrate[Piecewise[{{c[x], 0 <= x <= 1/2}, {0, True}}], {x, 0, 1}]

I think it should be the same as

Integrate[c[x], {x, 0, 1/2}]

Thanks

Answer 1

I can't think of an easy way to just Simplify a Piecewise integral. But here's something that might achieve what you're looking for (in a hopefully more general setting).

Define a "piecewise integral" function:

pIntegrate[p_, {var_, min_, max_}] := 
 Module[{funranges = 
    SortBy[{#[[1]], Sort@Cases[#[[2]], _?NumberQ]} & /@ p[var][[1]], Mean[#[[2]]] &]},

  (* Turn "x < a" into "-Infinity < x < a" *)
  If[Length[funranges[[1, 2]]] == 1, funranges[[1, 2]] = {-Infinity, funranges[[1, 2, 1]]}];
  If[Length[funranges[[-1, 2]]] == 1, funranges[[-1, 2]] = {funranges[[-1, 2, 1]], Infinity}];

  (* Only want the pieces of the function that are defined between the
     limits of integration *)
  funranges = Select[funranges, Or @@ IntervalMemberQ[Interval[{min, max}], #[[2]]] &];
  funranges[[;; , 2]] = Sort[Join[{min, max}, #[[2]]], Less][[{2, 3}]] & /@ funranges;

  (* Integrate the pieces and return their sum *)
  Total[Integrate[#[[1]], Flatten[{var, #[[2]]}]] & /@ funranges] 
 ]

Then applying it to your example:

f[x_] := Piecewise[{{c[x], 0 <= x <= 1/2}, {0, True}}]

pIntegrate[f, {x, 0, 1}]

But it can also do more complicated things. Suppose f has many pieces, which overhang the limits of integration in odd ways (and that the pieces aren't even presented in the right order):

f[x_] := Piecewise[{{a[x], 0 <= x < 1/2}, {c[x], x >= 1}, {b[x], 1/2 <= x < 1}, 
           {d[x], x < 0}, {0, True}}]

pIntegrate[f, {x, -5, 0.75}]

(Sadly, the output is reordered according to Mathematica's whim -- function names? -- and not according to the integration limits of the pieces. But I think there's little that can be done about that.)

I'm sure there must be a cleaner, more concise way of doing this. Also, it would be nice to be able to set the options for Integrate through pIntegrate. But it works, and perhaps it will inspire someone to do it improve on it.