Definite and indefinite integrations produce different results for same integrand

Question

Consider a function subjected to the following assumptions:

assum = 
  a0 ∈ Reals && a1 ∈ Reals && a1 > a0 && a0 > 0 && 
  m ∈ Integers && p ∈ Integers && m >= 0 && p >= 0; 
TmpD = Sin[m*Pi*(a - a0)/(a1 - a0)]*Sin[p*Pi*(a - a0)/(a1 - a0)]/a

Integrating TmpD using a definite integral

Assuming[assum && m == p, Simplify[Integrate[TmpD, {a, a0, a1}]]]

produces

 (* 1/2 (-I π Cos[(2 a0 p π)/(a0-a1)]+Cos[(2 a0 p π)/(a0-a1)] CosIntegral[(2 a0 p π)/(a0-a1)]-Cos[(2 a0 p π)/(a0-a1)] CosIntegral[-((2 a1 p π)/(a0-a1))]-Log[a0]+Log[a1]+Sin[(2 a0 p π)/(a0-a1)] SinIntegral[(2 a0 p π)/(a0-a1)]-Sin[(2 a0 p π)/(a0-a1)] SinIntegral[(2 a1 p π)/(a0-a1)])*)

while integrating using a definite integral

Assuming[assum && m == p, 
 Simplify[tmp = Integrate[TmpD, a]; (tmp /. a -> a1) - (tmp /. a -> a0)]]

produces a result without the additional complex part

(* 1/2 (Cos[(2 a0 p π)/(a0-a1)] CosIntegral[-((2 a0 p π)/(a0-a1))]-Cos[(2 a0 p π)/(a0-a1)] CosIntegral[-((2 a1 p π)/(a0-a1))]-Log[a0]+Log[a1]+Sin[(2 a0 p π)/(a0-a1)] SinIntegral[(2 a0 p π)/(a0-a1)]-Sin[(2 a0 p π)/(a0-a1)] SinIntegral[(2 a1 p π)/(a0-a1)]) *)

Why are the results different?

Answer 1

The results in Mathematica 11.0.1 (Windows) are the same. Try using global assumptions with $Assumptions for simplification with FullSimplify

assum = a0 \[Element] Reals && a1 \[Element] Reals && a1 > a0 && 
   a0 > 0 && m \[Element] Integers && p \[Element] Integers && 
   m >= 0 && p >= 0;
(*Make assumptions global*)
$Assumptions = assum;
(*Compute*)
TmpD = Sin[m*Pi*(a - a0)/(a1 - a0)]*Sin[p*Pi*(a - a0)/(a1 - a0)]/a;
out1 = Assuming[assum && m == p, 
   Simplify[Integrate[TmpD, {a, a0, a1}]]];
out2 = Assuming[assum && m == p, 
   Simplify[
    tmp = Integrate[TmpD, a]; (tmp /. a -> a1) - (tmp /. a -> a0)]];
(*Test without global assumptions*)
out1 - out2
(*Test with global assumptions*)
out1 - out2 // FullSimplify