I already looked into the smaller problems, but I just don't get the clean first derivative, because Mathematica wants to work with imaginary numbers. How do I best supress this here? I tried it with Refine and assumptions but nothing worked yet.

  f[e_, n_, a_, b_, c_] := Integrate[ 1/(b - e)
       Integrate[((s - c)/(b - c))^(n - 1), {s, c, y}]
      , {y, e, b}]
  D[f[e, n, a, b, c], e]
  • $\begingroup$ You can make ConditionalExpression go away with Normal, but keep in mind that the result may not be valid for parameter values that violate the condition you just dropped. Is there any reason why you insist to nest these two integrals and put them in a function, given that you still use exclusively symbolic parameters? $\endgroup$ – Szabolcs Nov 7 '16 at 14:20
  • $\begingroup$ You can also wrap your code with an Assuming command and set your parameters to real, if they are indeed real. $\endgroup$ – ivbc Nov 7 '16 at 14:35
  • $\begingroup$ I am not sure if i get your question right.If i get you right, then my answer is, that the 2nd integral runs from c till y and y is the running variable in the first integral. So they need to be nested, imo. $\endgroup$ – Paul Nov 7 '16 at 14:37

This worked for me:

Assuming[n>0, f[e_, n_, a_, b_, c_] :=Integrate[1/(b - e) 
Integrate[((s - c)/(b - c))^(n - 1), {s, c, y}], {y, e, b}];
D[f[e, n, a, b, c], e]]

Notice that you could have added information about other parameters as well, like this: a \[Element] Reals && n > 0.

I noticed you tried Assumptions and failed. Probably your command only applied inside the integration, while mine wraps the differentiation as well.

  • $\begingroup$ Thank you all. Using the framing Assuming made my day :-D $\endgroup$ – Paul Nov 7 '16 at 14:50

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