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I think the following integration result can be shown without Gamma.

Can anyone explain

  1. Why Mathematica Uses Gamma.
  2. How to simplify the result to eliminate Gamma.

enter image description here

code:

Simplify[Integrate[(x/c)^(2/(1 - b)), {x, 0, 1/b}], Assumptions -> {1 > b > 0, c > 0}] 
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    $\begingroup$ Please do not post images of your work. Please post your actual Mathematica code in the form of text that can be copied and pasted into a Mathematica notebook. Without such, it will be difficult to reproduce your problem and to experiment with possible solutions. $\endgroup$
    – m_goldberg
    Sep 20 '18 at 15:52
  • $\begingroup$ Your integral diverges for 0<b<1 and so Mathematica is returning the principal value. $\endgroup$
    – Carl Woll
    Sep 20 '18 at 15:56
  • $\begingroup$ @ Carl Woll, if I change 2/(1-b) to 1/(1-b), the integral is also diverging? but there is no Gamma in the result, could you elaborate more on it? $\endgroup$
    – aa7on
    Sep 20 '18 at 16:42
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    $\begingroup$ Try this: Assuming[{1 > b > 0, c > 0}, Integrate[(x/c)^(2/(1 - b)), {x, 0, 1/b}]] $\endgroup$ Sep 20 '18 at 16:45
  • $\begingroup$ @ Mariusz Iwaniuk, thanks $\endgroup$
    – aa7on
    Sep 20 '18 at 16:48
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Try this:

Integrate[(x/c)^(2/(1 - b)), {x, 0, 1/b}, 
 Assumptions -> {1 > b > 0, c > 0}]

(* ((-1 + b) (b c)^(2/(-1 + b)))/((-3 + b) b) *)

Have fun!

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