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I need to calculate this integral using Mathematica:

∫ t^z * ln(t) dt from x to y

with the assumptions that z is an integer and 0 < x < y


This is what I'm trying to do:

Integrate[t^z*ln (t), {t, x, y}, 
 Assumptions -> z ∈ Integers, Assumptions -> 0 < x < y]

And this is what I get:

ConditionalExpression[(ln (-x^(2 + z) + y^(2 + z)))/(
 2 + z), (Re[x/(x - y)] > 
     1 || (x != 0 && x^2 != x y && Re[x/(x - y)] < 0 && 
      Re[x/(-x + y)] > 0) || 
    x/(x - y) \[NotElement] 
     Reals) && ((x \[Element] 
       Reals && ((Im[x] <= 
           Im[y] && (Im[x] >= 0 || Im[y] <= 0 || 
            Im[y] Re[x] >= Im[x] Re[y])) || (Im[x] >= 
           Im[y] && (Im[x] <= 0 || Im[y] >= 0 || 
            Im[y] Re[x] <= Im[x] Re[y])))) || (Im[x] >= 
       0 && (Im[y] >= 
         0 || (Im[x] <= Im[y] && (Im[y] Re[x])/Im[x] <= 
           Re[y]))) || (Im[x] >= 
       Im[y] && (((Im[y] Re[x])/Im[x] <= 
           Re[y] && (Im[x] <= 0 || Im[y] >= 0 || 
            Im[y] Re[x] <= Im[x] Re[y])) || (Im[x] <= 0 && 
          Im[y] <= 0))) || (Im[x] <= 
       Im[y] && (((Im[y] Re[x])/Im[x] <= 
           Re[y] && (Im[y] <= 0 || 
            Im[y] Re[x] >= Im[x] Re[y])) || (Im[x] <= 0 && 
          Im[y] <= 0))))]

I know that ln(y^(2+z) - x^(2+z)) / (2 + z) is not the right answer, so what did I do wrong?

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closed as off-topic by RunnyKine, bobthechemist, Öskå, Sjoerd C. de Vries, m_goldberg Aug 11 at 17:36

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question arises due to a simple mistake such as a trivial syntax error, incorrect capitalization, spelling mistake, or other typographical error and is unlikely to help any future visitors, or else it is easily found in the documentation." – RunnyKine, bobthechemist, Öskå, Sjoerd C. de Vries, m_goldberg
If this question can be reworded to fit the rules in the help center, please edit the question.

    
Sorry, it was supposed to be the natural logarithm like RunnyKine said. This is the first time I'm using Mathematica. –  user19093 Aug 11 at 16:31

1 Answer 1

up vote 2 down vote accepted

Assuming by ln you mean the natural logarithm, in Mathematica this is entered as Log

Integrate[t^z Log[t], {t, x, y}, Assumptions -> {z ∈ Integers, 0 < x < y}]

Mathematica graphics

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