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Is there an analytical expression for the following integral?

$$\int_a^b \ln(x-a)\ln(x-c)\,\mathrm{d}x,$$

where $c < a < b$ are all real numbers.

I don't really need to know how this integral is done (though if you post that too it would be nice). I just need the final result.

Note: Please, post an understandable formula. Not just the code returned by a Computer Algebra System.

a06e
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  • The Maple code $$int(ln(x-a)*ln(x-c), x = a .. b, AllSolutions) $$ produces the answer in terms of the dilog function. – user64494 Jun 17 '14 at 19:04
  • @user64494 I don't have Maple. Can you post the result? – a06e Jun 17 '14 at 19:04
  • I run it on Mathematica, ANS: ConditionalExpression[ after that $2 a + 2 b + a Log[a - c] - b Log[a - c] - b Log[(b - c)/(a - c)] + c Log[(b - c)/(a - c)] + Log[-a + b] (a - b + (-a + b) Log[a - c] + (b - c) Log[(b - c)/(a - c)]) + (a - c) PolyLog[2, (a - b)/(a - c)],$ – L.K. Jun 17 '14 at 19:07
  • After that I don't know. – L.K. Jun 17 '14 at 19:09
  • @lavkush Please, see added note. Your second comment is unreadable! :) – a06e Jun 17 '14 at 19:12
  • @becko, Got it. It is conditional expression, which you written. So you are seeing the garbage looking condition, which I have written. Let me understand, then I will edit it. – L.K. Jun 17 '14 at 19:15
  • @lavkush Thanks. I think many of those conditions are unnecessary, since $c<a<b$. – a06e Jun 17 '14 at 19:22
  • related: http://math.stackexchange.com/q/839944/10063, http://math.stackexchange.com/q/840146/10063 – a06e Jun 28 '14 at 13:05

1 Answers1

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The answer done with Maple by $$int(ln(x-a)*ln(x-c), x = a .. b)\, assuming\, c<a$$ is $$\ln \left( -c+b \right) \ln \left( b-a \right) b-\ln \left( b-a \right) c\ln \left( -c+b \right) -\ln \left( b-a \right) a\ln \left( -c+a \right) +$$ $$\ln \left( b-a \right) c\ln \left( -c+a \right) +\ln \left( b-a \right) a-$$ $$\ln \left( b-a \right) b-\ln \left( -c+b \right) b+\ln \left( -c+b \right) c+{\it dilog} \left( { \frac {-c+b}{-c+a}} \right) a-$$ $${\it dilog} \left( {\frac {-c+b}{-c+a}} \right) c+\ln \left( -c+a \right) a-\ln \left( -c+a \right) c-2\,a+ 2\,b, $$ where the dilog function is decribed here. I think the answer is obtained through integration by parts.

user64494
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