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# Error in Mathematica 9.0.1 integral of mixed Exp and Power type functions?

Posted 11 years ago
 Hello,In the description below you can find an error, which occurs for a Rational function of exponent argument. Do I miss anything?Followed the instruction and attached the problem description below. Hope it is readable now. Still, in my opinion it looks not as good as the original code in Mathematica (e.g. fractions are without horizontal lines, no symbolic representation for 'Pi' and 'Infinity').1. Integrate the following Integrate[Exp[a x]/(1 + Exp[b x])^2, {x, -\[Infinity], \[Infinity]}, Assumptions -> 2 b > a && a > 0]to obtain((-a + b) \[Pi] Csc[(a \[Pi])/b])/b^2The resulting expression coincides with the known formula in integration books (e.g. Gradshtein & Ryshik)2. Now let us consider the more general integralIntegrate[Exp[a x]/(1 + Exp[b x])^c, {x, -\[Infinity], \[Infinity]},Assumptions -> b c > a && a > 0 && b > 0]to obtainHypergeometric2F1[a/b, c, (a + b)/b, -1]/a +Hypergeometric2F1[c, -(a/b) + c, 1 - a/b + c, -1]/(-a + b c)As internal check choose the previous c = 2 value.c = 2; FullSimplify[Hypergeometric2F1[a/b, c, (a + b)/b, -1]/a +Hypergeometric2F1[c, -(a/b) + c, 1 - a/b + c, -1]/(-a + b c)                    ]to obtain ((-a + b) \[Pi] Csc[(a \[Pi])/b])/b^2the latter concides with case 1.Now the error.3. Remove the assumptions in case 2 Integrate[Exp[a x]/(1 + Exp[b x])^c, {x, -\[Infinity], \[Infinity]}]to obtainConditionalExpression[-(Hypergeometric2F1[a/b, c, (a + b)/b, -1]/a) +                        Hypergeometric2F1[c, -(a/b) + c, 1 - a/b + c, -1]/( a - b c),(Re[(-1)^(-1/b)] >= 1 ||    Re[(-1)^(-1/b)] <= 0 || (-1)^(-1/b) \[NotElement]     Reals) && (Re[(-1)^(1/b)] >= 1 ||    Re[(-1)^(1/b)] <= 0 || (-1)^(1/b) \[NotElement] Reals) &&  Re[b] < 0 && Re[a - b c] < 0 && Re[a] > 0]The first term acquires now an additional minus signInsert c = 2 to compare with case 2 considered abovec = 2; FullSimplify[-(Hypergeometric2F1[a/b, c, (a + b)/b, -1]/a) +  Hypergeometric2F1[c, -(a/b) + c, 1 - a/b + c, -1]/(a - b c)]to obtain ((a - b) \[Pi] Csc[(a \[Pi])/b])/b^2It differs from case 2 by a minus sign!To summarize, the same integrals (cases 1 and 3), were obtained by two different paths and yield two different results.'Pi' and 'Infinity' didn't appear as symbolic when I pasted the code. It appeard as symbolic only after I pressed the "Publish" button. The fraction lines are still not horizontal as in original file.