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# Solve[] with an Integral

Posted 10 years ago
 Hello, How can I have this solve for 'c'? A 'c' of around 1 will satisfy the equation. a =. b = 1; c =. Solve[Integrate[(-(1/(-1 + E)) - E^a/(1 - E))^(1/(b*c)), {a, 0, 1}] == 1/2, c, Reals]  Attachments:
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Posted 10 years ago
 Hi Greg,the integral evaluates to: ConditionalExpression[ 2.71828^(0.458675/c) c Hypergeometric2F1[-(1./c), -(1./c), 1. - 1./c, 0.367879] - 2.71828^(-0.541325/c) c Hypergeometric2F1[-(1./c), -(1./c), 1. - 1./c, 1.], Re[1/c] > -1.] So suppose that Solve will not be able to find a non-numerical solution to your equation. If you plot the solution you find: Plot[{Evaluate[2.718281828459045^(0.4586751453870819/c) c Hypergeometric2F1[-(1./c), -(1./c), 1. - 1./c, 0.36787944117144233] - 2.718281828459045^(-0.541324854612918/c) c Hypergeometric2F1[-(1./c), -(1./c), 1. - 1./c, 1.]], 0.5}, {c, 0.1, 1.7}] You can find the solution by FindRoot[Evaluate[ 1. 2.718281828459045^(0.4586751453870819/c) c Hypergeometric2F1[-(1./c), -(1./c), 1. - 1./c, 0.36787944117144233] - 1. 2.718281828459045^(-0.541324854612918/c) c Hypergeometric2F1[-(1./c), -(1./c), 1. - 1./c, 1.]] == 0.5, {c, 1.4}] this gives {c -> 1.36961}.Cheers,Marco
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