Solve[] with an Integral

Questions
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1

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]
2

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}]

ScreenShot2015-02-20at08.23.33.png
ScreenShot2015-02-20at08.23.33.png369×237 5.8 KB

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