Re: Novice/Integral
- To: mathgroup at smc.vnet.net
- Subject: [mg79374] Re: Novice/Integral
- From: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>
- Date: Wed, 25 Jul 2007 02:22:50 -0400 (EDT)
- Organization: The Open University, Milton Keynes, UK
- References: <f84ih0$oie$1@smc.vnet.net>
Daniel wrote: > Hi all, > > I'm new at this, but I'm trying to evalute the integral like so, > > g = {g0, g1, g2, g3, g4} > fg[x_] = (1 - x)^g[[3]]*(1 + g[[4]]*x^g[[5]]) /. x -> E^(y) > IntX[m_, y_] = Integrate[fg[x], y] > > I'm hoping there's standard tricks for manipulating integrals so mathematica can win out. Any advice is appreciated. > > Thanks, > Danny You could tell Mathematica that the values of g are real and transform the exponentials into equivalent trigonometric functions. For instance, In[1]:= $Version Out[1]= "6.0 for Microsoft Windows (32-bit) (June 19, 2007)" In[2]:= g = {g0, g1, g2, g3, g4}; fg[x_] = (1 - x)^g[[3]]*(1 + g[[4]]*x^g[[5]]) /. x -> E^y; IntX[m_, y_] = Integrate[ExpToTrig[fg[x]], y, Assumptions -> {Element[g, Reals]}] Out[4]= ((1 - \[ExponentialE]^-y)^-g2 (1 - \[ExponentialE]^y)^ g2 Hypergeometric2F1[-g2, -g2, 1 - g2, \[ExponentialE]^-y])/g2 + ((\[ExponentialE]^y)^ g4 g3 Hypergeometric2F1[-g2, g4, 1 + g4, \[ExponentialE]^y])/g4 This works also with version prior to 6.0. In[1]:= $Version Out[1]= 5.2 for Microsoft Windows (June 20, 2005) In[2]:= g = {g0, g1, g2, g3, g4}; fg[x_] = (1 - x)^g[[3]]*(1 + g[[4]]*x^g[[5]]) /. x -> E^y; IntX[m_, y_] = Integrate[ExpToTrig[fg[x]], y, Assumptions -> {Element[g, Reals]}] Out[4]= y g2 (1 - E ) Hypergeometric2F1[-g2, -g2, 1 - g2, -y 1 E ] -- ---------------------------------------------------\ g2 -y g2 (1 - E ) y g4 y (E ) g3 Hypergeometric2F1[-g2, g4, 1 + g4, E ] + ------------------------------------------------ g4 Regards, Jean-Marc