Re: Why all the if's the answer

*To*: mathgroup at smc.vnet.net*Subject*: [mg70797] Re: [mg70783] Why all the if's the answer*From*: Bob Hanlon <hanlonr at cox.net>*Date*: Sat, 28 Oct 2006 05:21:31 -0400 (EDT)*Reply-to*: hanlonr at cox.net

f = -Log[y*(1 + Cos[alpha])] + Log[(b*c - b*y + c*y*Cos[alpha] + Sqrt[b^2*(c - y)^2 + c^2*y^2 + 2*b*c*(c - y)*y*Cos[alpha]])/c]; soln1=Integrate[y/c*f, {y, 0, c},GenerateConditions->False]; soln2=Simplify[soln1,cond={y>0 , c>0 , b> 0 , 0 < alpha<Pi/2}]/.b^2-2*c*Cos[alpha]*b+c^2:>x; soln3=FullSimplify[soln2,cond] (b*c*(Sqrt[x]*(c - b) + b*(b - c*Cos[alpha])*(Log[c] + Log[c - b*Cos[alpha] + Sqrt[x]] - Log[b*(-b + c*Cos[alpha] + Sqrt[x])])))/(2*x^(3/2)) where x = b^2-2*c*Cos[alpha]*b+c^2 Bob Hanlon ---- aaronfude at gmail.com wrote: > Suppose I start with : > > f = -Log[y*(1 + Cos[alpha])] + Log[(b*c - b*y + c*y*Cos[alpha] + > Sqrt[b^2*(c - y)^2 + c^2*y^2 + 2*b*c*(c - y)*y*Cos[alpha]])/c]; > > which I think is relatively innocent. I then integrate over y: > > Assuming[y > 0 && c > 0 && b > 0 && alpha > 0 && alpha < Pi/2, > Integrate[y/c*f, {y, 0, c}]] > > After a few hours get a humongous answer with many If's, etc. What else > can I assume or otherwise do to get an analytical answer? > > Many thanks in advance! > > Aaron Fude > -- Bob Hanlon hanlonr at cox.net