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)

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:
>
> 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?
>
>
> Aaron Fude
>

--

Bob Hanlon
hanlonr at cox.net

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