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MathGroup Archive 2006

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Re: Why all the if's the answer

  • To: mathgroup at smc.vnet.net
  • Subject: [mg70806] Re: Why all the if's the answer
  • From: "ben" <benjamin.friedrich at gmail.com>
  • Date: Sat, 28 Oct 2006 05:21:45 -0400 (EDT)
  • References: <ehs31u$oe4$1@smc.vnet.net>

Dear Aaron,

Cos[alpha]>0 is safer than 0<alpha<pi/2.

Look

In[550]:=
\!\(Integrate[y/c*f, {y, 0, c},
    Assumptions \[Rule] {Cos[alpha] > 0, Sin[alpha] > 0, b > 0, c > 0,
        b\^2 + c\^2 \[NotEqual] 2\ b\ c\ Cos[alpha]}]\)

Out[550]=
\!\(\(-\(\(1\/\(2\ \((b\^2 + c\^2 - 2\ b\ c\
Cos[alpha])\)\^\(3/2\)\)\)\((b\ \
c\ \((b\ \@\(b\^2 + c\^2 - 2\ b\ c\ Cos[alpha]\) -
            c\ \@\(b\^2 + c\^2 - 2\ b\ c\ Cos[alpha]\) + b\^2\ Log[b] -

            b\^2\ Log[c] -
            b\^2\ Log[\(c - b\ Cos[alpha] + \@\(b\^2 + c\^2 - 2\ b\ c\
\
Cos[alpha]\)\)\/\@\(b\^2 + c\^2 - 2\ b\ c\ Cos[alpha]\)] +
            b\^2\ Log[\(\(-b\) + c\ Cos[alpha] + \@\(b\^2 + c\^2 - 2\
b\ c\ \
Cos[alpha]\)\)\/\@\(b\^2 + c\^2 - 2\ b\ c\ Cos[alpha]\)] -
            b\ c\ Cos[
                alpha]\ \((Log[b] -
                  Log[\(c\ \((c - b\ Cos[alpha] + \@\(b\^2 + c\^2 - 2\
b\ c\ \
Cos[alpha]\))\)\)\/\@\(b\^2 + c\^2 - 2\ b\ c\ Cos[alpha]\)] +
                  Log[\(\(-b\) + c\ Cos[alpha] + \@\(b\^2 + c\^2 - 2\
b\ c\ \
Cos[alpha]\)\)\/\@\(b\^2 + c\^2 - 2\ b\ c\ Cos[alpha]\)])\))\))\)\)\)\)


In case Mathematica is in trouble, replacing constant subexps by new constants
is sometimes a good idea

feasy = -Log[A y] + Log[B + C y + Sqrt[D + E y + F y^2]].

Bye
Ben

aaronfude at gmail.com schrieb:

> 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


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