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- Newsgroups: sci.math.symbolic
- Path: sparky!uunet!utcsri!torn!watserv2.uwaterloo.ca!watdragon.uwaterloo.ca!daisy.uwaterloo.ca!gjfee
- From: gjfee@daisy.uwaterloo.ca (Greg Fee)
- Subject: definite integration
- Message-ID: <BxGIzL.386@watdragon.uwaterloo.ca>
- Sender: news@watdragon.uwaterloo.ca (USENET News System)
- Organization: University of Waterloo
- Date: Mon, 9 Nov 1992 16:26:08 GMT
- Lines: 157
-
- Re: the definite integral
-
- integrate(ln(1 - b*x/(a+c*x^2))/x, x=0..infinity)
-
- Here is another derivation of the closed form based on differentiation
- with respect to a parameter.
-
- |\^/| MAPLE V
- ._|\| |/|_. Copyright (c) 1981-1990 by the University of Waterloo.
- \ MAPLE / All rights reserved. MAPLE is a registered trademark of
- <____ ____> Waterloo Maple Software.
- | Type ? for help.
- > f := ln(1 - b*x/(a+c*x^2))/x;
- b x
- ln(1 - --------)
- 2
- a + c x
- f := ----------------
- x
-
- > M := a^(1/2)/c^(1/2);
- 1/2
- a
- M := ----
- 1/2
- c
-
- # Invert integrand about M to transform [M,infinity] to [0,M] .
- > f1 := -subs(x=M^2/x,f)*diff(M^2/x,x);
- b a
- ln(1 - --------------)
- / 2 \
- | a |
- c x |a + ----|
- | 2|
- \ c x /
- f1 := ----------------------
- x
-
- > g := normal(f+f1);
- 2
- - a - c x + b x
- ln(- ----------------)
- 2
- a + c x
- g := 2 ----------------------
- x
-
- # New upper limit of integration is M, but transform it to 1
- > g := normal(subs(x=M*x,g)*diff(M*x,x));
- 1/2 2 1/2 1/2
- - a c - a x c + b a x
- ln(- -------------------------------)
- 1/2 2
- c a (1 + x )
- g := 2 -------------------------------------
- x
-
- > g1 := op(1,x/2*g);
- 1/2 2 1/2 1/2
- - a c - a x c + b a x
- g1 := - -------------------------------
- 1/2 2
- c a (1 + x )
-
- > g1 := convert(g1,parfrac,x);
- b x
- g1 := 1 - ------------------
- 1/2 1/2 2
- a c (1 + x )
-
- # Let b/a^(1/2)/c^(1/2) = 2*sin(k)
- > ss := b=2*sin(k)*a^(1/2)*c^(1/2);
- 1/2 1/2
- ss := b = 2 sin(k) a c
-
- > h1 := subs(ss,g1);
- sin(k) x
- h1 := 1 - 2 --------
- 2
- 1 + x
-
- > h := 2/x*ln(h1);
- sin(k) x
- ln(1 - 2 --------)
- 2
- 1 + x
- h := 2 ------------------
- x
-
- > hk := normal(diff(h,k));
- cos(k)
- hk := - 4 -------------------
- 2
- 1 + x - 2 sin(k) x
-
- > hk := -4*cos(k)/(cos(k)^2+sin(k)^2+x^2-2*sin(k)*x);
- cos(k)
- hk := - 4 -----------------------------------
- 2 2 2
- cos(k) + sin(k) + x - 2 sin(k) x
-
- # integrate with resect to x between 0 and 1
- > ih := int(hk,x);
- 2 x - 2 sin(k)
- ih := - 4 arctan(1/2 --------------)
- cos(k)
-
- > nih := normal(ih);
- x - sin(k)
- nih := - 4 arctan(----------)
- cos(k)
-
- > ihx := eval(subs(x=1,nih)-subs(x=0,nih));
- 1 - sin(k) sin(k)
- ihx := - 4 arctan(----------) - 4 arctan(------)
- cos(k) cos(k)
-
- > a1 := eval(subs(sin(k)/cos(k)=tan(k),ihx));
- 1 - sin(k)
- a1 := - 4 arctan(----------) - 4 k
- cos(k)
-
- > a2 := subs(arctan((1-sin(k))/cos(k))=Pi/4-k/2,a1);
- a2 := - Pi - 2 k
-
- # Now integrate k back
- > a3 := int(a2,k);
- 2
- a3 := - Pi k - k
-
- # Find the missing constant
- > int(eval(subs(k=0,h)),x=0..1);
- 0
-
- > subs(k=0,a3);
- 0
-
- # Missing constant in a3 is 0 so a3 is correct
- # Put back original parameters
- > sk := solve(ss,{k});
- b
- sk := {k = arcsin(1/2 ---------)}
- 1/2 1/2
- a c
-
- > answer := subs(sk,a3);
- b b 2
- answer := - Pi arcsin(1/2 ---------) - arcsin(1/2 ---------)
- 1/2 1/2 1/2 1/2
- a c a c
-
- > answer := factor(answer);
- b / b \
- answer := - arcsin(1/2 ---------) |Pi + arcsin(1/2 ---------)|
- 1/2 1/2 | 1/2 1/2 |
- a c \ a c /
-