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Conference rusure::math

Title:Mathematics at DEC
Moderator:RUSURE::EDP
Created:Mon Feb 03 1986
Last Modified:Fri Jun 06 1997
Last Successful Update:Fri Jun 06 1997
Number of topics:2083
Total number of notes:14613

1105.0. "definite integral problem" by AITG::DERAMO (Daniel V. {AITG,ZFC}:: D'Eramo) Sat Aug 05 1989 01:36

[header lines deleted -- Dan]
        
In article <1989Aug1.031230.8237@alberta.uucp> robert@alberta.uucp (Robert Baron) writes:
>Could anyone help me out with the following integral?
>
>                +infinity
>                    /                  2
>                    |                 x
>                    | --------------------------------- d x
>                    |  (exp(c x) + k) (exp(- c x) + k)
>                    /
>                -infinity
>
>with c > 0, and 0 < k < 1.  Thanks.


Answer:

	[I've withheld this part, but eventually
        I will post it. :-) -- Dan]
        
Does any symbolic integration program recognize this one???

-- 
  Gerald A. Edgar          
  Department of Mathematics                     TS1871@OHSTVMA.bitnet
  The Ohio State University                     edgar@shape.mps.ohio-state.edu
  Columbus, OH 43210   ...!{att,pyramid}!osu-cis!shape.mps.ohio-state.edu!edgar
T.RTitleUserPersonal
Name
DateLines
1105.1AITG::DERAMODaniel V. {AITG,ZFC}:: D'EramoMon Aug 07 1989 23:4111
        Well, let's see.  The denominator is
        
        (exp(c x) + k) (exp(- c x) + k)
        	= exp(c x) exp(- c x) + exp(c x) k + k exp(- c x) + k^2
        	= 1 + k(exp(c x) + exp(- c x)) + k^2
        	= 1 + 2 k cosh(c x) + k^2
        
        Does anyone recognize any standard definite integrals
        with this form?  :-)
        
        Dan
1105.2ALLVAX::ROTHIf you plant ice you'll harvest windMon Aug 14 1989 19:3512
    It doesn't look too obvious... it's easy to get the residues but that
    doesn't work because the denominator has an essential singularity at
    infinity.  You can expand the denomenator in partial fractions in terms
    of exp, but that isn't recognizable either.

    You could easily numerically get a value because it's well behaved
    though.

    So, did someone just pull some amazing continued fraction expansion
    out of a hat for this one?

    - Jim
1105.3AITG::DERAMODaniel V. {AITG,ZFC}:: D'EramoMon Aug 14 1989 20:177
	Only the answer was given, with no indication of how it was
	obtained.  The answer wasn't that complicated.  One approach
	is to ignore .1 and break the fraction up into a sum of two
	fractions, one over exp(c x) + k and the other over exp(- c x) + k.
	Perhaps those simpler integrands can then be solved or looked up.

	Dan
1105.4ALLVAX::ROTHIf you plant ice you'll harvest windTue Aug 15 1989 13:3928
1105.5AITG::DERAMODaniel V. {AITG,ZFC}:: D'EramoTue Aug 15 1989 17:0321
	When I tried it I got

			 2
			x  dx
		--------------------- =
		  cx		  cx
		(e    + k)(1 + k e  )

		   2
		  x	   - k		1
		------ ( ------- + --------- )
		     2	  cx		  cx
		1 - k	 e   + k   1 + k e

	Our numerators are reversed, and you multiplied both sides
	of the identity by e^(cx) dx.

	Another USENET message came by with the indefinite integral.
	What's a polylog?

	Dan
1105.6polygamma functionALLVAX::ROTHIf you plant ice you'll harvest windWed Aug 16 1989 08:4537
1105.7both the definite and indefinite integralsAITG::DERAMODaniel V. {AITG,ZFC}:: D'EramoThu Aug 17 1989 19:59107
	Here is the answer; the following are edited and
	are not the complete USENET articles.

	Dan

Article 6567 of sci.math
Path: ryn.esg.dec.com!shlump.nac.dec.com!decuac!haven!ames!ll-xn!rp
From: rp@XN.LL.MIT.EDU (Richard Pavelle)
Newsgroups: sci.math,sci.math.symbolic
Subject: Re: A puzzling definite integral
Summary: MACSYMA can do this.
Message-ID: <1515@xn.LL.MIT.EDU>
Date: 8 Aug 89 20:59:12 GMT
References: <1989Aug1.031230.8237@alberta.uucp> <151@shape.mps.ohio-state.edu>
Organization: MIT Lincoln Laboratory, Lexington, MA
Lines: 49
Xref: ryn.esg.dec.com sci.math:6567 sci.math.symbolic:667

In article <151@shape.mps.ohio-state.edu>, edgar@gem.mps.ohio-state.edu
(Gerald Edgar) writes:
> In article <1989Aug1.031230.8237@alberta.uucp> robert@alberta.uucp (Robert Baron) writes:
> >Could anyone help me out with the following integral?
> >
> >                +infinity
> >                    /                  2
> >                    |                 x
> >                    | --------------------------------- d x
> >                    |  (exp(c x) + k) (exp(- c x) + k)
> >                    /
> >                -infinity
> >
> >with c > 0, and 0 < k < 1.  Thanks.
> 
> 
> Answer:
>                         2          2
>            2 (log k) (pi  + (log k)  )
>           ------------------------------
>                      2       3
>               3    (k  - 1) c
> 
> Does any symbolic integration program recognize this one???

With a change of variable, namely x=y/c, MACSYMA gives the answer
above. Now I am using an old version, circa 1985, so perhaps the newer
versions can do it directly. However, the indefinite integral involves
polylogs so I doubt it. 

-- 
Richard Pavelle         UUCP: ...ll-xn!rp
                        ARPANET: rp@XN.LL.MIT.EDU

Article 6632 of sci.math
Path: ryn.esg.dec.com!shlump.nac.dec.com!decuac!haven!rutgers!iuvax!uxc.cso.uiuc.edu!garcon!procyon!george
From: george@procyon (John George)
Newsgroups: sci.math,sci.math.symbolic
Subject: Re: A puzzling definite integral
Message-ID: <1762@garcon.cso.uiuc.edu>
Date: 13 Aug 89 23:52:50 GMT
References: <1989Aug1.031230.8237@alberta.uucp> <151@shape.mps.ohio-state.edu> <1515@xn.LL.MIT.EDU>
Reply-To: george@symcom.math.uiuc.edu (John George)
Organization: Dept. of Mathematics, Univ. of Illinois at Urbana-Champaign
Lines: 67
Xref: ryn.esg.dec.com sci.math:6632 sci.math.symbolic:675

In article <1515@xn.LL.MIT.EDU> rp@XN.LL.MIT.EDU (Richard Pavelle) writes:
>With a change of variable, namely x=y/c, MACSYMA gives the answer
>above. Now I am using an old version, circa 1985, so perhaps the newer
>versions can do it directly. However, the indefinite integral involves
>polylogs so I doubt it. 
>

Mathematica can do the indefinite integral;

In[1]:= Integrate[x^2/( (Exp[c x] + k) (Exp[-c x] + k) ), x]

                                c x                      c x
                     2         E                        E
               3    x  Log[1 + ----]   2 x PolyLog[2, -(----)]
              x                 k                        k
Out[1]= -((k (--- - ---------------- - ----------------------- + 
              3 k         c k                    2
                                                c  k
 
                            c x
                           E
            2 PolyLog[3, -(----)]
                            k               2
>           ---------------------)) / (1 - k )) + 
                     3
                    c  k
 
       3    2          c x                        c x
      x    x  Log[1 + E    k]   2 x PolyLog[2, -(E    k)]
>    (-- - ------------------ - ------------------------- + 
      3            c                        2
                                           c
 
                        c x
        2 PolyLog[3, -(E    k)]          2
>       -----------------------) / (1 - k )
                   3
                  c

However, Mathematica could not find the limits for the definite integral.

--John C. George