Conference rusure::math

        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

    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

	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

�	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.

   Right, that's what I meant by making a partial fraction expansion
   in terms of exp(c*x)... I got something like this

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


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


    I don't have maple up and runnng on my workstation right now, nor a
    good table of integrals handy.  But it doesn't look that familiar,
    unless it has something to do with Laplace transforms.

    - Jim

	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

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

    Ah... you must mean a polygammma (or psi) function.  These are
    expressed in terms of logarithmic derivatives of the gamma function
    and can be used to telescope sums whose terms are rational functions of
    the summation index.

    define

		    d		       d
	psi[1](z) = -- ln(gamma(z)) = (-- gamma(z)) / gamma(z)
		    dz		       dz	

		    d
	psi[n](z) = -- psi[n-1](z)
		    dz

    Then
		       n	         1
	psi[n](z) = (-) n! SUM(k>=0) ---------
				     (k + z)^n

    This has the integral representation (due to Gauss probably)

	   n             q^(n-1) exp(-z q)
	(-)  INT(0,inf) ------------------ dq
			   1 - exp(-q)

    So you could massage the pieces in the prior notes into this form
    and get the answer.

    I knew I'd seen something like that somewhere.  Look in AMS55, or also
    Morse and Fesbach, Mathematical Methods of Physics Vol I, under the gamma
    function.

    - Jim

	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: [email protected] (Richard Pavelle)
Newsgroups: sci.math,sci.math.symbolic
Subject: Re: A puzzling definite integral
Summary: MACSYMA can do this.
Message-ID: <[email protected]>
Date: 8 Aug 89 20:59:12 GMT
References: <[email protected]> <[email protected]>
Organization: MIT Lincoln Laboratory, Lexington, MA
Lines: 49
Xref: ryn.esg.dec.com sci.math:6567 sci.math.symbolic:667

In article <[email protected]>, [email protected]
(Gerald Edgar) writes:
> In article <[email protected]> [email protected] (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: [email protected]

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: <[email protected]>
Date: 13 Aug 89 23:52:50 GMT
References: <[email protected]> <[email protected]> <[email protected]>
Reply-To: [email protected] (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 <[email protected]> [email protected] (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