| 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
|
| 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
|