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Re: Integrate (a curious result)

  • To: mathgroup at smc.vnet.net
  • Subject: [mg74679] Re: Integrate (a curious result)
  • From: "dimitris" <dimmechan at yahoo.com>
  • Date: Sat, 31 Mar 2007 01:29:18 -0500 (EST)
  • References: <esls78$q0v$1@smc.vnet.net><etb5a5$k14$1@smc.vnet.net>

This is from an old post of Peter Pein and has been covered completely by me
in this thread.

Integrate[Log[Sin[x]^2]*Tan[x], {x, 0, Pi}] (*wrong*)
Integrate::idiv: Integral of Log[Sin[x]^2]*Tan[x] does not converge on
\
{x,0,Pi}.
Integrate[Log[Sin[x]^2]*Tan[x], {x, 0, Pi}] (*correct*)

Integrate[Log[Sin[x]^2]*Tan[x], {x, 0, Pi/2, Pi}]
0

Trying to figure out why the default setting for the iterator of
Integrate
gives the warning message I take the antidrivative of the integrand:

F[x_] = Integrate[Log[Sin[x]^2]*Tan[x], x]
Plot[Re[F[x]], {x, 0, Pi}]
Plot[Im[F[x]], {x, 0, Pi}]

Log[Sec[x/2]^2]^2 + 2*Log[Sec[x/2]^2]*Log[(1/2)*Cos[x]*Sec[x/2]^2] +
Log[Sec[x/2]^2]*Log[Sin[x]^2] -
  2*Log[Sec[x/2]^2]*Log[-1 + Tan[x/2]^2] - Log[Sin[x]^2]*Log[-1 +
Tan[x/2]^2] + Log[Tan[x/2]^2]*Log[-1 + Tan[x/2]^2] + 2*PolyLog[2,
(1/2)*Sec[x/2]^2] + PolyLog[2, Cos[x]*Sec[x/2]^2] + PolyLog[2, -Tan[x/
2]^2]

Due to the jump discontinuity at x=Pi/2, the correct application of
the Newton-Leibniz formula
in [0,Pi] is

FullSimplify[(Limit[F[x], x -> Pi, Direction -> 1] - Limit[F[x], x ->
Pi/2, Direction -> -1]) +
   (Limit[F[x], x -> Pi/2, Direction -> 1] - Limit[F[x], x -> 0,
Direction -> -1])]
0

and of course give the correct result.

If Mathematica will not detect the jump discontinuity he may return

FullSimplify[Limit[F[x], x -> Pi, Direction -> 1] - Limit[F[x], x ->
0, Direction -> -1]]
4*I*Pi*Log[2]

which it doesn't. So something other happen.
Does anyone has any ideas?

Note also that strangely

Block[{Message}, Integrate[Log[Sin[x]^2]*Tan[x], {x, 0, Pi}]]
Integrate[Log[Sin[x]^2]*Tan[x], {x, 0, Pi}]

I said "strangely" because I was waiting Infinity as below for a
really
divergent integral

Integrate::idiv: Integral of BesselJ[0, x]*Csc[x] does not converge on
\
{0,Infinity}.
Integrate[BesselJ[0, x]*Csc[x], {x, 0, Infinity}]

Block[{Message}, Integrate[BesselJ[0, x]/Sin[x], {x, 0, Infinity}]]
Infinity

Any comments?


Thanks
Dimitris

=CF/=C7 dimitris =DD=E3=F1=E1=F8=E5:
> >   Dimitris, this is a marvelous solution to my problem. I really apprec=
ia=
> te
> > your help. I will now see if I can solve all my other (similar) integra=
ls=
>  using the same trick.
>
> It gives me great pleasure that you found so helpful my response.
>
> > Timing is not really the big issue if I get results in a reasonable amo=
un=
> t of time.
>
> Note that you must clarify to your mind what is actually a "reasonable
> amount of time".
> For example on my machine I do believe that the almost 17 seconds for
> evaluating the
> integral with my trick is more than just reasonable time.
>
> Note e.g. a very interesting example
>
> Integrate[BesselJ[0, x]/Sqrt[1 + x^2], {x, 0, Infinity}]//Timing
> N@%[[2]]
> NIntegrate[BesselJ[0, x]/Sqrt[1 + x^2], {x, 0, Infinity}, Method -> \
> Oscillatory]
>
> {4.186999999999999*Second, BesselI[0, 1/2]*BesselK[0, 1/2]}
> 0=2E9831043098467616
> 0=2E9831043098470272
>
> Quit
>
> Integrate[BesselJ[0, x]/Sqrt[1 + x^2], {x, 0, Infinity},
> GenerateConditions -> False]//Timing
> {0.31299999999999994*Second, BesselI[0, 1/2]*BesselK[0, 1/2]}
>
> If you know indepently that the integral converges why leave
> Mathematica to do the "dirty job"
> causing waste of your invaluable time?
>
> > Also the references you cited are quite interesting, because they give =
so=
> me insight
> > what might go on inside Mathematica concerning integration.
>
> Here is a more complete list.
> Note that following the moderator's advice I will avoid the use of
> link references.
>
> "Indefinite and Definite Integration" by Kelly Roach (1992)
> "The evaluation of Bessel functions via G-function identities" by
> Victor Adamchik (1995)
> "Definite Integration in Mathematica 3.0" by the same author
> "Symbolic Definite Integration" by Daniel Lichtblau
>
> Check also
>
> Manuel Bronstein's Symbolic Integration 1 and some references therein.
>
> > I also agree with you that it is my task as a programmer to "help" Math=
em=
> atica
> > and guide it into the direction I want it to go. Sometimes this is an "=
ar=
> t"...
>
> It is an a combination of art, experience, reading, thinking (and
> guessing sometimes!).
>
> Let look an integral appeared in an old post Peter Pein.
>
> $Version
> 5=2E2 for Microsoft Windows (June 20, 2005)
>
> Integrate[Log[Sin[x]^2]*Tan[x], {x, 0, Pi}]
> Integrate::idiv: Integral of Log[Sin[x]^2]*Tan[x] does not converge on
> \
> {x,0,Pi}.
> Integrate[Log[Sin[x]^2]*Tan[x], {x, 0, Pi}]
>
> Of course the integral is zero as you can figure out from the plot of
> the integrand
> in integration range or noticing that f[x]+f[Pi-x]=0 where
>
> f[x_] := Log[Sin[x]^2]*Tan[x]
>
> Also a direct numerical integration verify our statement.
>
> Tr@(NIntegrate[Log[Sin[x]^2]*Tan[x],
> {x,#[[1]],#[[2]]},WorkingPrecision->
>           50,PrecisionGoal->40] &/@Partition[Range[0,Pi,Pi/2],2,1])
> 0``39.9568146932663
>
> It is interesting to see what I got from another CAS (due to the
> policy
> of the forum I can't mention its name).
>
> int(log(sin(x)^2)*tan(x),x=0..Pi);
> undefined
>
> int(log(sin(x)^2)*tan(x),x=0..Pi/2);
> -Infinity
>
> int(log(sin(x)^2)*tan(x),x=Pi/2..Pi);
> Infinity
>
> On the contrary Mathematica may give a wrong value for the
> definite integral in the whole range (0,Pi) but
>
> Integrate[Log[Sin[x]^2]*Tan[x], {x, 0, Pi/2}]
> -(Pi^2/12)
>
> Integrate[Log[Sin[x]^2]*Tan[x], {x, Pi/2, Pi}]
> Pi^2/12
>
> Their sum give zero as it should be!
>
> What is happening here? I will try to figure out (I am not quite sure
> and as you may notice
> I have a recent post mentioning this integral).
>
> Note first that Mathematica evaluates correctly an indefinite integral
> of Log[Sin[x]^2]*Tan[x]
>
> F[x_] = Integrate[f[x], x]
> Log[Sec[x/2]^2]^2 + 2*Log[Sec[x/2]^2]*Log[(1/2)*Cos[x]*Sec[x/2]^2] +
> Log[Sec[x/2]^2]*Log[Sin[x]^2] -
>   2*Log[Sec[x/2]^2]*Log[-1 + Tan[x/2]^2] - Log[Sin[x]^2]*Log[-1 +
> Tan[x/2]^2] + Log[Tan[x/2]^2]*Log[-1 + Tan[x/2]^2] + 2*PolyLog[2,
> (1/2)*Sec[x/2]^2] + PolyLog[2, Cos[x]*Sec[x/2]^2] + PolyLog[2, -Tan[x/
> 2]^2]
>
> Simplify[D[F[x], x] == f[x]]
> True
>
> The obtained antiderivative has real and imaginary part
>
> Plot[Re@F[x], {x, 0, Pi}]
> Plot[Im@F[x], {x, 0, Pi}]
>
> Helping now Mathematica we can take the correct answer
>
> FullSimplify[(Limit[F[x], x -> Pi, Direction -> 1] - Limit[F[x], x ->
> Pi/2, Direction -> -1]) +
>    (Limit[F[x], x -> Pi/2, Direction -> 1] - Limit[F[x], x -> 0,
> Direction -> -1])]
> 0
>
> The problem is the vanishing of the integrand at Pi/2.
>
> Indeed using the following undocumentated setting of Integrate (which
> is well documentated for NIntegrate) Mathematica gives again the
> correct
> answer with less effort
>
> Integrate[Log[Sin[x]^2]*Tan[x], {x, 0, Pi/2, Pi}]
> 0
>
> Note that if we didn't take notice of Pi/2 and applied blindly the
> fundamental
> theorem of calculus we would get incorrectly
>
> FullSimplify[Limit[F[x], x -> Pi, Direction -> 1] - Limit[F[x], x ->
> 0, Direction -> -1]]
> 4*I*Pi*Log[2]
>
> That is a complex value for an integral whose integrand is real in the
> integration range.
> If Mathematica would simply return this complex value, I would say
> that it failed
> to detect the discontinuity of the antiderivative at Pi/2. But it
> returns a warning
> message for divergence, so...I GIVE UP!
>
> Note also that Integrate has been already from version 3.0 very
> powerful.
> But since version 5.0 symbolic integration has become one of the most
> important advantages of Mathematica over other CAS.
>
> But as Daniel Lichtblau pointed out in his article: The powerfulness
> has the price
> of slower speed (usually) and some level of platform dependent
> behavior (less usually)
> than in the past.
>
> Indeed in the original thread regarding this integral there was the
> following response
> from Bob Hanlon:
>
> >>Works in my version:
>
> >>$Version
> >>5.2 for Mac OS X (June 20, 2005)
>
> >>f[x_]:=Log[Sin[x]^2]Tan[x];
>
> >>Integrate[f[x],{x,0,Pi}]
> >>0
>
> Anyway other possible ways of working under Mathematica 5.2 for
> Windows include
>
> Tr @( Integrate[f[x], {x, #1[[1]], #1[[2]]}] &  /@
> Partition[Range[0, Pi, Pi/2], 2, 1] )
> 0
>
> Integrate[f[x], {x, 0, z}, Assumptions -> Inequality[Pi/2, Less, z,
> LessEqual, Pi]]
> Limit[%, z -> Pi, Direction -> 1]
> -(Pi^2/3) - 2*I*Pi*Log[2] + Log[2]^2 + 4*I*Pi*Log[Sec[z/2]] + 4*Log[-
> (Cos[z]/(1 + Cos[z]))]*Log[Sec[z/2]] + 4*Log[Sec[z/2]]^2 -
>   4*Log[Sec[z/2]]*Log[(-Cos[z])*Sec[z/2]^2] + 4*Log[Sec[z/
> 2]]*Log[Sin[z]] - 2*Log[(-Cos[z])*Sec[z/2]^2]*Log[Sin[z]] +
>   2*Log[(-Cos[z])*Sec[z/2]^2]*Log[Tan[z/2]] + 2*PolyLog[2, 1/(1 +
> Cos[z])] + PolyLog[2, Cos[z]*Sec[z/2]^2] +
>   PolyLog[2, -Tan[z/2]^2]
> 0
>
> Assuming[a >= 1, Integrate[f[a*x], {x, 0, Pi}]]
> Limit[%, a -> 1]
>
> (1/(3*a))*(-Pi^2 - 6*I*Pi*Log[2] + 3*Log[2]^2 + 6*Log[Cos[a*Pi]/(1 +
> Cos[a*Pi])]*Log[Sec[(a*Pi)/2]^2] +
>    3*Log[Sec[(a*Pi)/2]^2]^2 + 3*Log[Sec[(a*Pi)/2]^2]*Log[Sin[a*Pi]^2]
> - 6*Log[Sec[(a*Pi)/2]^2]*Log[-1 + Tan[(a*Pi)/2]^2] -
>    3*Log[Sin[a*Pi]^2]*Log[-1 + Tan[(a*Pi)/2]^2] + 3*Log[Tan[(a*Pi)/
> 2]^2]*Log[-1 + Tan[(a*Pi)/2]^2] +
>    6*PolyLog[2, 1/(1 + Cos[a*Pi])] + 3*PolyLog[2, Cos[a*Pi]*Sec[(a*Pi)/
> 2]^2] + 3*PolyLog[2, -Tan[(a*Pi)/2]^2])
> 0
>
> Integrate[f[x], {x, 0, Pi}, GenerateConditions -> False]
> 0
>
> You think and act properly. That is the correct attitude!
>
> > Nevertheless, in the present situation I do not really understand why M=
at=
> hematica wants
> > me to do that rather trivial variable transformation, which is at the h=
ea=
> rt of your solution.
> > The integrand is still a rather complicated rational function of the sa=
me=
>  order. The form
> > of the integrand did not really change
> > substantially as it is the case with some other ingenious substitutions=
 o=
> ne uses in order to
> > do some complicated looking integrals "by hand".
>
> I agree with you. But I can't provide you more insight than I did
> provide you in my original post.
> I think helping in this way Mathematica makes the application of
> partial fraction decomposition
> which is the heart of Risch Algorithm more easily for Integrate. If we
> could understand all of
> Mathematica's Integrate implementation algorithm we will probably work
> at WRI!
>
> I agree also that the the variable transformation is rather trivial.
> But who tells we must not think
> simple things first in order to assist CASs?
>
> Of course we can think and less trivial valiable transformations!
> For example in above mentioned integral:
>
> integrand = f[x]*dx /. x -> ArcSin[Sqrt[u]] /. dx ->
> D[ArcSin[Sqrt[u]], u]
> Integrate[integrand, {u, 0, Sin[Pi]^2}]
>
> Log[u]/(2*(1 - u))
> 0
>
> And what is more important as Peter Pein show in his original post in
> this way
> we can get a more simplified indefinite integral than Mathematica and
> is real in the
> whole integration range.
>
> inintegrand = f[x]*dx /. x -> -ArcSin[Sqrt[u]] /. dx -> D[-
> ArcSin[Sqrt[u]], u]
> Integrate[integrand, {u, 0, Sin[z]^2}, Assumptions -> 0 < z < Pi]
> Plot[%, {z, 0, Pi}]
> (%% /. z -> Pi) - (%% /. z -> 0)
> (D[%%%, z] // Simplify) /. z -> x
>
> Log[u]/(2*(1 - u))
> (1/12)*(-Pi^2 + 6*PolyLog[2, Cos[z]^2])
> (*plot to be displayed*)
> 0
> Log[Sin[x]^2]*Tan[x]
>
>
> > I think the fact that we are forced to such tricks shows that the Mathe=
ma=
> tica integrator
> > is still a bit "immature" in special cases, as also the very interestin=
g =
> article by D. Lichtblau,
> > which you cite, seems to indicate. So all this is probably perfectly kn=
ow=
> n to the
> > Mathematica devellopers. And I hope the next Mathematica version has al=
l =
> this "ironed out"??
>
> Reading again the article.
> Bear in your mind that the subject of the symbolic (indefinite and
> definite) integration
> is perhaps the most serious challenge for all CASs and of course
> Mathematica.
> You can't accuse Mathematica of immaturity even in special cases. If
> you think
> that your integral is a trivial one for a CAS (you don't certainly
> want to see what output I
> got for another CAS!) note Mathematica almost fifteen (yes only 15!)
> ago prior to the quite
> revolutionary version 3.0 it will return -3/2 for the following
> integral
>
> Integrate[1/x^2, {x, -1, 2}]
>
> that is, it would have failed to detect at first stage the
> nonintegarble singularity at x=0 and
> would have returned at second stage a negative value for a possitive
> integrand in the integration
> range.
>
> A lot of evolution has taken place. Don't you agree?
>
>
> Kind Regards
> Dimitris
>
>
>
> =CF/=C7 Michael Weyrauch =DD=E3=F1=E1=F8=E5:
> > Hello,
> >
> >   Dimitris, this is a marvelous solution to my problem. I really apprec=
ia=
> te
> > your help. I will now see if I can solve all my other (similar) integra=
ls=
>  using the same trick.
> > Timing is not really the big issue if I get results in a reasonable amo=
un=
> t of time.
> >
> > Also the references you cited are quite interesting, because they give =
so=
> me insight
> > what might go on inside Mathematica concerning integration.
> >
> > I also agree with you that it is my task as a programmer to "help" Math=
em=
> atica
> > and guide it into the direction I want it to go. Sometimes this is an "=
ar=
> t"...
> >
> > Nevertheless, in the present situation I do not really understand why M=
at=
> hematica wants
> > me to do that rather trivial variable transformation, which is at the h=
ea=
> rt of your solution.
> > The integrand is still a rather complicated rational function of the sa=
me=
>  order. The form
> > of the integrand did not really change
> > substantially as it is the case with some other ingenious substitutions=
 o=
> ne uses in order to
> > do some complicated looking integrals "by hand".
> >
> > I think the fact that we are forced to such tricks shows that the Mathe=
ma=
> tica integrator
> > is still a bit "immature" in special cases, as also the very interestin=
g =
> article by D. Lichtblau,
> > which you cite, seems to indicate. So all this is probably perfectly kn=
ow=
> n to the
> > Mathematica devellopers. And I hope the next Mathematica version has al=
l =
> this "ironed out"??
> >
> > Many thanks again,  Michael



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