Integrate question
Oskar Itzinger
Integrate[x/(3 x^2 - 1)^3,{x,0,1}]
doesn't converge on [0,1].
However, Mathematica 2.1 under Windows gives the corrrect answer, (1/16).
When did Mathematica lose the ability to do said integral?
Thanks.
DrMajorBob
Plot[x/(3 x^2 - 1)^3, {x, 0, 1}]
and you'll see that the integrand is unbounded on both sides of 1/Sqrt[3].
The pole is of order 3, so no, the integral does NOT converge.
It's possible, of course, that negative areas on one side cancel positive
on the other, so that the following limit might be finite:
epsilon =.
Integrate[x/(3 x^2 - 1)^3, {x, 0, 1/Sqrt[3] - epsilon}] +
Integrate[x/(3 x^2 - 1)^3, {x, 1/Sqrt[3] + epsilon, 1}] // Simplify
Limit[%, epsilon -> 0]
(-32 Sqrt[3] + 144 epsilon - 216 epsilon^3 +
81 epsilon^5)/(144 epsilon (4 - 3 epsilon^2)^2)
-\[Infinity]
But, as you see, it isn't. Here's a partial verification:
Table[
Integrate[x/(3 x^2 - 1)^3, {x, 0, 1./Sqrt[3] - 10^-n}] +
Integrate[x/(3 x^2 - 1)^3, {x, 1./Sqrt[3] + 10^-n, 1}], {n, 1, 10}]
{-0.181712, -2.34348, -23.9916, -238.35, -257.148, 2.12135*10^6,
2.11647*10^9, 1.85626*10^12, 6.40027*10^14, 1.02864*10^16}
and here's another:
Table[
Quiet@NIntegrate[x/(3 x^2 - 1)^3, {x, 0, 1./Sqrt[3] - 10^-n}] +
Quiet@NIntegrate[x/(3 x^2 - 1)^3, {x, 1./Sqrt[3] + 10^-n, 1}], {n,
1, 12}]
{-0.181712, -2.34349, -23.9938, -240.5, -2405.57, -24058.2, -242718., \
-4.5604*10^6, -2.39253*10^9, -2.15502*10^12, -2.15478*10^15, \
-2.15494*10^18}
Even if this limit WERE finite, that wouldn't imply convergence for the
original integral, since there's no reason to use those particular limits
on the left and right (equal gaps around the pole). Indeed, we can make
the integral anything we want, by choosing limits accordingly.
If we want the integral to be Pi, and epsilon1 > 0 is chosen so that the
right integral is equal to 10^n (any n), we can chose epsilon2 > 0 for the
left integral to make it equal to Pi - 10^n, which makes the sum equal to
Pi. A sequence of epsilon1, epsilon2 values chosen this way (both
converging to zero) make the limit equal to Pi.
So... versions 5.2 and 6 are correct; version 2.1 was wrong.
Bobby
--
Bhuvanesh
Bhuvanesh,
Wolfram Research
Daniel Lichtblau
> Mathematica 5.2 under IRIX complains that
>
> Integrate[x/(3 x^2 - 1)^3,{x,0,1}]
>
> doesn't converge on [0,1].
>
> However, Mathematica 2.1 under Windows gives the corrrect answer, (1/16).
>
> When did Mathematica lose the ability to do said integral?
>
> Thanks.
There is a nonintegrable singularity at 1/Sqrt[3]. You can check this
via Series.
In[6]:= Series[x/(3*x^2 - 1)^3, {x,1/Sqrt[3],2}]
1 1 1
Out[6]= -------------------- - ---------------------------- + ---------- -
1 3 1 2 32 Sqrt[3]
72 (-(-------) + x) 48 Sqrt[3] (-(-------) + x)
Sqrt[3] Sqrt[3]
1 1 2
5 (-(-------) + x) 9 Sqrt[3] (-(-------) + x)
Sqrt[3] Sqrt[3] 1 3
> ------------------ + --------------------------- + O[-(-------) + x]
128 256 Sqrt[3]
To get a regularized integral you can use GenerateConditions->False.
In[7]:= Integrate[x/(3*x^2 - 1)^3, {x,0,1}, GenerateConditions->False]
1
Out[7]= --
16
I realize this is an overloaded option and expect that in some future
version there will be a cleaner way to request regularized results of
definite integration.
Daniel Lichtblau
Wolfram Research
David W.Cantrell
> Mathematica 5.2 under IRIX complains that
>
> Integrate[x/(3 x^2 - 1)^3,{x,0,1}]
>
> doesn't converge on [0,1].
Actually, that's good! The integral doesn't converge. The integrand has a
pole at x = 1/Sqrt[3].
> However, Mathematica 2.1 under Windows gives the corrrect answer, (1/16).
The answer 1/16 is correct in the sense of Cauchy principal value. But I
think that the behavior of version 5.2 is better, that is, if an integral
doesn't converge in the usual sense, complain that it doesn't converge,
rather than giving a Cauchy principal value which was not specifically
requested.
> When did Mathematica lose the ability to do said integral?
Of course you can get the Cauchy principal value easily in version 5.2.
The quick and dirty method is to get the antiderivative and then just use
Newton-Leibniz:
In[3]:= Integrate[x/(3 x^2 - 1)^3, x]
Out[3]= -(1/(12*(-1 + 3*x^2)^2))
In[4]:= (% /. x - > 1) - (% /. x -> 0)
Out[4]= 1/16
However, it would have been reasonable to expect that setting the option
PrincipalValue to True would even more easily give the desired value. But
that would be "reasonable to expect" only if one had not carefully read the
documentation for PrincipalValue:
"Setting PrincipalValue->True gives finite answers for integrals that had
simple pole divergences with PrincipalValue->False"
Note the crucial word "simple". The pole at x = 1/Sqrt[3] in not simple,
and thus, after having read the documentation, it should not be a surprise
that
Integrate[x/(3 x^2 - 1)^3, {x, 0, 1}, PrincipalValue -> True]
also causes version 5.2 to complain that the integral diverges.
I consider the restriction of PrincipalValue to simple poles to be
unfortunate. Is it still that way in version 6? If so, I hope that the
utility of PrincipalValue will be broadened in some future version.
David W. Cantrell
Jean-Marc Gulliet
FWIW, Mathematica for Windows 5.2 as well as 6.0.1 cannot do it either,
although it does not seem that hard to get the correct answer (the
indefinite integral and the limits are evaluated in a breeze).
In[1]:= int = Integrate[x/(3 x^2 - 1)^3, x]
Out[1]= -(1/(12 (-1 + 3 x^2)^2))
In[2]:= Limit[int, x -> 1] - int /. x -> 0
Out[2]= 1/16
Regards,
--
Jean-Marc
Jean-Marc Gulliet
> FWIW, Mathematica for Windows 5.2 as well as 6.0.1 cannot do it either,
> although it does not seem that hard to get the correct answer (the
> indefinite integral and the limits are evaluated in a breeze).
>
> In[1]:= int = Integrate[x/(3 x^2 - 1)^3, x]
>
> Out[1]= -(1/(12 (-1 + 3 x^2)^2))
>
> In[2]:= Limit[int, x -> 1] - int /. x -> 0
>
> Out[2]= 1/16
Sorry to have posted this nonsense (I was too quick to answer without
thinking). Indeed, Mathematica 2.1 is wrong (as I was in my previous
reply) in claiming that the definite integral is equal to 1/16, for
the integrand has a non removable discontinuity at 1/Sqrt[3] (about
0.55) and each integral are divergent on the interval {0,1/Sqrt[3]}
(-Infinity) and {1/Sqrt[3],1} (Infinity).
In[1]:= Plot[x/(3 x^2 - 1)^3, {x, 0, 1}]
Solve[12 (-1 + 3 x^2)^2 == 0, x]
% // N
Integrate[x/(3 x^2 - 1)^3, {x, 0, 1/Sqrt[3]}]
Integrate[x/(3 x^2 - 1)^3, {x, 1/Sqrt[3], 1}]
Out[1]= [ ... graphic deleted ... ]
Out[2]= {{x -> -(1/Sqrt[3])}, {x -> -(1/Sqrt[3])}, {x -> 1/Sqrt[
3]}, {x -> 1/Sqrt[3]}}
Out[3]= {{x -> -0.57735}, {x -> -0.57735}, {x -> 0.57735}, {x ->
0.57735}}
During evaluation of In[1]:= Integrate::idiv: Integral of x/(-1+3 \
x^2)^3 does not converge on {0,1/Sqrt[3]}. >>
Out[4]= Integrate[x/(-1 + 3*x^2)^3, {x, 0, 1/Sqrt[3]}]
During evaluation of In[1]:= Integrate::idiv: Integral of x/(-1+3 \
x^2)^3 does not converge on {1/Sqrt[3],1}. >>
Out[5]= Integrate[x/(-1 + 3*x^2)^3, {x, 1/Sqrt[3], 1}]
Sorry for the confusion,
--
Jean-Marc
Andrzej Kozlowski
On 16 Oct 2007, at 16:28, Oskar Itzinger wrote:
> Mathematica 5.2 under IRIX complains that
>
> Integrate[x/(3 x^2 - 1)^3,{x,0,1}]
>
> doesn't converge on [0,1].
>
> However, Mathematica 2.1 under Windows gives the corrrect answer,
> (1/16).
>
> When did Mathematica lose the ability to do said integral?
>
> Thanks.
>
>
>
The reason is that Mathematica 2.1 was wrong and Mathematica 5.2 is
much more careful and right. What Mathematica 2.1 did here was simply:
Subtract @@ (Integrate[x/(3 x^2 - 1)^3, x] /. {{x -> 1}, {x -> 0}})
1/16
in other words, it applied the Newton-Leibnitz rule in a mindless
way. Later versions are more intelligent and see that the singularity at
=CE=B1 = Last[x /. Solve[3*x^2 - 1 == 0, x]]
1/Sqrt[3]
One can also see this graphically (of course!):
Plot[x/(3 x^2 - 1)^3, {x, 0, 1}]
the integral still might exist in the sense of Cauchy PrincipalValue
but we see that it does not:
Integrate[x/(3*x^2 - 1)^3, {x, 0, 1}, PrincipalValue -> True]
Integrate::idiv:Integral of x/(3 x^2-1)3 does not converge on {0,1}. >>
Integrate[x/(3*x^2 - 1)^3, {x, 0, 1}, PrincipalValue -> True]
If you still don't beleive it, you can do it "by hand":
int = FullSimplify[Integrate[x/(3*x^2 - 1)^3,
{x, 0, 1/Sqrt[3] - =CE=B5}] +
Integrate[x/(3*x^2 - 1)^3,
{x, 0, 1/Sqrt[3] + =CE=B5}], =CE=B5 > 0]
(9*=CE=B5^2*(3*=CE=B5^4 - 8*=CE=B5^2 + 5) - 4)/
(18*(3*=CE=B5^3 - 4*=CE=B5)^2)
Limit[int, =CE=B5 -> 0]
-=E2=88=9E
Andrzej Kozlowski
Oskar Itzinger
"wrong"?
/oskar
"Andrzej Kozlowski" <ak...@mimuw.edu.pl> wrote in message
news:ff4hpp$k0e$1...@smc.vnet.net...
David W.Cantrell
> "Oskar Itzinger" <os...@opec.org> wrote:
> > Mathematica 5.2 under IRIX complains that
> >
> > Integrate[x/(3 x^2 - 1)^3,{x,0,1}]
> >
> > doesn't converge on [0,1].
>
> Actually, that's good! The integral doesn't converge. The integrand has a
> pole at x = 1/Sqrt[3].
>
> > However, Mathematica 2.1 under Windows gives the corrrect answer,
> > (1/16).
>
> The answer 1/16 is correct in the sense of Cauchy principal value.
Sorry!! That's false, as at least two respondents have already pointed out.
Although regularizing the integral gives 1/16, as mentioned by Daniel
Lichtblau, that is not a Cauchy principal value. Consequently, the next two
times I mention "Cauchy principal value", it should actually have been
"regularized value" instead.
> But I
> think that the behavior of version 5.2 is better, that is, if an integral
> doesn't converge in the usual sense, complain that it doesn't converge,
> rather than giving a Cauchy principal value which was not specifically
> requested.
>
> > When did Mathematica lose the ability to do said integral?
>
> Of course you can get the Cauchy principal value easily in version 5.2.
> The quick and dirty method is to get the antiderivative and then just use
> Newton-Leibniz:
>
> In[3]:= Integrate[x/(3 x^2 - 1)^3, x]
>
> Out[3]= -(1/(12*(-1 + 3*x^2)^2))
>
> In[4]:= (% /. x - > 1) - (% /. x -> 0)
>
> Out[4]= 1/16
Unfortunately, I had overlooked the even easier way, mentioned by Daniel,
for getting the regularized value: using GenerateConditions -> False as an
option in the definite integral.
David W. Cantrell
Oskar Itzinger
Integrate can evaluate definite integrals whenever the correct result can be
found by taking limits
of the indefinite form at the endpoints.
?
/oskar
"Andrzej Kozlowski" <ak...@mimuw.edu.pl> wrote in message
news:ff4hpp$k0e$1...@smc.vnet.net...
>
DrMajorBob
indef[x_] = Integrate[x/(3 x^2 - 1)^3, x]
indef[1] - indef[0]
-(1/(12 (-1 + 3 x^2)^2))
1/16
But that's meaningless, isn't it, since "indef" isn't continuous on the
range?
Plot[indef[x], {x, 0, 1}]
Bobby
On Wed, 17 Oct 2007 02:55:32 -0500, Daniel Lichtblau <da...@wolfram.com>
wrote:
> Oskar Itzinger wrote:
>> Mathematica 5.2 under IRIX complains that
>>
>> Integrate[x/(3 x^2 - 1)^3,{x,0,1}]
>>
>> doesn't converge on [0,1].
>>
>> However, Mathematica 2.1 under Windows gives the corrrect answer,
>> (1/16).
>>
>> When did Mathematica lose the ability to do said integral?
>>
>> Thanks.
>
--
DrMajorBob
No, I don't think so.
Integrate[x/(3 x^2 - 1)^3, {x, 0, 1}, PrincipalValue -> True]
Integrate::idiv: Integral of x/(<<1>>)^3 does not converge on {0,1}. \
>>
Integrate[x/(-1 + 3 x^2)^3, {x, 0, 1}, PrincipalValue -> True]
Testing it numerically, I get:
cauchy[epsilon_?Positive] :=
Integrate[x/(3 x^2 - 1)^3, {x, 0, Sqrt[1/3] - epsilon}] +
Integrate[x/(3 x^2 - 1)^3, {x, Sqrt[1/3] + epsilon, 1}]
cauchy /@ (3.^-Range[20])
{-0.0233868, -0.158072, -0.588358, -1.8865, -5.78329, -17.4737, \
-52.526, -157.163, -457.048, -978.052, 7680.08, 309458., 8.64827*10^6,
2.33306*10^8, 6.21267*10^9, 1.60783*10^11, 3.83616*10^12,
7.35457*10^13, 8.58974*10^14, 4.49786*10^15}
Bobby
On Wed, 17 Oct 2007 02:58:36 -0500, David W.Cantrell
<DWCan...@sigmaxi.net> wrote:
> "Oskar Itzinger" <os...@opec.org> wrote:
>> Mathematica 5.2 under IRIX complains that
>>
>> Integrate[x/(3 x^2 - 1)^3,{x,0,1}]
>>
>> doesn't converge on [0,1].
>
> Actually, that's good! The integral doesn't converge. The integrand has a
> pole at x = 1/Sqrt[3].
>
>> However, Mathematica 2.1 under Windows gives the corrrect answer,
>> (1/16).
>
> The answer 1/16 is correct in the sense of Cauchy principal value. But I
> think that the behavior of version 5.2 is better, that is, if an integral
> doesn't converge in the usual sense, complain that it doesn't converge,
> rather than giving a Cauchy principal value which was not specifically
> requested.
>
>> When did Mathematica lose the ability to do said integral?
>
> Of course you can get the Cauchy principal value easily in version 5.2.
> The quick and dirty method is to get the antiderivative and then just use
> Newton-Leibniz:
>
> In[3]:= Integrate[x/(3 x^2 - 1)^3, x]
>
> Out[3]= -(1/(12*(-1 + 3*x^2)^2))
>
> In[4]:= (% /. x - > 1) - (% /. x -> 0)
>
> Out[4]= 1/16
>
> However, it would have been reasonable to expect that setting the option
> PrincipalValue to True would even more easily give the desired value. But
> that would be "reasonable to expect" only if one had not carefully read
> the
> documentation for PrincipalValue:
>
> "Setting PrincipalValue->True gives finite answers for integrals that had
> simple pole divergences with PrincipalValue->False"
>
> Note the crucial word "simple". The pole at x = 1/Sqrt[3] in not simple,
> and thus, after having read the documentation, it should not be a
> surprise
> that
>
> Integrate[x/(3 x^2 - 1)^3, {x, 0, 1}, PrincipalValue -> True]
>
> also causes version 5.2 to complain that the integral diverges.
>
> I consider the restriction of PrincipalValue to simple poles to be
> unfortunate. Is it still that way in version 6? If so, I hope that the
> utility of PrincipalValue will be broadened in some future version.
>
> David W. Cantrell
>
>
--
Kevin J. McCann
Kevin
DrMajorBob wrote:
> Testing it numerically, I get:
>
> cauchy[epsilon_?Positive] :=
> Integrate[x/(3 x^2 - 1)^3, {x, 0, Sqrt[1/3] - epsilon}] +
> Integrate[x/(3 x^2 - 1)^3, {x, Sqrt[1/3] + epsilon, 1}]
> cauchy /@ (3.^-Range[20])
>
> {-0.0233868, -0.158072, -0.588358, -1.8865, -5.78329, -17.4737, \
> -52.526, -157.163, -457.048, -978.052, 7680.08, 309458., 8.64827*10^6,
> 2.33306*10^8, 6.21267*10^9, 1.60783*10^11, 3.83616*10^12,
> 7.35457*10^13, 8.58974*10^14, 4.49786*10^15}
>
> Bobby
--
Kevin J. McCann
Research Associate Professor
JCET/Physics
Physics Building
University of Maryland, Baltimore County
1000 Hilltop Circle
Baltimore, MD 21250
Oskar Itzinger
u=3 x^2 - 1
and integrating
(1/6)/u^3 on [-1,2]
be acceptable?
/oskar
"Oskar Itzinger" <os...@opec.org> wrote in message
news:ff77b9$ncr$1...@smc.vnet.net...
Andrzej Kozlowski
answer:
Integrate[1/6/u^3, {u, -1, 2}]
Integrate::idiv:Integral of 1/u3 does not converge on {-1,2}. >>
Integrate[1/(6*u^3), {u, -1, 2}]
What else did you expect? It's the same and therefore 1/16 it is
equally "unacceptable" or "non-acceptable" as an answer as before.
The integral does not converge in any usual mathematical sense : as
Riemann or Lebesgue integral or even in the sense of "Principal
value". However, that does not mean that 1/16 is not "acceptable"
answer for any purpose anyone can imagine. It just does not represent
any of the quantities people normally use integrals for, such as an
area, or an average value etc. For example, you consider the
divergent series 1-1+1-1+1 -...., then there is a context in which
the answer 1/2 is an or rather the "acceptable" value of this sum,
but in most usual contexts this is a divergent series and thus its
sum has no value. There are some situations where taking 1/2 as the
answer can be useful.
Even something like 1+1 = 5 is acceptable, if you are doing
arithmetic modulo 3. All mathematics makes sense (or does not as the
case may be) in some precisely defined context. Without a context you
are just playing with meaningless symbols.
Andrzej Kozlowski
On 19 Oct 2007, at 18:09, Oskar Itzinger wrote:
> Would setting
>
> u=3 x^2 - 1
>
> and integrating
>
> (1/6)/u^3 on [-1,2]
>
> be acceptable?
>
> /oskar
>
> "Oskar Itzinger" <os...@opec.org> wrote in message
> news:ff77b9$ncr$1...@smc.vnet.net...
DrMajorBob
Bobby
On Fri, 19 Oct 2007 04:09:09 -0500, Oskar Itzinger <os...@opec.org> wrote:
> Would setting
>
> u=3 x^2 - 1
>
> and integrating
>
> (1/6)/u^3 on [-1,2]
>
> be acceptable?
>
> /oskar
>
> "Oskar Itzinger" <os...@opec.org> wrote in message
> news:ff77b9$ncr$1...@smc.vnet.net...
--
DrMajorBob
> "wrong"?
Because it was wrong. The integral doesn't converge.
The Newton-Leibnez rule (second fundamental theorem of integral calculus)
doesn't apply, since the antiderivative isn't equal to the integrand on
the range. That condition fails at the pole (1/Sqrt[3]), where the
antiderivative is neither continuous nor bounded.
Integrate didn't "notice" the theorem doesn't apply, so it used it anyway.
Bobby
On Thu, 18 Oct 2007 03:44:22 -0500, Oskar Itzinger <os...@opec.org> wrote:
> Ok, Mathematica 2.1 was "mindless" here, but why was it considered to be
> "wrong"?
>
> /oskar
>
> "Andrzej Kozlowski" <ak...@mimuw.edu.pl> wrote in message
> news:ff4hpp$k0e$1...@smc.vnet.net...
>>
>> The reason is that Mathematica 2.1 was wrong and Mathematica 5.2 is
>> much more careful and right. What Mathematica 2.1 did here was simply:
>>
>> Subtract @@ (Integrate[x/(3 x^2 - 1)^3, x] /. {{x -> 1}, {x -> 0}})
>> 1/16
>>
>> in other words, it applied the Newton-Leibnitz rule in a mindless
>> way.
>
>
>
>
David W.Cantrell
> It represents the same mathematica object, so you will get the same
> answer:
>
> Integrate[1/6/u^3, {u, -1, 2}]
> Integrate::idiv:Integral of 1/u3 does not converge on {-1,2}. >>
> Integrate[1/(6*u^3), {u, -1, 2}]
>
> What else did you expect? It's the same and therefore 1/16 it is
> equally "unacceptable" or "non-acceptable" as an answer as before.
Of course, you are correct in saying that 1/16 is unacceptable for
Integrate[1/6/u^3, {u, -1, 2}] since nothing other than Mathematica's
standard integration was specified. But curiously, there is a sense (see
below) in which 1/16 is now the correct answer, even though it wasn't for
the original integral in terms of x.
> The integral does not converge in any usual mathematical sense : as
> Riemann or Lebesgue integral or even in the sense of "Principal
> value".
No, it _does_ converge in the sense of Cauchy principal value. Of course,
as I noted before, PrincipalValue in the current version of Mathematica
doesn't handle anything but _simple_ poles. But the integral in terms of
u converges to 1/16 in the sense of Cauchy principal value, although the
original integral in terms of x did not:
In[11]:= Simplify[Assuming[0 < e < 1,
Integrate[1/(6*u^3), {u, -1, -e}] + Integrate[1/(6*u^3), {u, e, 2}]]]
Out[11]= 1/16
David
Andrzej Kozlowski
in my original response) before writing it was not true in the Cauchy
principal value sense. But this actually confirms the other point I
made: that one cannot exclude the possibility that there is a sense
in which the answer 1/16 is "acceptable". However, and this is the
key point, the fact that such a sense exists is only a curiosity
unless this "sense" fits the intended purpose of this "integration".
This is why I am usually not very impressed when someone asks a
question to which, in the intended context, the answer is "no" and
then someone writes to point out that there is some other context in
which the answer is "yes". There is a countless number of situations
of this kind and it is easy to multiply examples. You can take almost
any theorem of elementary analysis and find that in some more
abstract context it is not valid. The first example that comes to my
mind (just as I am writing this): the familiar and even "obvious"
fact that a smooth function is continuous is not true in the context
of analysis on locally convex spaces more general than Banach spaces
(the Frolicher-Kriegl theory). But this is completely irrelevant to
anyone who never deals with anything other than finite dimensional
euclidean spaces. Similarly, the fact that this particular integral
(after the change of variable) has Cauchy Principal value 1/16 is
probably irrelevant - "probably" because the OP never told us what
sort of "integral" he had in mind.
Andrzej Kozlowski
On 20 Oct 2007, at 21:22, David W. Cantrell wrote:
> [Message also posted to: comp.soft-sys.math.mathematica]
DrMajorBob
> made: that one cannot exclude the possibility that there is a sense
> in which the answer 1/16 is "acceptable". However, and this is the
> key point, the fact that such a sense exists is only a curiosity
> unless this "sense" fits the intended purpose of this "integration".
I gotta agree with Andrzej, and I'll repeat my earlier comment that you
can make the integral anything you want by approaching the pole
(in)appropriately from left and right.
If you want the result to be Pi, choose epsilon1 so that the right
integral is Pi/2 + n and epsilon2 so that the left integral is Pi/2 - n.
e1 and e2 are not equal, but they both tend to zero as n -> Infinity, and
the partial integral is Pi for every n.
A version of this is precisely what happened in the change of variables:
unequal epsilons (original problem) mapped to equal epsilons (new
problem), so the new problem has a finite Cauchy principal value when the
original did not. I suspect there's a change of variables that
accomplishes the same trick for ANY integrand whose antiderivative tends
to +Infinity on one side of the pole and -Infinity on the other.
Integrals that aren't invariant (even as to the question of convergence)
under a change of variables?
Yikes. Have fun with that.
Bobby
On Sat, 20 Oct 2007 21:21:08 -0500, Andrzej Kozlowski <ak...@mimuw.edu.pl>
wrote:
>>> It represents the same mathematica object, so you will get the sam=
e
>>> answer:
>>>
>>> Integrate[1/6/u^3, {u, -1, 2}]
>>> Integrate::idiv:Integral of 1/u3 does not converge on {-1,2}. >>
>>> Integrate[1/(6*u^3), {u, -1, 2}]
>>>
>>> What else did you expect? It's the same and therefore 1/16 it is
>>> equally "unacceptable" or "non-acceptable" as an answer as before.
>>
>> Of course, you are correct in saying that 1/16 is unacceptable for
>> Integrate[1/6/u^3, {u, -1, 2}] since nothing other than Mathematica's=
>> standard integration was specified. But curiously, there is a sense
>> (see
>> below) in which 1/16 is now the correct answer, even though it
>> wasn't for
>> the original integral in terms of x.
>>
>>> The integral does not converge in any usual mathematical sense : as
>>> Riemann or Lebesgue integral or even in the sense of "Principal
>>> value".
>>
>> No, it _does_ converge in the sense of Cauchy principal value. Of
>> course,
>> as I noted before, PrincipalValue in the current version of
>> Mathematica
>> doesn't handle anything but _simple_ poles. But the integral in
>> terms of
>> u converges to 1/16 in the sense of Cauchy principal value,
>> although the
>> original integral in terms of x did not:
>>
>> In[11]:= Simplify[Assuming[0 < e < 1,
>> Integrate[1/(6*u^3), {u, -1, -e}] + Integrate[1/(6*u^3), {u, e, 2}]]]=
>>
>> Out[11]= 1/16
>>
>> David
>>
>>
>>> On 19 Oct 2007, at 18:09, Oskar Itzinger wrote:
>>>
>>>> Would setting
>>>>
>>>> u=3 x^2 - 1
>>>>
>>>> and integrating
>>>>
>>>> (1/6)/u^3 on [-1,2]
>>>>
>>>> be acceptable?
>>>>
>>>> /oskar
>>>>
>>>> "Oskar Itzinger" <os...@opec.org> wrote in message
>>>> news:ff77b9$ncr$1...@smc.vnet.net...
>>>>> Hmm, from Mathematica 2.1 Help:
>>>>>
>>>>> Integrate can evaluate definite integrals whenever the correct
>>>>> result can be found by taking limits
>>>>> of the indefinite form at the endpoints.
>>>>>
>>>>> ?
>>>>>
>>>>> /oskar
>>>>>
>>>>> "Andrzej Kozlowski" <ak...@mimuw.edu.pl> wrote in message
>>>>> news:ff4hpp$k0e$1...@smc.vnet.net...
>>>>>>
>>>>>> On 16 Oct 2007, at 16:28, Oskar Itzinger wrote:
>>>>>>
>>>>>>> Mathematica 5.2 under IRIX complains that
>>>>>>>
>>>>>>> Integrate[x/(3 x^2 - 1)^3,{x,0,1}]
>>>>>>>
>>>>>>> doesn't converge on [0,1].
>>>>>>>
>>>>>>> However, Mathematica 2.1 under Windows gives the corrrect answer=
,
>>>>>>> (1/16).
>>>>>>>
>>>>>>> When did Mathematica lose the ability to do said integral?
>>>>>>>
>>>>>>> Thanks.
>>>>>>>
>>>>>>>
>>>>>>>
>>>>>>
>>>>>>
>>>>>> The reason is that Mathematica 2.1 was wrong and Mathematica
>>>>>> 5.2 is
>>>>>> much more careful and right. What Mathematica 2.1 did here was
>>>>>> simply:
>>>>>>
>>>>>> Subtract @@ (Integrate[x/(3 x^2 - 1)^3, x] /. {{x -> 1}, {x ->
>>>>>> 0}})
>>>>>> 1/16
>>>>>>
>>>>>> in other words, it applied the Newton-Leibnitz rule in a mindless=
>>>>>> way. Later versions are more intelligent and see that the
>>>>>> singularity at
>>>>>> =CE=B1 = Last[x /. Solve[3*x^2 - 1 == 0, x]]
>>>>>> 1/Sqrt[3]
>>>>>>
>>>>>> One can also see this graphically (of course!):
>>>>>>
>>>>>> Plot[x/(3 x^2 - 1)^3, {x, 0, 1}]
>>>>>>
>>>>>>
>>>>>> the integral still might exist in the sense of Cauchy
>>>>>> PrincipalValue
>>>>>> but we see that it does not:
>>>>>>
>>>>>> Integrate[x/(3*x^2 - 1)^3, {x, 0, 1}, PrincipalValue -> True]
>>>>>> Integrate::idiv:Integral of x/(3 x^2-1)3 does not converge on
>>>>>> {0,1}. >>
>>>>>> Integrate[x/(3*x^2 - 1)^3, {x, 0, 1}, PrincipalValue -> True]
>>>>>>
>>>>>> If you still don't beleive it, you can do it "by hand":
>>>>>>
>>>>>> int = FullSimplify[Integrate[x/(3*x^2 - 1)^3,
>>>>>> {x, 0, 1/Sqrt[3] - =CE=B5}] +
>>>>>> Integrate[x/(3*x^2 - 1)^3,
>>>>>> {x, 0, 1/Sqrt[3] + =CE=B5}], =CE=B5 > 0]
>>>>>> (9*=CE=B5^2*(3*=CE=B5^4 - 8*=CE=B5^2 + 5) - 4)/
>>>>>> (18*(3*=CE=B5^3 - 4*=CE=B5)^2)
>>>>>>
>>>>>> Limit[int, =CE=B5 -> 0]
>>>>>> -=E2=88=9E
>>>>>>
>>>>>>
>>>>>> Andrzej Kozlowski
>
>
>
-- =
David W.Cantrell
> > principal value sense. But this actually confirms the other point I
> > made: that one cannot exclude the possibility that there is a sense
> > in which the answer 1/16 is "acceptable". However, and this is the
> > key point, the fact that such a sense exists is only a curiosity
> > unless this "sense" fits the intended purpose of this "integration".
>
> I gotta agree with Andrzej,
The way you say that makes it seem as though you think I disagree with
Andrzej about his "key point". But of course I don't, and never have.
> and I'll repeat my earlier comment that you
> can make the integral anything you want by approaching the pole
>
> (in)appropriately from left and right.
But in dealing with Cauchy principal value, one is not at liberty to have
two epsilons which approach the pole differently. Of course, if you do what
you indicate below, you can get any value you wish. Perhaps that could be
called Bobby's arbitrary value method, but that's not Cauchy principal
value. If the latter exists, it is unique.
> If you want the result to be Pi, choose epsilon1 so that the right
> integral is Pi/2 + n and epsilon2 so that the left integral is Pi/2 - n.
> e1 and e2 are not equal, but they both tend to zero as n -> Infinity, and
> the partial integral is Pi for every n.
>
> A version of this is precisely what happened in the change of variables:
> unequal epsilons (original problem) mapped to equal epsilons (new
> problem), so the new problem has a finite Cauchy principal value when the
> original did not. I suspect there's a change of variables that
> accomplishes the same trick for ANY integrand whose antiderivative tends
> to +Infinity on one side of the pole and -Infinity on the other.
>
> Integrals that aren't invariant (even as to the question of convergence)
> under a change of variables?
Then perhaps you would prefer using the regularized integral, mentioned by
Daniel Lichtblau, since its value is invariant.
> Yikes. Have fun with that.
I can't ever remember _using_ either a Cauchy principal value or a
regularized integral myself. I was merely trying to assist the OP in
providing a sense in which the value of the integral could be said to be
1/16. That seemed to be what he wanted. Whether such a sense fits his
intended purpose is a matter which must be left to the OP; he didn't
provide enough information for us to make that assessment for him.
For better or worse, it seems that regularized integrals are used in
physics (quantum mechanics, in particular) etc.
David