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MathGroup Archive 2007

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Re: Diferent solution of integral in versions 4 and 5...

  • To: mathgroup at smc.vnet.net
  • Subject: [mg73574] Re: Diferent solution of integral in versions 4 and 5...
  • From: "dimitris" <dimmechan at yahoo.com>
  • Date: Thu, 22 Feb 2007 04:32:02 -0500 (EST)
  • References: <ergr6r$jjg$1@smc.vnet.net>

Hi Victor.


f[k3_, x3_, k12_] = Exp[(-I)*k3*x3]/(k12^2 + k3^2)^2;

Let sum up a litlle the results.

Here is the result by Mathematica 5.2

math52 = Integrate[f[k3, x3, k12], {k3, -Infinity, Infinity},
Assumptions -> k12 != 0 && Im[k12] == 0 && Im[x3] == 0]
(2*Pi^(3/2)*MeijerG[{{1/2}, {1/4, 3/4}}, {{0, 1/2, 3/2}, {1/4, 3/4}},
(-(1/4))*k12^2*x3^2]*Sign[k12])/k12^3

Note that it isn't need to specify in Assumptions that Im[k3]==0,
since Mathematica is able to understand that the integration take
place in the real axis from the integration range

Here are the results (obtained by you; I don't have version 5 and 4 to
check them) of Mathematica 5
and 4 respectively.

math5 = (Abs[k12]*MeijerG[{{1/2}, {}}, {{0, 1/2, 3/2}, {}}, (-
(1/4))*k12^2*x3^2])/(k12^4*Sqrt[Pi]);

math4 = ((1 + Abs[k12]*Abs[x3])/Abs[k12]^2)*Exp[(-Abs[k12])*Abs[x3]];

Let check them before proceeding (Always check and justify the
analytic results, especially of definite integration, obtained by
Mathematica; the same holds true for all CAS as well! Oh well, with
Mathematica you must be LESS worry but this is another story!)

First we create a list of rules for k12 and x3 which will be used
pretty soon.

lst = (Thread[{k12, x3} -> #1] & ) /@ Table[{Random[Real, {0.5, 5}],
Random[Real, {0.5, 5}]}, {10}]
{{k12 -> 4.836012434059756, x3 -> 1.161391367338516}, {k12 ->
2=2E697618802043507, x3 -> 3.948782942915272},
  {k12 -> 0.5879722475631715, x3 -> 1.4585935450614775}, {k12 ->
3=2E8264510032098684, x3 -> 1.1377544670274133},
  {k12 -> 0.9423075629493529, x3 -> 4.504649319447406}, {k12 ->
3=2E3469082016444465, x3 -> 2.9410714608957313},
  {k12 -> 3.1668742774509693, x3 -> 3.87255878871341}, {k12 ->
0=2E5213112463283281, x3 -> 1.8340283782021},
  {k12 -> 4.096904065330009, x3 -> 3.497403976629866}, {k12 ->
1=2E3543039427927996, x3 -> 3.826924805414662}}

Now let examine what math52, math5 and math4 give if applied to lst:

math52/. lst
{-17.627352220954013 + 0.0036566566980830674*I, -32662.497107944495 +
0=2E009984691140152617*I,
  2.594045382153368 + 3.151577613312229*I, -7.310810775791415 +
0=2E009716274939033773*I,
  -424.8182654466529 + 0.6667159424606566*I, -6978.9252658160885 +
0=2E005708751570477014*I,
  -118049.32688610257 + 0.005294998992999599*I, 1.2662681012532613 +
4=2E8074082644294185*I,
  -508552.26461332565 + 0.0020739497895243112*I, -471.31406786105373 +
0=2E18242159506445238*I}

math5/.lst

{0.00033424880976802655 - 0.003656656698083018*I,
0=2E00002204703944537411 - 0.009984691140181724*I,
  6.08903032843357 - 3.15157761331223*I, 0.0019303959098684953 -
0=2E009716274939034273*I,
  0.14118541053919034 - 0.66671594246068*I, 0.000024120240009942828 -
0=2E005708751570954801*I,
  3.0957487047021305*^-6 - 0.005294998998484296*I, 8.336634749717106 -
4=2E8074082644294185*I,
  2.1016174687768414*^-7 - 0.0020739497901484326*I,
0=2E021942603754409404 - 0.18242159506442307*I}

math4 /. lst
{0.0010290521899598626, 0.000037862647332005684, 2.279213916290803,
0=2E0047024335618241495, 0.08469594552759933,
  0.00005139319883990879, 6.241123267010999*^-6, 2.7667377224052694,
5=2E4674898692854*^-7, 0.018918400987402058}

Totally disagreement!!! A quite frustating situation, doesn't it?

Anyway here is the respective results of numerical integration

Block[{Message}, NIntegrate[Evaluate[f[k3, x3, k12] /. lst], {k3, -
Infinity, Infinity}, PrecisionGoal -> 6, MaxRecursion -> 12,
   SingularityDepth -> 1000]]
{0.00033424875490353437 + 3.1397800895452747*^-10*I,
0=2E00002204423954521238 + 4.170455622275269*^-9*I,
  6.0890306296175325 + 0.*I, 0.0019303955699897565 -
1=2E5964951099584796*^-10*I, 0.1411854390564662 -
1=2E7694179454963432*^-16*I,
  0.00002412891996607358 - 2.0422526252599225*^-9*I,
3=2E0956500991739846*^-6 + 2.4734685548805655*^-19*I,
  8.336634912627977 + 2.2659825404947043*^-17*I, 2.036360125508489*^-7
- 1.5526604938345225*^-9*I,
  0.021942610153701376 + 1.47764087388718*^-9*I}

Note also that

Chop[Block[{Message}, Integrate[f[k3, x3, k12] /. lst, {k3, -Infinity,
Infinity}]]]
{0.00033424880976003354, 0.000022047039154231812, 6.089030328433574,
0=2E001930395909868988, 0.14118541053917477,
  0.000024120245640529974, 3.095649730303139*^-6, 8.336634749717108,
2=2E0962934124184866*^-7, 0.02194260375448867}


CONSEQUENTLY, we figure out that:

1) No version of Mathematica give the correct (symbolic) answer for
the general case;
that is no arithmetic values for the parameters.
2) Version 5.2 gives right results substituting arithmetic values for
parameters.
3) The returning result of 5.2 for the general case is totally wrong.
However the returning result of 5 for the general case is correct,
PROVIDING
you drop out the buggy imaginary part!

Further I try to get a simpler analytic result using FullSimplify and
FunctionExpand with failed
results. For example

(FunctionExpand[#1, k12 != 0 && Im[k12] == 0 && Im[x3] == 0] & )
[math5]
(Abs[k12]*MeijerG[{{1/2}, {}}, {{0, 1/2, 3/2}, {}}, (-
(1/4))*k12^2*x3^2])/(k12^4*Sqrt[Pi])

So, in your position I will use as a symbolic RESULT for your
integration the following function

ff=Re[(Abs[k12]*MeijerG[{{1/2}, {}}, {{0, 1/2, 3/2}, {}}, (-
(1/4))*k12^2*x3^2])/(k12^4*Sqrt[Pi])]

(*check*)

ff /. lst
{0.00033424880976802655, 0.00002204703944537411, 6.08903032843357,
0=2E0019303959098684953, 0.14118541053919034,
  0.000024120240009942828, 3.0957487047021305*^-6, 8.336634749717106,
2=2E1016174687768414*^-7, 0.021942603754409404}


There are many helpful and more experienced than me persons in this
forum like Paul Abbott, Daniel Lichtblau, Andrzej Kozlowski and many
others which may provide with more insight!


Kind Regards
Dimitris


=CF/=C7 vromero =DD=E3=F1=E1=F8=E5:
> Hello,
> I tried the following integral:
>
> Integrate[Exp[-I k3 x3]/(k12^2 + k3^2)^2,{k3,-Infinity,Infinity},
> Assumptions -> k12!=0 && Im[k12] == 0 && Im[k3] == 0 && Im[x3] =
== 0]
>
> The result in version 5 is
>
> (Abs[k12]*MeijerG[{{1/2}, {}}, {{0, 1/2, 3/2}, {}},
> (-(1/4))*k12^2*x3^2])/(k12^4*Sqrt[Pi])
>
> But in version 4 is
>
> (1 + Abs[k12] Abs[x3]) / Abs[k12]^2 * Exp[- Abs[k12] Abs[x3]]
>
> How can I get last result from version 5?
>
> Thanks in advance...
>
> Victor Romero



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