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Re: Calculating a simple integral

  • To: mathgroup at smc.vnet.net
  • Subject: [mg131075] Re: Calculating a simple integral
  • From: "Nasser M. Abbasi" <nma at 12000.org>
  • Date: Mon, 10 Jun 2013 04:09:03 -0400 (EDT)
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  • Reply-to: nma at 12000.org

On 6/9/2013 3:26 AM, dsmirnov90 at gmail.com wrote:
> If there is a way to calculate with Mathematica the following integral:
>
> in = -((-1 + Cos[kz])/(kz^2 (kr^2 + kz^2)^2 (kz^2 - 4 \[Pi]^2)^2))
> Integrate[in, {kz, -Infinity, Infinity}, Assumptions -> kr > 0]
>
> Another system calculates the same integral instantly. :)
>
> Thanks for any suggestions.
>

You did not say which version you used.

I do get result from 9.01, but it took about 2 minutes

--------------------------
ClearAll["`*"];
in = -((-1 + Cos[kz])/(kz^2 (kr^2 + kz^2)^2 (kz^2 - 4 \[Pi]^2)^2));
Assuming[kr > 0, Integrate[in, {kz, -Infinity, Infinity}]]

-(1/(32 kr^5 \[Pi]^3 (kr^2+4 \[Pi]^2)^3))(-3 kr^7-
28 kr^5 \[Pi]^2+112 kr^2 \[Pi]^4+192 \[Pi]^6+4 kr \[Pi]^(5/2)
  (5 kr^2+4 \[Pi]^2) MeijerG[{{1/2,1},{}},{{-(1/2),1/2,1},{0}},
-((I kr)/2),1/2]+4 kr \[Pi]^(5/2) (5 kr^2+4 \[Pi]^2)
MeijerG[{{1/2,1},{}},{{-(1/2),1/2,1},{0}},(I kr)/2,1/2]+
8 kr^3 \[Pi]^(5/2) MeijerG[{{1/2,1},{}},{{-(1/2),1,3/2},{0}},
-((I kr)/2),1/2]+32 kr \[Pi]^(9/2) MeijerG[{{1/2,1},{}},
{{-(1/2),1,3/2},{0}},-((I kr)/2),1/2]+8 kr^3 \[Pi]^(5/2)
  MeijerG[{{1/2,1},{}},{{-(1/2),1,3/2},{0}},(I kr)/2,1/2]+
32 kr \[Pi]^(9/2) MeijerG[{{1/2,1},{}},{{-(1/2),1,3/2},{0}},(I kr)/2,1/2])
--------------------------------------

But something does not seem to be right. Can't evaluate any of
the MeijerG functions in the result for any kr value. For
example

N[MeijerG[{{1/2, 1}, {}}, {{-(1/2), 1/2, 1}, {0}}, -((I kr)/2), 1/
    2] /. kr -> 0]

gives error

MeijerG::hdiv: MeijerG[{{0.5,1.},{}},{{-0.5,0.5,1.},{0.}},0.,0.5]
does not exist. Arguments are not consistent. >>


--Nasser







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