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Re: Calculating a simple integral

  • To: mathgroup at smc.vnet.net
  • Subject: [mg131076] Re: Calculating a simple integral
  • From: Peter Klamser <klamser at googlemail.com>
  • Date: Mon, 10 Jun 2013 04:09:23 -0400 (EDT)
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  • References: <20130609083209.86C8769D8@smc.vnet.net>

in=-((-1+Cos[kz])/(kz^2 (kr^2+kz^2)^2 (kz^2-4 \[Pi]^2)^2))//ExpandAll

bring two terms in a sum:

t1=1/(kr^4 kz^6+2 kr^2 kz^8+kz^10-8 kr^4 kz^4 \[Pi]^2-16 kr^2 kz^6
\[Pi]^2-8 kz^8 \[Pi]^2+16 kr^4 kz^2 \[Pi]^4+32 kr^2 kz^4 \[Pi]^4+16
kz^6 \[Pi]^4)

t2=Cos[kz]/(kr^4 kz^6+2 kr^2 kz^8+kz^10-8 kr^4 kz^4 \[Pi]^2-16 kr^2
kz^6 \[Pi]^2-8 kz^8 \[Pi]^2+16 kr^4 kz^2 \[Pi]^4+32 kr^2 kz^4
\[Pi]^4+16 kz^6 \[Pi]^4)

int$int1 = \[Integral]t1 \[DifferentialD]kz

brings you a solution that should be treated with Limit twice:

Limit[int$int1, kz -> \[Infinity]] // ComplexExpand

and

Limit[int$int1, kz -> -\[Infinity]] // ComplexExpand

The same procedure with t2.

Kind regards from Peter

2013/6/9  <dsmirnov90 at gmail.com>:
> 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.
>



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