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Topic: Calculating a simple integral
Replies: 3   Last Post: Jun 10, 2013 4:05 AM

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Dr. Peter Klamser

Posts: 23
Registered: 6/11/11
Re: Calculating a simple integral
Posted: Jun 10, 2013 4:02 AM
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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


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

The same procedure with t2.

Kind regards from Peter

2013/6/9 <>:
> 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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