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. >
- References:
- Calculating a simple integral
- From: dsmirnov90@gmail.com
- Calculating a simple integral