Re: Calculating a simple integral
- To: mathgroup at smc.vnet.net
- Subject: [mg131083] Re: Calculating a simple integral
- From: Bob Hanlon <hanlonr357 at gmail.com>
- Date: Mon, 10 Jun 2013 04:11:43 -0400 (EDT)
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- References: <20130609083209.86C8769D8@smc.vnet.net>
This is far from instantly, $Version "9.0 for Mac OS X x86 (64-bit) (January 24, 2013)" in = ((1 - Cos[kz])/ (kz^2 (kr^2 + kz^2)^2 (kz^2 - 4 Pi^2)^2)); in2 = (in*(1 + Cos[kz]) // Simplify)/ (1 + Cos[kz]) // Simplify (2*Sin[kz/2]^2)/(kz^2*(kr^2 + kz^2)^2*(kz^2 - 4*Pi^2)^2) in (and in2) is an even function of kz in == (in /. kz -> -kz) == in2 == (in2 /. kz -> -kz) // Simplify True Clear[f] Timing[f[kr_?Positive] = 2*Integrate[in2, {kz, 0, Infinity}, Assumptions -> kr > 0]] { 285.13703099999997903069015592336654663086`8.47565353\ 646281, (1/(32*kr^5*Pi^3*(kr^2 + 4*Pi^2)^3))* (I*(-3*I*kr^7 - 28*I*kr^5*Pi^2 + 8*Pi^(5/2)*(5*kr^2 + 4*Pi^2)* MeijerG[{{1, 1, 3/2}, {}}, {{1, 1, 3/2}, {0, 1/2}}, -((I*kr)/2), 1/2] - 8*Pi^(5/2)*(5*kr^2 + 4*Pi^2)* MeijerG[{{1, 1, 3/2}, {}}, {{1, 1, 3/2}, {0, 1/2}}, (I*kr)/2, 1/2] + 16*kr^2*Pi^(5/2)* MeijerG[{{1, 1, 3/2}, {}}, {{1, 3/2, 2}, {0, 1/2}}, -((I*kr)/2), 1/2] + 64*Pi^(9/2)*MeijerG[{{1, 1, 3/2}, {}}, {{1, 3/2, 2}, {0, 1/2}}, -((I*kr)/2), 1/2] - 16*kr^2*Pi^(5/2)*MeijerG[{{1, 1, 3/2}, {}}, {{1, 3/2, 2}, {0, 1/2}}, (I*kr)/2, 1/2] - 64*Pi^(9/2)*MeijerG[{{1, 1, 3/2}, {}}, {{1, 3/2, 2}, {0, 1/2}}, (I*kr)/2, 1/2]))} LogPlot[f[kr] // Chop, {kr, .1, 3}, Frame -> True, Axes -> False, PlotRange -> All] Bob Hanlon On Sun, Jun 9, 2013 at 4:32 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. > >
- References:
- Calculating a simple integral
- From: dsmirnov90@gmail.com
- Calculating a simple integral