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.
>
>

```

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