Re: Calculating a simple integral

• To: mathgroup at smc.vnet.net
• Subject: [mg131098] Re: Calculating a simple integral
• From: "Brambilla Roberto Luigi (RSE)" <Roberto.Brambilla at rse-web.it>
• Date: Tue, 11 Jun 2013 02:31:31 -0400 (EDT)
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```I have for semplicity rewritten your integral as

f[x_,a_]:= Sqrt[2]Sin[x/2]/(x(x^2+a^2)(x+2Pi)(x-2Pi))
(*this function is NOT singular in 0,2Pi,-2Pi*)

The numerical integration is readily obtained

kr=0.09;

int1=2 NIntegrate[f[kz,kr]^2,{kz,0,Infinity}]
(*the integrand is even*)

0.691084

Also the indefinite integral is obtained in a couple of minutes (ver.8)

int2=Integrate[f[kz,kr]^2,{kz,0,Infinity}]=....

The result is a long expression that can be rewritten as

foo1[x_,y_]:=Im[MejerG[{{1,1,3/2},{}},{{1,1,3/2},{0,1/2}},x,y]]

foo2[x_,y_]:=Im[MejerG[{{1,1,3/2},{}},{{1,3/2,2},{0,1/2}},x,y]]

A=3 kr^7+28 kr^5 Pi2+

+16 Pi^5/2(5 kr^2+4Pi^2)*foo1[I kr/2,1/2]+

+32Pi^5/2(kr^2+4Pi^2)*foo2[I kr/2,1/2]

249.281

B=32 kr^5 Pi^3 (kr^2+4PI^2)

360.71

int2=A/B

0.691084

Bye Roberto

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