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Topic: The Charlwood Fifty
Replies: 52   Last Post: Jun 24, 2013 10:24 PM

 Messages: [ Previous | Next ]
 Nasser Abbasi Posts: 6,677 Registered: 2/7/05
Re: The Charlwood Fifty
Posted: Jun 9, 2013 10:25 AM

On 6/9/2013 4:58 AM, Albert Rich wrote:

fyi;

This is another hard integral posted at the Mathematica newsgroup
last night.

I do not know if you like to add it to your test suite as Rubi4 and
Mathematica 9.01 have some hard time with it.

ClearAll["`*"];
in = -((-1 + Cos[z])/(z^2 (r^2 + z^2)^2 (z^2 - 4 \[Pi]^2)^2));
Assuming[r > 0, Int[in, {z, -Infinity, Infinity}]]

Not evaluated. Maple 17 does it fast:

restart;
integrand:=-((-1 + cos(z))/(z^2*(r^2 + z^2)^2 *(z^2 - 4*Pi^2)^2));
int(integrand, z=-infinity..infinity) assuming r>0;

-(1/32)*(64*Pi^6*sinh(r)*r-64*Pi^6*cosh(r)*r+16*Pi^4*sinh(r)*
r^3-16*Pi^4*cosh(r)*r^3+192*Pi^6*sinh(r)-192*Pi^6*cosh(r)-
128*Pi^6*r+112*Pi^4*sinh(r)*r^2-112*Pi^4*cosh(r)*r^2-96*Pi^4*
r^3-28*Pi^2*r^5-3*r^7+192*Pi^6+112*Pi^4*r^2)/(r^5*(4*Pi^2+r^2)^3*Pi^3)

Mathematica takes about 2-3 minutes and gives this:

ClearAll["`*"];
in = -((-1 + Cos[z])/(z^2 (r^2 + z^2)^2 (z^2 - 4 Pi^2)^2));
Assuming[r > 0, Integrate[in, {z, -Infinity, Infinity}]]

-(1/(128 \[Pi]^4 r^5 (4 \[Pi]^2 + r^2)^3))(768 \[Pi]^7 +
448 \[Pi]^5 r^2 - 92 \[Pi]^3 r^5 - 11 \[Pi] r^7 +
40 I \[Pi]^2 r^5 CosIntegral[2 \[Pi]] +
2 I r^7 CosIntegral[2 \[Pi]] -
40 I \[Pi]^2 r^5 ExpIntegralEi[-2 I \[Pi]] -
2 I r^7 ExpIntegralEi[-2 I \[Pi]] +
16 \[Pi]^(7/2)
r (4 \[Pi]^2 +
5 r^2) MeijerG[{{1/2, 1}, {}}, {{-(1/2), 1/2, 1}, {0}}, -((I r)/
2), 1/2] +
16 \[Pi]^(7/2)
r (4 \[Pi]^2 +
5 r^2) MeijerG[{{1/2, 1}, {}}, {{-(1/2), 1/2, 1}, {0}}, (I r)/2,
1/2] + 128 \[Pi]^(11/2)
r MeijerG[{{1/2, 1}, {}}, {{-(1/2), 1, 3/2}, {0}}, -((I r)/2), 1/
2] + 32 \[Pi]^(7/2)
r^3 MeijerG[{{1/2, 1}, {}}, {{-(1/2), 1, 3/2}, {0}}, -((I r)/2),
1/2] + 128 \[Pi]^(11/2)
r MeijerG[{{1/2, 1}, {}}, {{-(1/2), 1, 3/2}, {0}}, (I r)/2, 1/
2] + 32 \[Pi]^(7/2)
r^3 MeijerG[{{1/2, 1}, {}}, {{-(1/2), 1, 3/2}, {0}}, (I r)/2, 1/
2] + 40 \[Pi]^2 r^5 SinIntegral[2 \[Pi]] +
2 r^7 SinIntegral[2 \[Pi]])

--Nasser