
Re: Calculating a simple integral
Posted:
Jun 10, 2013 4:05 AM


On 9 Jun 2013, at 10:32, dsmirnov90@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. >
Which version of Mathematica are you using?
Mathematica does quite quickly calculate answers to this integral for numerical values of kr. For example, for kr=1 I get:
Integrate[(Cos[x]  1)/(x^2*(x^2  4*Pi^2)^2*(x^2 + 1)^2), {x, Infinity, Infinity}]
(3*E  28*E*Pi^2 + 16*(8 + E)*Pi^4 + 64*(4 + E)*Pi^6)/(32* E*(Pi + 4*Pi^3)^3)
Numerically this gives:
N[%]
0.00049113
which agrees with the value returned by NIntegrate, so it should be correct. The general case takes a lot longer but there is still an answer:
Integrate[(Cos[x]  1)/(x^2*(x^2  4*Pi^2)^2*(x^2 + a^2)^2), {x, Infinity, Infinity}, Assumptions > a > 0]
(1/(128*a^5*Pi^4*(a^2 + 4*Pi^2)^3))*(11*a^7*Pi  92*a^5*Pi^3 + 448*a^2*Pi^5 + 768*Pi^7 + 2*I*a^7*CosIntegral[2*Pi] + 40*I*a^5*Pi^2*CosIntegral[2*Pi]  2*I*a^7*ExpIntegralEi[2*I*Pi]  40*I*a^5*Pi^2*ExpIntegralEi[2*I*Pi] + 16*a*Pi^(7/2)*(5*a^2 + 4*Pi^2)*MeijerG[{{1/2, 1}, {}}, {{(1/2), 1/2, 1}, {0}}, ((I*a)/2), 1/2] + 16*a*Pi^(7/2)*(5*a^2 + 4*Pi^2)* MeijerG[{{1/2, 1}, {}}, {{(1/2), 1/2, 1}, {0}}, (I*a)/2, 1/2] + 32*a^3*Pi^(7/2)* MeijerG[{{1/2, 1}, {}}, {{(1/2), 1, 3/2}, {0}}, ((I*a)/2), 1/2] + 128*a*Pi^(11/2)*MeijerG[{{1/2, 1}, {}}, {{(1/2), 1, 3/2}, {0}}, ((I*a)/2), 1/2] + 32*a^3*Pi^(7/2)*MeijerG[{{1/2, 1}, {}}, {{(1/2), 1, 3/2}, {0}}, (I*a)/2, 1/2] + 128*a*Pi^(11/2)* MeijerG[{{1/2, 1}, {}}, {{(1/2), 1, 3/2}, {0}}, (I*a)/2, 1/2] + 2*a^7*SinIntegral[2*Pi] + 40*a^5*Pi^2*SinIntegral[2*Pi])
I have no idea if this is correct or not and don't see how this could be useful. What sort of answer does the other system give? And why do you think this is a "simple" integral? (There might be a way to evaluate it using the calculus of residues but probably it needs some clever trick since the obvious approaches don't seem to work.)
Andrzej Kozlowski

