```Date: Jun 10, 2013 4:02 AM
Author: Andrzej Kozlowski
Subject: Re: Calculating a simple integral

On 9 Jun 2013, at 21:45, Dmitry Smirnov <dsmirnov90@gmail.com> wrote:> Thanks a lot for all your advises!>> Unfortunately none of them works. Because:> 1) I have to calculate integral symbolicaly, not numerically.> 2) It can't be divided into two parts because each of them diverges. However, the whole expression is always finite. For example at the point kz=2*Pi both numerator and denominator turn into zero.The function has removable singularities at 2Pi and -2Pi. It has a pole of order 2 at I and at -I and it is easy to compute the residues there. Unfortunately this is not enough to compute the integral since the modulus of Cos[z]-1 grows as z becomes large, so the standard trick of residue calculus (integrating over a half-circle) can't be used.AK>> Finally I have taken this one and some similar integrals in other system and saved the results into file.>> Thanks again for the efforts!>>> 2013/6/9 Andrzej Kozlowski <akozlowski@gmail.com>>> 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>>>
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