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Re: Calculating a simple integral

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  • Subject: [mg131077] Re: Calculating a simple integral
  • From: Andrzej Kozlowski <akozlowski at gmail.com>
  • Date: Mon, 10 Jun 2013 04:09:43 -0400 (EDT)
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  • References: <20130609083209.86C8769D8@smc.vnet.net> <8984F0D9-AD1E-4ECB-914A-1E3E88498517@mimuw.edu.pl> <CAOg3Lcf+dYe0jq40b8UUf3gucO+6rZxvsm1PfwMSzw1VDpejnw@mail.gmail.com>

On 9 Jun 2013, at 21:45, Dmitry Smirnov <dsmirnov90 at 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 at gmail.com>
>
> On 9 Jun 2013, at 10:32, 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.
> >
>
>
>
> 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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