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Topic: Calculating a simple integral
Replies: 5   Last Post: Jun 10, 2013 4:05 AM

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 Andrzej Kozlowski Posts: 226 Registered: 1/29/05
Re: Calculating a simple integral
Posted: Jun 10, 2013 4:02 AM

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
>
> 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
>
>
>

Date Subject Author
6/10/13 Andrzej Kozlowski
6/10/13 Andrzej Kozlowski
6/10/13 dsmirnov90@gmail.com
6/10/13 dsmirnov90@gmail.com