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Re: Calculating a simple integral

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  • Subject: [mg131084] Re: Calculating a simple integral
  • From: Roland Franzius <roland.franzius at>
  • Date: Mon, 10 Jun 2013 04:12:03 -0400 (EDT)
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  • References: <kp1e82$80h$>

Am 09.06.2013 10:26, schrieb dsmirnov90 at
> 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. :)

Did you check the numerics?

The expression is a square of

f[x_,a_] = Sin[x/2]((x (x - 2 Pi)(x + 2 Pi)) /(x^2+a^2)

This function looks like a multiple Sinc function with a canceling of 
zeroes of the numerator function sin(x/2) at x = 0 and +- 2pi by the 
denominator polynomial.

Mathematica sticks in Nirwana with the definite integral but 
accomplishes to find the indefinite integral

F[x_,a_] = Integrate[f[x,a]^2,x]

But, of course, with doubious imaginary parts and a definitely wrong
Limit[ F[x,a]-F[-x,a],x->oo]

despite the fact, that the derivative of F wrt to x is correct.

Another hint here, that the new developments in finding definite 
integrals are probably full of errors implemented.

When setting
$Assumptions=x \in Reals && a>0
I got an errormessage by NIntegrate, that x > 1/4096  :-(

But the FourierTransform package is working

FL[a_]= Limit[ Sqrt[2 Pi] FourierTansform[f[x,a]^2,x,k ], k->0]

Compare numerically eg



NIntegrate[ f[x,0.05`20]^2,{x,-oo,oo},WorkingPrecision->20]

Hope ist helps.


Roland Franzius

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