Re: Calculating a simple integral
- To: mathgroup at smc.vnet.net
- Subject: [mg131084] Re: Calculating a simple integral
- From: Roland Franzius <roland.franzius at uos.de>
- Date: Mon, 10 Jun 2013 04:12:03 -0400 (EDT)
- Delivered-to: l-mathgroup@mail-archive0.wolfram.com
- Delivered-to: l-mathgroup@wolfram.com
- Delivered-to: mathgroup-outx@smc.vnet.net
- Delivered-to: mathgroup-newsendx@smc.vnet.net
- References: <kp1e82$80h$1@smc.vnet.net>
Am 09.06.2013 10:26, schrieb dsmirnov90 at gmail.com:
> 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
FL[0.05`20]
and
NIntegrate[ f[x,0.05`20]^2,{x,-oo,oo},WorkingPrecision->20]
Hope ist helps.
--
Roland Franzius