
Re: Calculating a simple integral
Posted:
Jun 10, 2013 4:04 AM


Am 09.06.2013 10:26, schrieb dsmirnov90@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

