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)
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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