Re: Calculation of a not so simple integral
- To: mathgroup at smc.vnet.net
- Subject: [mg131253] Re: Calculation of a not so simple integral
- From: Roland Franzius <roland.franzius at uos.de>
- Date: Fri, 21 Jun 2013 06:00:50 -0400 (EDT)
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Am 15.06.2013 10:15, schrieb Roland Franzius: > (*Partial Fractions decomposition, Fourier Integrals > > The problem diagnostics: > > Calculate > > Integrate[ Sin[x/2]^2 / x^2 / (x^2-4*Pi^2 )^2 / (x^2 + a^2)^2 , > {x,-oo,oo}, Assumptions-> a>0] > > The integrand is nonnegative, has no poles on the real line and decays > rapidly ~ x^-10 as x->+-oo > > Mathematica version 6+8 are failing to give a result in a reasonable > time of calculation. > Time constraint does not seem to work > > The indefinite integral is calculated readily with an obscure complex > result, the primitive function giving completely useless limits for > x->0,+-oo > > The FourierTransform with Limit k->0 seems to give a correct result > compared to numerical integration for values of > a -> 1/20 > > *) Remains to be noted that suddenly today in the morning my Mathematica 8 (local install with license server) seems to have been updated over the internet with respect to this integral. While yesterday the result was MeiersG, when I came back after 2 hours, suddenly we have amuch faster and slightly better result In[14]:= ires=Timing[ Integrate[Evaluate[Sin[x/2]^2 f[x,2\[Pi],a]],{x,-\[Infinity],\[Infinity]},Assumptions->0<a<1]] Out[14]= {64.865, ( 3 a^7 + 28 a^5 \[Pi]^2 + 16 a^2 (-7 + 6 a) \[Pi]^4 + 64 (-3 + 2 a) \[Pi]^6)/(64 a^5 \[Pi]^3 (a^2 + 4 \[Pi]^2)^3)} still wrong, the exponential terms are missing. But at least for a>1 we get the algebraically correct result. In[31]:= iresg=Timing[ Integrate[Evaluate[Sin[x/2]^2 f[x,2\[Pi],a]],{x,-\[Infinity],\[Infinity]},Assumptions->1<a<\[Infinity]]] Out[31]= {63.477, ( E^-a (16 a^2 (7 + a) \[Pi]^4 + 64 (3 + a) \[Pi]^6 + E^a (3 a^7 + 28 a^5 \[Pi]^2 + 16 a^2 (-7 + 6 a) \[Pi]^4 + 64 (-3 + 2 a) \[Pi]^6)))/(64 a^5 (a^2 \[Pi] + 4 \[Pi]^3)^3)} With Assumptions -> a>0, Integrate produces both results as a conditional expression. It Contains the difference of two imaginary Log's. So I conclude there seem to exists problems with the partial fractions decomposition engine and in the simplification of complex Log's and Atan's. Of course this is a nontrivial matter because it involves complex contour integrals which cannot be desribed by the endpoints only. -- Roland Franzius