Re: Calculation of a not so simple integral

*To*: mathgroup at smc.vnet.net*Subject*: [mg131239] Re: Calculation of a not so simple integral*From*: "Dr. Wolfgang Hintze" <weh at snafu.de>*Date*: Thu, 20 Jun 2013 04:45:52 -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*: <kpmns8$h52$1@smc.vnet.net>

On 17 Jun., 12:19, Alexei Boulbitch <Alexei.Boulbi... at iee.lu> wrote: > (*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 > > Hi, Roland, > > I checked your expression, it has a pole on the x axis. Check this > > Sin[x/2]^2 / x^2 / (x^2-4*Pi^2 )^2 / (x^2 + a^2)^2//StandardForm > > Could it be the case, that you have just written it down in a wrong way? > > Alexei BOULBITCH, Dr., habil. > IEE S.A. > ZAE Weiergewan, > 11, rue Edmond Reuter, > L-5326 Contern, LUXEMBOURG > > Office phone : +352-2454-2566 > Office fax: +352-2454-3566 > mobile phone: +49 151 52 40 66 44 > > e-mail: alexei.boulbi... at iee.lu Let me explain what I meant with my short remark. The integral can be calculated completely in Mathematica with the result fwh[a_] = (3*a^7*E^a + 28*a^5*E^a*Pi^2 - 112*a^2*(-1 + E^a)*Pi^4 + 16*a^3*(1 + 6*E^a)*Pi^4 - 192*(-1 + E^a)*Pi^6 + 64*a*(1 + 2*E^a)*Pi^6)/(E^a*(64*a^5*Pi^3*(a^2 + 4*Pi^2)^3)) Numerical example In[62]:= fwh[0.1] Out[62]= 0.25188490534423524 As described in my contribution the aforementioned thread, the main idea is decomposition into partial fractions. The point is, however, that other methods lead to wrong results Method 1: Calculate the integral directly using Integrate -> fi[a] Without Assumptions on a it gives a complicated expression involving functions MeierG. With the Assumptions->a>0 the result contains only elementary functions and "looks good". But it is wrong, as the numerical value at a=0.1 is -141.075. Method 2: Calculate the indefinite integral and taking limits -> fl[a] Leads to another well-looking results containing only elementary functions. But it is also wrong giving fl[0.1] = - 161.232 (which is diferent from the wrong result fi[a]). Method 3: Make decomposition into partial fractions and integrate term by term This is the method I proposed and decribed in detail earlier (I don't repat it here to save space). This leads to the correct result above. Why do I consider this solution correct? Well, mainly because it is in agreement with the numerical integration using NIntegrate . Actually, defining In[27]:= fn[aa_] := NIntegrate[in /. a -> aa, {x, -Infinity, Infinity}] we find In[28]:= fn[0.1] Out[28]= 0.2518853035259861 in good agreement with fwh[0.1]. The Plot shows the agreement over a broad range of values: In[32]:= Plot[{fwh[a]/fn[a]}, {a, 0.1, 5}, PlotRange -> {1 - 0.1, 1 + 0.1}] What do we learn from this study? First of all, that Mathematica's Integrate-Procedure is very far from being perfect (as was pointed out already by many users on many occasions). It can have the very unpleasant feature to give "good-looking" results which are competely wrong. What I strongly suggest is to perform an "obligatory" numerical check of the result using NIntegrate. Best regards, Wolfgang