Re: Calculating a simple integral
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- Subject: [mg131146] Re: Calculating a simple integral
- From: "Dr. Wolfgang Hintze" <weh at snafu.de>
- Date: Fri, 14 Jun 2013 05:01:15 -0400 (EDT)
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On 13 Jun., 08:33, "Dr. Wolfgang Hintze" <w... at snafu.de> wrote: > On 11 Jun., 08:23, Andrzej Kozlowski <akozlow... at gmail.com> wrote: > > > > > > > No, it's similar to: > > > Integrate[(1 - > > Cos[x])/(x^2*(x^2 - 4*Pi^2)^2), {x, -Infinity, Infinity}] > > > 3/(32*Pi^3) > > > On 10 Jun 2013, at 10:11, djmpark <djmp... at comcast.net> wrote: > > > > Doesn't this have a singularity at 2 Pi that produces non-convergence? It's > > > similar to: > > > > Integrate[1/x^2, {x, \[Epsilon], \[Infinity]}, > > > Assumptions -> \[Epsilon] > 0] > > > > 1/\[Epsilon] > > > > That diverges as epsilon -> 0. > > > > Are you sure you copied the integral correctly? > > > > David Park > > > djmp... at comcast.net > > >http://home.comcast.net/~djmpark/index.html > > > > From: dsmirno... at gmail.com [mailto:dsmirno... 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. :) > > > > Thanks for any suggestions. > > Sorry, but I made indeed a calculation error! > Correcting it the partial fraction decomposition leads to Dmitry's > result. > Furthermore, calculating first the indefinite integral and then taking > limits leads to a false result. > Direct calculation of the integral leads to MeierG functions which are > useless because we cannot enter any numerical value. > So, rather than provding the correct result Mathematica comes up with > different false result depending on the method used, and we cannot tel > which one is correct without "research" work. > Summarizing, I need to restate my criticism of Mathematica with > respect to integration (I'm using version 8). > > Regards, > Wolfgang Sorry once more. Here comes a final, compact and correct derivation Complete compact calculation of the integral. The main idea is decomposition into partial fractions. The integrand is In[1]:= in = -((-1 + Cos[z])/(z^2*(-4*Pi^2 + z^2)^2*(r^2 + z^2)^2)); Decomposing into partial fractions gives In[2]:= Apart[1/(z^2*(-4*Pi^2 + z^2)^2*(r^2 + z^2)^2)] Out[2]= 1/(16*Pi^4*r^4*z^2) + 1/(64*Pi^4*(4*Pi^2 + r^2)^2*(-2*Pi + z)^2) + (-28*Pi^2 - 3*r^2)/(128* Pi^5*(4*Pi^2 + r^2)^3*(-2*Pi + z)) + 1/(64*Pi^4*(4*Pi^2 + r^2)^2*(2*Pi + z)^2) + (28*Pi^2 + 3*r^2)/(128* Pi^5*(4*Pi^2 + r^2)^3*(2*Pi + z)) - 1/(r^2*(4*Pi^2 + r^2)^2*(r^2 + z^2)^2) + (-4*Pi^2 - 3*r^2)/(r^4*(4*Pi^2 + r^2)^3*(r^2 + z^2)) Frequently, treating a list of terms is better than the sum in order to see what's going on in detail In[3]:= ls = List @@ % Out[3]= {1/(16*Pi^4*r^4*z^2), 1/(64*Pi^4*(4*Pi^2 + r^2)^2*(-2*Pi + z)^2), (-28*Pi^2 - 3*r^2)/(128* Pi^5*(4*Pi^2 + r^2)^3*(-2*Pi + z)), 1/(64*Pi^4*(4*Pi^2 + r^2)^2*(2*Pi + z)^2), (28*Pi^2 + 3*r^2)/(128* Pi^5*(4*Pi^2 + r^2)^3*(2*Pi + z)), -(1/(r^2*(4*Pi^2 + r^2)^2*(r^2 + z^2)^2)), (-4*Pi^2 - 3*r^2)/(r^4*(4*Pi^2 + r^2)^3*(r^2 + z^2))} Now integrating the list gives In[4]:= t = Timing[Table[ Integrate[(1 - Cos[z])*ls[[k]], {z, -Infinity, Infinity}, Assumptions -> r > 0], {k, 1, Length[ls]}]]; t[[1]] is = t[[2]] During evaluation of In[4]:= Integrate::idiv:Integral of (-1+Cos[z])/ (2 \[Pi]-z) does not converge on {-\[Infinity],\[Infinity]}. >> During evaluation of In[4]:= Integrate::idiv:Integral of (-1+Cos[z])/ (2 \[Pi]+z) does not converge on {-\[Infinity],\[Infinity]}. >> Out[5]= 4.867 Out[6]= {1/(16*Pi^3*r^4), 1/(64*Pi^3*(4*Pi^2 + r^2)^2), Integrate[((-28*Pi^2 - 3*r^2)*(1 - Cos[z]))/(128* Pi^5*(4*Pi^2 + r^2)^3*(-2*Pi + z)), {z, -Infinity, Infinity}, Assumptions -> r > 0], 1/(64*Pi^3*(4*Pi^2 + r^2)^2), Integrate[((28*Pi^2 + 3*r^2)*(1 - Cos[z]))/(128* Pi^5*(4*Pi^2 + r^2)^3*(2*Pi + z)), {z, -Infinity, Infinity}, Assumptions -> r > 0], (Pi*(1 - E^r + r))/(E^ r*(2*r^5*(4*Pi^2 + r^2)^2)), ((-1 + E^(-r))* Pi*(4*Pi^2 + 3*r^2))/(r^5*(4*Pi^2 + r^2)^3)} Oops, there appear divergences in the third and fifth term. Ok, let's calculate the sum of both instead In[65]:= Integrate[(1 - Cos[z])*(ls[[3]] + ls[[5]]), {z, -Infinity, Infinity}] Out[65]= 0 Fine, the terms cancel each other. Now, collecting the results In[72]:= (is[[#1]] & ) /@ {1, 2, 4, 6, 7} Out[72]= {1/(16*Pi^3*r^4), 1/(64*Pi^3*(4*Pi^2 + r^2)^2), 1/(64*Pi^3*(4*Pi^2 + r^2)^2), (Pi*(1 - E^r + r))/(E^ r*(2*r^5*(4*Pi^2 + r^2)^2)), ((-1 + E^(-r))* Pi*(4*Pi^2 + 3*r^2))/(r^5*(4*Pi^2 + r^2)^3)} Summing up and simplifying gives my (wh) final result In[73]:= fwh[r_] = Simplify[Plus @@ %] Out[73]= (28*E^r*Pi^2*r^5 + 3*E^r*r^7 + 64*Pi^6*(3 + r + E^r*(-3 + 2*r)) + 16*Pi^4*r^2*(7 + r + E^r*(-7 + 6*r)))/(E^ r*(32*Pi^3*r^5*(4*Pi^2 + r^2)^3)) In[74]:= fwh[1.] Out[74]= 0.0004911297292720861 Dmitry's result is slightly more lengthy In[71]:= fs[ r_] := (-192*Pi^6 + 128*Pi^6*r - 112*Pi^4*r^2 + 96*Pi^4*r^3 + 28*Pi^2*r^5 + 3*r^7 + 16*Pi^4*(4*Pi^2*(3 + r) + r^2*(7 + r))*Cosh[r] - 16*Pi^4*(4*Pi^2*(3 + r) + r^2*(7 + r))*Sinh[r])/(32*Pi^3* r^5*(4*Pi^2 + r^2)^3) But both are identical In[76]:= FullSimplify[fwh[r] == fs[r]] Out[76]= True I can also comfirm that the result is in agreement with numerical calculations using NIntegrate (as was stated earlier by Dmitry). Regards, Wolfgang