Re: Calculating a simple integral

*To*: mathgroup at smc.vnet.net*Subject*: [mg131124] Re: Calculating a simple integral*From*: "Dr. Wolfgang Hintze" <weh at snafu.de>*Date*: Thu, 13 Jun 2013 02:37:15 -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*: <7586978.94875.1370767085206.JavaMail.root@m06>

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. I have found a rather simple exact expression for the the integral in question in = -((-1 + Cos[z])/(z^2 (r^2 + z^2)^2 (z^2 - 4 \[Pi]^2)^2)) ix = Integrate[in, {z, -Infinity, Infinity}, Assumptions -> r > 0] = (32*Pi^4*(r^2 + 4*(r^4*(4 + r) + Pi^2*(1 + 4*r^3))) - E^r*(-3*r^7 + 12*r^9 + 32*Pi^2*r^5*(-1 + 5*r^2) + 128*Pi^6*(1 - r + 4*r^3) + 16*Pi^4*r^2*(2 + r*(-7 + 4*r*(8 + 9*r)))))/(E^ r*(64*r^5*(4*Pi^3 + Pi*r^2)^3)) which only contains elementary functions (no cumbersome and error- prone MeierG). Derivation ~~~~~~~~ My starting point was Andrzej's observation that a simplified version of the integrand is easily integrable: Indeed, let in2 = -((-1 + Cos[z])/(z^2 (z^2 - 4 \[Pi]^2)^2)); then Integrate[in2, {z, -Infinity, Infinity}, Assumptions -> r > 0] 3/(32*Pi^3) Now we decompose the term which completes the integrand, ir = 1/(r^2 + z^2)^2; into partial fractions: ir1 = 1/((I*r + z)^2*((-I)*r + z)^2); Simplify[ir == ir1] True Apart[ir1]; ip = List @@ % {-(1/(4*r^2*((-I)*r + z)^2)), -(I/(4* r^3*((-I)*r + z))), -(1/ (4*r^2*(I*r + z)^2)), I/(4*r^3*(I*r + z))} Integrating directly a term involving a quadratic denominator leads to the "unpleasant" MeierG-functions we would like to avoid. But we can get rid of the second power by diferentiating (like 1/x' = -1/x^2) as follows g = 1/((-I)*r + z); ip[[1]] == (I/(4*r^2))*D[g, r] True Hence we need to calculate just one integral (which takes 77 seconds on my PC) ig = Integrate[in2*g, {z, -Infinity, Infinity}, Assumptions -> r > 0] (I*(20*Pi^2*r^3 + 3*r^5 + 32*Pi^4*(-1 + E^(-r) + r)))/(32* Pi^3*(4*Pi^2*r + r^3)^2) Now the sums of the first and the third term of ip leads to ix0 = Simplify[(I/(4*r^2))*D[ig*4*r^2, r] - (I/(4*r^3))*ig] (- 16*Pi^4*r^2*(-2*(1 + 16*r^2 + 4*r^3) + E^r*(2 - 7*r + 32*r^2 + 36*r^3)))/(E^ r*(128*Pi^3*r^5*(4*Pi^2 + r^2)^3)) Because this quantity is real the final result is twice this expression (ix = 2 ix0). End of derivation. The idea is clear, but I'm not sure that no handling error has crept in. So please check my result. BTW the function ix(r) reminds me of some potential between two particles ... Regards, Wolfgang