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Re: Calculating a simple integral


I believe that oscillating function as (Cos[z]-1) allows one to calculate
this integral by the residues. Anyway you can check numerically that the
correct answer is following:

-1/32/((kr^2 + 4*Pi^2)^3*kr^5*Pi^3)*(192*Pi^6*Sinh[kr] +
   16*Pi^4*Sinh[kr]*kr^3 + 64*Pi^6*Sinh[kr]*kr - 192*Pi^6*Cosh[kr] -
   112*Pi^4*Cosh[kr]*kr^2 + 112*Pi^4*Sinh[kr]*kr^2 -
   16*Pi^4*Cosh[kr]*kr^3 - 64*Pi^6*Cosh[kr]*kr + 192*Pi^6 - 3*kr^7 -
   28*kr^5*Pi^2 - 96*kr^3*Pi^4 - 128*kr*Pi^6 + 112*kr^2*Pi^4)




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