# Calculate the following integral with assumptions?

Posted 1 year ago
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 Can this integral be approached in a different way? Mathematica 11.2 spits out the input. $Assumptions = p1 > 0 && Element[p1, Reals]$Assumptions = p4 > 0 && Element[p4, Reals] $Assumptions = p5 > 0 && Element[p5, Reals]$Assumptions = r1 > 0 && Element[r1, Reals] Integrate[Pi*(Sin[p1*r1]/(p1*r1))*((Log[(p1 - p3)^2/(p1 + p3)^2]*Log[(p1 - p4)^2/(p1 + p4)^2])/(p3*p4)), {p1, a, b}] (* Integrate[(Pi*Log[(p1 - p3)^2/(p1 + p3)^2]*Log[(p1 - p4)^2/(p1 + p4)^2]*Sin[p1*r1])/(p1*p3*p4*r1), {p1, a, b}]*) 
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Posted 1 year ago
 If You put all Your variables equal to 1, the the equation equals locally, p1=1, infinity. Numerically You can integrate 0..1 and 1..., but not 0...1.1. Therefore with other real and positive variables the integral is expected to approach infinity.
Posted 1 year ago
 Thanks. Is there a way to "regularize" the integral in Mathematica, including "blocking" the singular points?
Posted 1 year ago
 At least NIntegrate has Exclusions option, works like this: NIntegrate[1/Sqrt[Sin[x^2 + y]], {x, -2, 4}, {y, -2, 4}, Exclusions -> (Sin[x^2 + y] == 0)] (* 30.4933 - 29.9073 I *) 
Posted 1 year ago
 \$Assumptions is not cumulative. The way you give the assumptions, only the last one is kept. Also, with p1 > 0 && Element[p1, Reals], declaring p1 to be real is superfluous: p1>0 implies already that p1 is real. It would be more meaningful to assume `0
Posted 1 year ago
 Thank you Giancula and Sam.
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