# Calculate the following integral with assumptions?

Posted 10 months 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 10 months 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 10 months ago
 Thanks. Is there a way to "regularize" the integral in Mathematica, including "blocking" the singular points?
Posted 10 months 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 10 months 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 10 months ago
 Thank you Giancula and Sam.
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