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Calculate the following integral with assumptions?

GROUPS:

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}]*)
POSTED BY: Arny Toynbee
Answer
9 days 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 BY: Tanel Telliskivi
Answer
9 days ago

Thanks. Is there a way to "regularize" the integral in Mathematica, including "blocking" the singular points?

POSTED BY: Arny Toynbee
Answer
6 days 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 BY: Sam Carrettie
Answer
6 days 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<a<b.

POSTED BY: Gianluca Gorni
Answer
6 days ago

Thank you Giancula and Sam.

POSTED BY: Arny Toynbee
Answer
5 days ago

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