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I want to evaluate the following integral:

Integrate[Sin[θ]^(D1 - Nc - 1)/(A Cos[θ] - I ϵ)^(N1 - Nc), {θ, 0, π}, 
Assumptions -> A > 0 && ϵ > 0]

and then take the limit $\epsilon\to0$:

Limit[%, ϵ -> 0]

However, the first integral takes eternity to produce an answer.

Workaround I tried:

I tried using a general variable $B$ instead of $i\epsilon$ to perform the integral. The following integral evaluates to a Gauss hypergeometric function under some conditions on the paramters:

Integrate[Sin[θ]^(D1 - Nc - 1)/(A Cos[θ] - B)^(N1 - Nc), {θ, 0, π}]

ConditionalExpression[ 2^(-1 + D1 - Nc) (-A - B)^(-N1 + Nc) Gamma[(D1 - Nc)/2]^2 Hypergeometric2F1Regularized[(D1 - Nc)/2, N1 - Nc, D1 - Nc, (2 A)/( A + B)], (Re[B/A] >= 1 || Re[B/A] <= -1 || B/A [NotElement] Reals) && Re[A + B] < 0 && Re[D1] > Re[Nc]]

Then if I take the limit $B\to 0$:

Limit[%, B -> 0]

it gives me

((-A)^(-N1 + Nc) Sqrt[[Pi]] Gamma[(D1 - Nc)/2] Hypergeometric2F1[(D1 - Nc)/2, N1 - Nc, D1 - Nc, 2])/Gamma[1/2 (1 + D1 - Nc)]

I think the above answer is the correct answer I was looking for (based on various checks with special values of $D,N_c,N_1$. But the condition $Re[A+B]<0$ means the answer may not work for the case $A>0,B=iϵ,ϵ>0$. Is there anyway to check this?

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You can use AsymptoticIntegrate:

AsymptoticIntegrate[
    Sin[θ]^(D1 - Nc - 1)/(A Cos[θ] - I ε)^(N1 - Nc),
    {θ, 0, π},
    {ε, 0, 1},
    Assumptions -> A>0 && ε>0
] /. ε->0

((-1)^-N1 ((-1)^N1 + (-1)^Nc) A^(-N1 + Nc) Gamma[(D1 - Nc)/2] Gamma[1/2 (1 - N1 + Nc)])/(2 Gamma[1/2 (1 + D1 - N1)])

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  • $\begingroup$ This is amazing! $\endgroup$ – Subho Jun 26 '19 at 15:35

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