# Limit of an integral and hypergeometric function

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?

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)])

• This is amazing! – Subho Jun 26 '19 at 15:35