Consider an improper integral with a pole on the integration contour at say z=1,
I=∫∞−∞dz e−z2z−1+iϵ, ϵ>0.
Let f(z)=e−z2z−1+iϵ
then
∑residues inside Γ=0=∮Γf(z)=I+(∫Γϵ+∫Γ∞)f(z),
where the total contour is Γ≡(−R,R)+Γϵ+Γ∞ with R→∞.
Thus
I=−(∫Γϵ+∫Γ∞)f(z).
The contour Γϵ is a semicircle centered about z=1 of radius ϵ. Its contribution is given by
∫Γϵdz f(z)=i(θ2−θ1) Res(f;z=1)=−iπe.
Evaluating (1) in Mathematica and taking the ϵ→0 limit gives
I=e(ϵ+i)2(−πerfi(1−iϵ)+log(−1+iϵ)+log(iϵ+i)−2iπ)⟶−π(erfi(1)+i)e (as ϵ→0).
Thus apparently,
∫Γ∞dzf(z)=2πie+erfi(1)e.
Can anyone derive this contribution from the semicircle at infinity? I.e. is (2) correct and how about generalizations of (1) to integrals of the form
I=∫∞−∞dz zne−z2(z−a+iϵ)(z−b−iϵ), ϵ>0,a,b∈R,n∈N.
Note, erfi is defined as
erfi(z)≡erf(iz)/i with the familiar error function.
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