Sunday, 5 May 2019

integration - Arc contribution in intiinftynftymathrmdzfracez2z1

Consider an improper integral with a pole on the integration contour at say z=1,



I=dz ez2z1+iϵ,     ϵ>0.
Let f(z)=ez2z1+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(θ2θ1) Res(f;z=1)=iπe.



Evaluating (1) in Mathematica and taking the ϵ0 limit gives
I=e(ϵ+i)2(πerfi(1iϵ)+log(1+iϵ)+log(iϵ+i)2iπ)π(erfi(1)+i)e   (as ϵ0).



Thus apparently,
Γdzf(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 znez2(za+iϵ)(zbiϵ),     ϵ>0,a,bR,nN.







Note, erfi is defined as
erfi(z)erf(iz)/i with the familiar error function.

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