Question #dce86

Aug 11, 2015

${\lim}_{e \psi \rightarrow {0}^{+}} {\int}_{-} {R}^{R} \frac{1}{w + i e \psi} \mathrm{dw} = - \pi i$

Explanation:

Contour integral:
${\oint}_{C} f \left(z\right) = {\int}_{-} {R}^{R} f \left(w\right) \mathrm{dw} + {\int}_{S} f \left(z\right) \mathrm{dz} = 2 \pi i \sum \text{Res} f \left(z\right)$
C anti-clockwise
Note that if C is clockwise, then there is a minus sign in front of $2 \pi i \sum \text{Res} f \left(z\right)$.

$f \left(z\right) = \frac{1}{z + i e \psi}$
Poles: $z = - i e \psi$
$\text{Res} f \left(- i e \psi\right) = {\lim}_{z \rightarrow - i e \psi} \left(z + i e \psi\right) f \left(z\right) = 1$

The path of C is a straight line from -R to R followed by a semi-circle from R to -R in the lower half-plane (since $e \psi > 0$). Now the direction of the contour is clockwise:
${\lim}_{e \psi \rightarrow {0}^{+}} {\int}_{-} {R}^{R} f \left(w\right) \mathrm{dw} = - 2 \pi i \text{Res} f \left(- i e \psi\right) - {\lim}_{e \psi \rightarrow {0}^{+}} {\int}_{S} f \left(z\right) \mathrm{dz}$

Take: $z = A {e}^{i \theta}$
${\lim}_{e \psi \rightarrow {0}^{+}} {\int}_{S} f \left(z\right) \mathrm{dz} = {\lim}_{e \psi \rightarrow {0}^{+}} {\int}_{- \pi}^{0} \frac{A i {e}^{i \theta}}{A {e}^{i \theta} + i e \psi} d \theta = {\int}_{- \pi}^{0} i d \theta = \pi i$

So: ${\lim}_{e \psi \rightarrow {0}^{+}} {\int}_{-} {R}^{R} \frac{1}{w + i e \psi} \mathrm{dw} = - \pi i$