Contour integral:
# oint_C f(z) = int_-R^R f(w) dw + int_S f(z) dz = 2pi i sum "Res" f(z) #
C anti-clockwise
Note that if C is clockwise, then there is a minus sign in front of # 2pi i sum"Res" f(z) #.
# f(z) = 1/(z+i epsi) #
Poles: # z = -i epsi #
# "Res" f(-i epsi) = lim_(z rarr -i epsi) (z+i epsi) f(z) = 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 #epsi > 0#). Now the direction of the contour is clockwise:
# lim_(epsi rarr 0^+) int_-R^R f(w) dw = - 2pi i "Res" f(-i epsi) - lim_(epsi rarr 0^+) int_S f(z) dz #
Take: #z = Ae^(itheta) #
# lim_(epsi rarr 0^+) int_S f(z) dz = lim_(epsi rarr 0^+)int_(-pi)^0 (Aie^(itheta))/(Ae^(itheta)+i epsi)d theta = int_(-pi)^0 id theta = pi i #
So: # lim_(epsi rarr 0^+) int_-R^R 1/(w+i epsi) dw = - pi i #
