# Integral of 1/z using different paths

by jmcelve
Tags: 1 or z, integral, paths
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,310 If, as I would read this, you are integrating on the straight line path from (0, -i) to (a, -i) to (a, i) to (0, i), you would not have to do anything about z= 0 because you never go anywhere near z= 0. On the path from (0, -i) to (a, -i) you can take z= x- i so that dz= dx and the integral is $\int_0^a (x- i)^{-1}dx$. On the path from (a, -i) to (a, i) you can take z= a+ iy so that dz= idy and the integral is $\int_{-1}^1 (a+iy)^{-1}dy$. Finally, on the path from (a, i) to (0, i), you can take z= x+ i so that dz= dx and the integral is $\int_1^0 (x+ i)^{-1}dx$.