Find the complex integral along vertices of square
Answers
If you don't know the Cauchy Formula or the Residue theorem, you can split it into four separate integrals, one along each side of the square. You should get the expression below
I=∫1−1i1+iydy+∫−11dxx+i+∫−11i−1+iydy+∫1−1dxx−iI=∫−11i1+iydy+∫1−1dxx+i+∫1−1i−1+iydy+∫−11dxx−i
Each of those four integrals evaluates to iπ/2iπ/2, giving 2πi2πi.
These integrals follow from the fact that if z=x+iyz=x+iy, dz=dxdz=dx or dz=idydz=idy (respectively) on the horizontal and vertical edges of the square.
Alternately, you can combine the integrals pairwise, e.g. the two integrals with respect to x can be combined into
∫1−12ix2+1dx∫−112ix2+1dx
which evaluates to to iπiπ, as does the other pair of integrals with respect to y.
Added: One easy way to evaluate each of the four integrals above is to multiply by the complex conjugate of the denominator, so you make your denominator real. Then you'll only deal with real integrals (essentially). Eg.
∫1−1dxx−i=∫1−1dxx−ix+ix+i=∫1−1x+ix2+1dx=0(odd integral) +itan−1x|+1−1=i(π/4−(−π/4))=iπ/2