4444.sagews

# Integrate e^z on the line y=x from 0 to 1+i

f(z) = e^z #define the integrand

c(t) = t+I*t # parametrize the line. c(t):[0,1] -> line in C.

start = 0
stop = 1
N = 1000    
#c(t) maps the interval [start, stop] onto the curve where f(z) is to be integrated. N is number of points to use in approximation of integral.

line_points = srange(float(start), float(stop), (stop-start)/N,include_endpoint=True)  
#N+1 points on [start, stop]
z = map(c,line_points)                               
#N+1 points on curve
#The points are denoted z[0], z[1], ..., z[N]

sum(f(z[i])*(z[i+1]-z[i]) for i in range(0,N-1))  # approximate integral of e^z dz on line

0.470416979755 + 2.28222422966*I

dc(t)=c.diff(t)  # derivative of c(t)
answer=integrate(f(c(t))*dc(t),(t,start,stop))   # integrate using parametrization
answer

e^(I + 1) - 1

float(answer.real_part())+float(answer.imag_part())*I
#check if answer from parametrization agrees with approximation by taking its floating-point value

0.468693939916 + 2.28735528718*I