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should, according to Cauchy's integral formula, yield 1/4. Numerical calculation respects this answer, but symbolic calculation returns 0.
EDIT 1
It seems that when we take the first half of the range values of t, l1 = (-I*Integral(2*I*pi*exp(6*I*pi*t)/(exp(2*I*pi*t) - 1/2), (t, 0, 1/2))/(2*pi))
we have that this portion of the curve is correct. If we take the second half, l2 = (-I*Integral(2*I*pi*exp(6*I*pi*t)/(exp(2*I*pi*t) - 1/2), (t, 0, 1/2))/(2*pi)),
we have that the real part is negated;
for i in (l2.doit().n(),l2.n(),l1.doit().n(),l1.n()):
print(i)
The code
should, according to Cauchy's integral formula, yield 1/4. Numerical calculation respects this answer, but symbolic calculation returns 0.
EDIT 1
It seems that when we take the first half of the range values of
t
,l1 = (-I*Integral(2*I*pi*exp(6*I*pi*t)/(exp(2*I*pi*t) - 1/2), (t, 0, 1/2))/(2*pi))
we have that this portion of the curve is correct. If we take the second half,
l2 = (-I*Integral(2*I*pi*exp(6*I*pi*t)/(exp(2*I*pi*t) - 1/2), (t, 0, 1/2))/(2*pi)),
we have that the real part is negated;
prints the following:
EDIT 2
This seems to occur regardless of the algorithm used, with the exception of manual.
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