- #1
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Homework Statement
Use suitable contours in the complex plane and the residue theorem to show that
integral from -infinity to +infinity of [1/(1+(x^4))] dx=pi/(sqrt(2))
Fix R > 1, and consider the counterclockwise-oriented contour C consisting of the upper half circle of radius R centered at the origin (call it γ) and the real axis for x in [-R, R].
So, ∫c dz/(1 + z^4) = ∫γ dz/(1 + z^4) + ∫(x = -R to R) dx/(1 + x^4).
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(i) Compute ∫c dz/(1 + z^4) by the Residue Theorem.
The integrand has two singularities in the upper half plane (where C is located):
1 + z^4 = 0 ==> z^4 = -1 = e^(πi) ==> z = e^(πi/4), e^(3πi/4); both are simple poles.
Computing their residues:
At z = e^(πi/4), we have
lim(z→e^(πi/4)) (z - e^(πi/4))/(1 + z^4)
= lim(z→e^(πi/4)) 1/(4z^3)
= lim(z→e^(πi/4)) z/(4z^4)
= e^(πi/4)/(4 * -1)
= (-1/4) e^(πi/4).
Similarly, the residue at z = e^(3πi/4) equals (-1/4) e^(3πi/4).
Therefore, ∫c dz/(1 + z^4) = 2πi [(-1/4) e^(πi/4) + (-1/4) e^(3πi/4)] = π/√2.
but why 2πi [(-1/4) e^(πi/4) + (-1/4) e^(3πi/4)] = π/√2?