(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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?

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# Use suitable contours in the complex plane and the residue theorem to show that

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