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

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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?

 

[micromass]

Well, try to write e^{\pi i/4} in it's usual form:

e^{\pi i 4}=\cos(\pi /4)+i\sin(\pi /4)

work that out abd substitute that in. Do the same with e^{2\pi i/4}. You'll see that it equals \pi/\sqrt{2}.

 

[lineintegral1]

Expand your result using Euler's Formula and be dazzled. ;) Complex contour integration is a truly beautiful application of complex analysis, as is seen in the result of this integral.