It has residue
(1/3!) * lim(z→i) (d^3/dz^3) [(z - i)^4 * 1/(1 + z^2)^4]
= (1/6) * lim(z→i) (d^3/dz^3) 1/(z+i)^4
= (1/6) * lim(z→i) (-4)(-5)(-6)/(z + i)^7
= -20/(2i)^7
= 5/(32i)
Hence, ∫(-∞ to ∞) dx/(1 + x^2)^4 = 2πi * (5/(32i)) = 5π/16.
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4) The same contour applies; we use the following modification (with Jordan's Lemma):
∫(-∞ to ∞) e^(ix) f(x) dx = 2πi * (sum of residues of e^(iz) f(z) in the upper half plane).
Here, f(z) = 1/[(z^2 + π^2)(z^2 + π^2/4)], whose singularities in the upper half plane
are z = πi and z = πi/2, both of which are simple poles.
Computing residues of e^(iz)/[(z^2 + π^2)(z^2 + π^2/4)]:
(i) At z = πi:
lim(z→πi) (z - πi) * e^(iz)/[(z^2 + π^2)(z^2 + π^2/4)]
= lim(z→πi) e^(iz)/[(z + πi)(z^2 + π^2/4)]
= e^(-π)/[(2πi)(-3π^2/4)]
= 2e^(-π)/(-3iπ^3).
(ii) At z = πi/2:
lim(z→πi/2) (z - πi/2) * e^(iz)/[(z^2 + π^2)(z^2 + π^2/4)]
= lim(z→πi/2) e^(iz)/[(z^2 + π^2)(z + πi/2)]
= 4e^(-π/2)/(3iπ^3).
Hence, ∫(-∞ to ∞) e^(ix) dx/[(x^2 + π^2)(x^2 + π^2/4)]
= 2πi [2e^(-π)/(-3iπ^3) + 4e^(-π/2)/(3iπ^3)]
= [8e^(-π/2) - 4e^(-π)] / (3π^2)
Taking real parts yields
∫(-∞ to ∞) cos x dx/[(x^2 + π^2)(x^2 + π^2/4)] = [8e^(-π/2) - 4e^(-π)] / (3π^2).
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(Double checked with Wolfram Alpha.)
I hope this helps!