help with some contour integrals?

1) Fix R > 1, and consider the semicircular contour C consisting of [-R, R] and Γ, the upper semicircle with radius R centered at the origin.

So, ∫c z^3 e^(iz) dz/(z^4 + 1) = ∫(-R to R) x^3 e^(ix) dx/(x^4 + 1) + ∫Γ z^3 e^(iz) dz/(z^4 + 1).

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Note that f(z) = z^3 e^(iz)/(z^4 + 1) has simple poles when z^4 + 1 = 0.

==> z = e^(πi(1+2k)/4), with k = 0, 1 (k = 2, 3 are outside C).

Residues (for k = 0, 1):

lim(z→e^(πi(1+2k)/4)) (z - e^(πi(1+2k)/4)) * z^3 e^(iz)/(z^4 + 1)

= lim(z→e^(πi(1+2k)/4)) z^3 e^(iz) * lim(z→e^(πi(1+2k)/4)) (z - e^(πi(1+2k)/4))/(z^4 + 1)

= lim(z→e^(πi(1+2k)/4)) z^3 e^(iz) * lim(z→e^(πi(1+2k)/4)) 1/(4z^3)

= e^(3πi(1+2k)/4)) e^(ie^(πi(1+2k)/4)) * 1/(4e^(3πi(1+2k)/4))

= (1/4) e^(ie^(πi(1+2k)/4)).

By the Residue Theorem,

∫c z^3 e^(iz) dz/(z^4 + 1)

= 2πi [(1/4) e^(ie^((1+0)πi/4)) + (1/4) e^(ie^(πi(1+2)/4))]

= (πi/2) [e^(-1/√2 + i/√2)) + e^(-1/√2 - i/√2))]

= (πi/2) e^(-1/√2) [e^(i/√2) + e^(-i/√2)]

= πi e^(-1/√2) cos(1/√2).

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Next, parameterizing Γ via z = Re^(it) for t in [0, π] yields

|∫Γ z^3 e^(iz) dz/(z^4 + 1)|

= |∫(t = 0 to π) R^3 e^(3it) e^(iRe^(it)) * iRe^(it) dt / (R^4 e^(4it) + 1)|

≤ ∫(t = 0 to π) |R^3 e^(3it) e^(iRe^(it)) * iRe^(it) dt / (R^4 e^(4it) + 1)|

≤ ∫(t = 0 to π) R^4 |e^(iRe^(it))| dt / (R^4 - 1), since R > 1

= ∫(t = 0 to π) R^4 |e^(R(-sin t + i cos t))| dt / (R^4 - 1)

= (R^4/(R^4 - 1)) * ∫(t = 0 to π) e^(-R sin t) dt

< (R^4/(R^4 - 1)) * (π/R), by Jordan's Inequality

→ 0, as R→∞.

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So, as R→∞

πi e^(-1/√2) cos(1/√2) = ∫(-∞ to ∞) x^3 e^(ix) dx/(x^4 + 1) + 0

==> πi e^(-1/√2) cos(1/√2) = ∫(-∞ to ∞) x^3 (cos x + i sin x) dx/(x^4 + 1)

==> π e^(-1/√2) cos(1/√2) = ∫(-∞ to ∞) x^3 sin x dx/(x^4 + 1), equating imag. parts

==> ∫(0 to ∞) x^3 sin x dx/(x^4 + 1) = (π/2) e^(-1/√2) cos(1/√2), since integrand is even.

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2) f(z) = e^(πiz) /(z^3 - 1) has a singularity at z = 1, which is on the real axis.

Fix R > 1, and consider the semicircular contour C consisting of [-R, 1 - ε] , Cε (the semicircular contour clockwise with radius ε centered at 0), [1 + ε, R] and Γ (the upper semicircle with radius R centered at the origin).

So, ∫c e^(πiz) dz/(z^3 - 1) = ∫(-R to 1-ε) e^(πix) dx/(x^3 - 1) + ∫Cε e^(πiz) dz/(z^3 - 1)

+ ∫(1+ε to R) e^(πix) dx/(x^3 - 1) + ∫Γ e^(πiz) dz/(z^3 - 1).

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The only singularity of f(z) = e^(πiz)/(z^3 - 1) inside C is when z = e^(2πi/3), the cube root of unity above the real axis, which yields residue

lim(z→e^(2πi/3)) (z - e^(2πi/3)) * e^(πiz)/(z^3 - 1)

= lim(z→e^(2πi/3)) e^(πiz) * lim(z→e^(2πi/3)) (z - e^(2πi/3))/(z^3 - 1)

= lim(z→e^(2πi/3)) e^(πiz) * lim(z→e^(2πi/3)) 1/(3z^2), by L'Hopital's Rule

= e^(πi e^(2πi/3)) * 1/(3e^(4πi/3))

= e^(-πi/2 - π√3/2)) * (1/3) * e^(2πi/3)

= (-i/3) e^(-π√3/2) (-1/2 + i√3/2).

So, the Residue Theorem yields

∫c e^(πiz) dz/(z^3 - 1) = 2πi * (-i/3) e^(-π√3/2) (-1/2 + i√3/2) = (π/3) e^(-π√3/2) (-1 + i√3).

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Next, parameterizing Γ via z = Re^(it) for t in [0, π] yields

|∫Γ e^(πiz) dz/(z^3 - 1)|

= |∫(t = 0 to π) e^(πiRe^(it)) * iRe^(it) dt/(R^3 e^(3it) - 1)|

≤ ∫(t = 0 to π) |e^(πiRe^(it))| * R dt / (R^3 - 1), since R > 1

= ∫(t = 0 to π) e^(-πR sin t) * R dt / (R^3 - 1)

< (R/(R^3 - 1)) * ∫(t = 0 to π) e^0 dt, since sin t > 0 on (0, π)

= πR/(R^3 - 1)

→ 0, as R→∞.

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Next, parameterizing Cε via z = 1 + εe^(it) for t in [0, π] (with opposite orientation) yields

∫Cε e^(πiz) dz/(z^3 - 1)

= - ∫(t = 0 to π) e^(πi(1 + εe^(it))) * iεe^(it) dt / [(1 + εe^(it))^3 - 1)]

= - ∫(t = 0 to π) e^(πi(1 + εe^(it))) * iεe^(it) dt / [ε^3 e^(3it) + 3ε^2 e^(2it) + 3εe^(it)]

= - ∫(t = 0 to π) e^(πi(1 + εe^(it))) * i dt / [ε^2 e^(2it) + 3ε e^(it) + 3]

Letting ε→0+, we obtain

- ∫(t = 0 to π) e^(πi) * i dt / (0 + 3) = -πi e^(πi)/3 = πi/3.

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So, letting R→∞ and ε→0+ yields

(π/3) e^(-π√3/2) (-1 + i√3) = ∫(-∞ to 1) e^(πix) dx/(x^3 - 1) + πi/3 + ∫(1 to ∞) e^(πix) dx/(x^3 - 1) + 0

==> (π/3) e^(-π√3/2) (-1 + i√3) - πi/3 = ∫(-∞ to ∞) e^(πix) dx/(x^3 - 1).

Equating imaginary parts yields

∫(-∞ to ∞) sin(πx) dx/(x^3 - 1) = (π/3) [√3 e^(-π√3/2) - 1].

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I hope this helps!