Evaluate the following improper integral using the residue theorem-complex analysis

∫(0 to ∞) [sin^3(x)/(x^3)]dx=3π/8
Thanks for your help

First of all, note that
sin³x = [(1/(2i)) (e^(ix) - e^(-ix))]³
........= (-1/(8i)) (e^(3ix) - e^(-3ix) - 3e^(ix) + 3e^(-ix))
........= (1/4) [3(e^(ix) - e^(-ix))/(2i) - (e^(3ix) - e^(-3ix))/(2i)]
........= Im ((1/4) (3e^(ix) - e^(3ix))).

So, consider ∫c (3e^(iz) - e^(3iz)) dz/(4z³), where C is the closed c-clockwise contour consisting of Γ, the upper half of |z| = R (with R > ε > 0), and the x-axis with x in [-R, 0 - ε] U [0 + ε, R], and an upper half semicircular arc γ with radius ε at z = 0.

Then, ∫c (3e^(iz) - e^(3iz)) dz/(4z³) = ∫Γ (3e^(iz) - e^(3iz)) dz/(4z³) + ∫(-R to -ε) (3e^(ix) - e^(3ix)) dx/(4x³)
+ ∫γ (3e^(iz) - e^(3iz)) dz/(4z³) + ∫(ε to R) (3e^(ix) - e^(3ix)) dx/(4x³).

(i) By Cauchy's Theorem, ∫c (3e^(iz) - e^(3iz)) dz/(4z³) = 0, since the singularities of the integrand are not inside C.

(ii) lim(R→∞) |∫Γ (3e^(iz) - e^(3iz)) dz/(4z³)| = 0 by Jordan's Lemma.

(iii) lim(ε→0+) ∫γ (3e^(iz) - e^(3iz)) dz/(4z³)
= lim(ε→0+) -∫(t = 0 to π) [3e^(iεe^(it)) - e^(3iεe^(it)] * iεe^(it)/(4ε³ e^(3it))
via z = εe^(it) with opposite orientation
= lim(ε→0+) -i ∫(t = 0 to π) [3e^(iεe^(it)) - e^(3iεe^(it)] dt/(4ε² e^(2it))

Using the exponential series:
lim(ε→0+) -i ∫(t = 0 to π) [3(iεe^(it) + (iεe^(it))²/2! + ...) - (3iεe^(it) + (3iεe^(it))²/2! + ...)] dt/(4ε²e^(2it))
= lim(ε→0+) -i ∫(t = 0 to π) (3ε²e^(2it) + ...) dt/(4ε²e^(2it))
= lim(ε→0+) -i ∫(t = 0 to π) (3/4 + [ε terms]) dt
= -3πi/4.
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So letting R→∞ and ε→0+ yields
0 = 0 + ∫(-∞ to ∞) (3e^(ix) - e^(3ix)) dx/(4x³) + (-3πi/4)
==> ∫(-∞ to ∞) (3e^(ix) - e^(3ix)) dx/(4x³) = 3πi/4
==> ∫(-∞ to ∞) sin³x dx/x³ = 3π/4, by equating imaginary parts
==> 2 ∫(0 to ∞) sin³x dx/x³ = 3π/4, since the integrand is even
==> ∫(0 to ∞) sin³x dx/x³ = 3π/8.

I hope this helps!

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