Evaluate \(\int _ { 0 } ^ { \infty } { \frac { \sin ^ { 3 } x } { x ^ { 3 } } } d x\)
Solution
Set \[ F(z)=\frac{3e^{iz}-e^{3iz}-2}{z^3}. \] For \(0<\varepsilon<R\), use the upper-half-plane contour consisting of \([-R,-\varepsilon]\), a clockwise upper semicircle of radius \(\varepsilon\) about \(0\), \([\varepsilon,R]\), and the large upper semicircle of radius \(R\).
1. Near \(z=0\),\[ F(z) = \frac3z +4i +O(z). \]
The exponential expansions give \[ e^{iz} = 1+iz-\frac{z^2}{2}-\frac{i z^3}{6}+O(z^4) \] and \[ e^{3iz} = 1+3iz-\frac{9z^2}{2}-\frac{9i z^3}{2}+O(z^4). \] Therefore \[ 3e^{iz}-e^{3iz}-2 = 3z^2+4iz^3+O(z^4). \] Divide by \(z^3\).
2. The small indented semicircle contributes\[ -3\pi i \]as \(\varepsilon\to0\).
By step 1, \[ F(z)=\frac3z+H(z), \] where \(H\) is bounded near \(0\). Parametrize the clockwise upper semicircle by \[ z=\varepsilon e^{i\theta}, \qquad \pi\geq\theta\geq0. \] Then \[ \int\frac3z\,dz = \int_\pi^0 3i\,d\theta = -3\pi i. \] The integral of \(H\) tends to zero because its arc has length \(\pi\varepsilon\).
3. The integral over the large upper semicircle tends to zero as\(R\to\infty\).
In the upper half-plane, \[ \abs{e^{iz}}\leq1 \qquad\text{and}\qquad \abs{e^{3iz}}\leq1. \] Hence on \(\abs{z}=R\), \[ \abs{F(z)} \leq \frac{3+1+2}{R^3} = \frac6{R^3}. \] The large arc has length \(\pi R\), so \[ \abs{ \int_{\text{large arc}}F(z)\,dz } \leq \frac{6\pi}{R^2} \longrightarrow0. \]
4. The symmetric indented real-axis integrals satisfy\[ \lim_{\substack{R\to\infty\\ \varepsilon\to0^+}} \left( \int_{-R}^{-\varepsilon}F(x)\,dx + \int_{\varepsilon}^{R}F(x)\,dx \right) = 3\pi i. \]
The indentation removes the only singularity of \(F\), so the function is holomorphic inside the indented contour. Cauchy’s theorem gives total contour integral zero. Thus the real-axis portions are the negatives of the two arc integrals. Steps 2 and 3 give the displayed limit.
5. For real \(x\neq0\),\[ \operatorname{Im}F(x) = 4\frac{\sin^3x}{x^3}. \]
The triple-angle identity \[ \sin(3x)=3\sin x-4\sin^3x \] gives \[ 3\sin x-\sin(3x)=4\sin^3x. \] Taking the imaginary part of the numerator of \(F(x)\) and dividing by \(x^3\) gives the claim.
6. The whole-line integral satisfies\[ 4\int_{-\infty}^{\infty} \frac{\sin^3x}{x^3}\,dx = 3\pi. \]
The function \(\sin^3x/x^3\) extends continuously across \(0\) with value \(1\), and it is absolutely integrable at infinity because it is bounded by \(1/\abs{x}^3\) there. Hence taking imaginary parts in step 4 and using step 5 gives the ordinary improper integral in the display.
7. The requested value is\[ \boxed{ \int_0^{\infty} \frac{\sin^3x}{x^3}\,dx = \frac{3\pi}{8}. } \]
The function \(\sin^3x/x^3\) is even. Therefore step 6 becomes \[ 8\int_0^{\infty} \frac{\sin^3x}{x^3}\,dx = 3\pi. \] Divide by \(8\).
8. Q.E.D.
Step 7 is the requested evaluation.