show that $\int_{0}^{\infty} \frac {\sin^3(x)}{x^3}dx=\frac{3\pi}{8}$

Solution 1:

Let $$f(y) = \int_{0}^{\infty} \frac{\sin^3{yx}}{x^3} \mathrm{d}x$$ Then, $$f'(y) = 3\int_{0}^{\infty} \frac{\sin^2{yx}\cos{yx}}{x^2} \mathrm{d}x = \frac{3}{4}\int_{0}^{\infty} \frac{\cos{yx} - \cos{3yx}}{x^2} \mathrm{d}x$$ $$f''(y) = \frac{3}{4}\int_{0}^{\infty} \frac{-\sin{yx} + 3\sin{3yx}}{x} \mathrm{d}x$$ Therefore, $$f''(y) = \frac{9}{4} \int_{0}^{\infty} \frac{\sin{3yx}}{x} \mathrm{d}x - \frac{3}{4} \int_{0}^{\infty} \frac{\sin{yx}}{x} \mathrm{d}x$$

Now, it is quite easy to prove that $$\int_{0}^{\infty} \frac{\sin{ax}}{x} \mathrm{d}x = \frac{\pi}{2}\mathop{\mathrm{signum}}{a}$$

Therefore, $$f''(y) = \frac{9\pi}{8} \mathop{\mathrm{signum}}{y} - \frac{3\pi}{8} \mathop{\mathrm{signum}}{y} = \frac{3\pi}{4}\mathop{\mathrm{signum}}{y}$$ Then, $$f'(y) = \frac{3\pi}{4} |y| + C$$ Note that, $f'(0) = 0$, therefore, $C = 0$. $$f(y) = \frac{3\pi}{8} y^2 \mathop{\mathrm{signum}}{y} + D$$ Again, $f(0) = 0$, therefore, $D = 0$.

Hence, $$f(1) = \int_{0}^{\infty} \frac{\sin^3{x}}{x^3} = \frac{3\pi}{8}$$

Solution 2:

Use Parseval's theorem:

$$\int_{-\infty}^{\infty} dx \, f(x) g^*(x) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} dk \, F(k) G^*(k)$$

where $f$, $g$ and $F$, $G$ are respective Fourier transform pairs, e.g.,

$$F(k) = \int_{-\infty}^{\infty} dx \, f(x) \, e^{i k x}$$

etc. If $f(x) = \sin{x}/x$, then

$$F(k) = \begin{cases} \pi & |k| \le 1\\0 & |k| \gt 1 \end{cases}$$

Further, if $g(x) = \sin^2{x}/x^2$, then

$$G(k) = \begin{cases}\pi \left (1-\frac{|k|}{2} \right ) & |k| \le 2 \\ 0& |k| \gt 2\end{cases}$$

Then

$$\int_{-\infty}^{\infty} dx \,\frac{\sin^3{x}}{x^3} = \frac{1}{2 \pi} \int_{-1}^1 dk \, \pi^2 \left (1-\frac{|k|}{2} \right ) = \pi - \frac{\pi}{2} \int_0^1 dk \,k = \pi-\frac{\pi}{4}$$

Therefore

$$\int_{0}^{\infty} dx \,\frac{\sin^3{x}}{x^3} = \frac{3 \pi}{8}$$

ADDENDUM

You can also use contour integration techniques. For the integral

$$\int_0^{\infty} dt \frac{\sin^3{ \pi t}}{(\pi t)^3} \cos{u t}$$

I have derived a complete solution to the problem of its evaluation here using both contour integral techniques as well as the convolution theorem. You will see that the results agree for $u=0$ by a simple rescaling of the integral.

Solution 3:

In this answer, the more general integral $$ \int_0^\infty\left(\frac{\sin(x)}{x}\right)^n\,\mathrm{d}x $$ is calculated.

Your integral is that integral for $n=3$.

A Different Way

In a fashion similar to this answer, we will use the equation $$ \frac{\mathrm{d}^2}{\mathrm{d}x^2}\frac{\sin^3(kx)}{k^3} =\frac{9\sin(3kx)-3\sin(kx)}{4k}\tag{1} $$ and the series for $0\lt x\le\pi$, $$ \sum_{k=1}^\infty\frac{\sin(kx)}{k}=\frac{\pi-x}{2}\tag{2} $$ Using $(2)$, we get $$ \begin{align} \sum_{k=1}^\infty\frac{9\sin(3kx)-3\sin(kx)}{4k} &=\frac94\frac{\pi-3x}{2}-\frac34\frac{\pi-x}{2}\\ &=\frac{3\pi}{4}-3x\tag{3} \end{align} $$ Integrating from $0$ twice to back out the derivatives taken in $(1)$ yields $$ \sum_{k=1}^\infty\frac{\sin^3(kx)}{k^3}=\frac{3\pi}{8}x^2-\frac12x^3\tag{4} $$ Set $x=1/n$ and multiply by $n^2$ to get $$ \sum_{k=1}^\infty\frac{\sin^3(k/n)}{k^3/n^3}\frac1n=\frac{3\pi}{8}-\frac1{2n}\tag{5} $$ and $(5)$ is a Riemann sum for $$ \int_0^\infty\frac{\sin^3(x)}{x^3}\,\mathrm{d}x=\frac{3\pi}{8}\tag{6} $$

Solution 4:

Well, there's a lot of creative answers but it seems that no one bothered to put here the "follow your nose" one, so I'll add it here for completeness.

Recall the identity: $$\sin^3x = \frac{3 \sin x - \sin(3x)}{4}.$$

So we have: $$\int_{-\infty}^{+\infty} \left(\frac{\sin x}{x}\right)^3\,{\rm d}x = \int_{-\infty}^{\infty} \frac{3 \sin x - \sin (3x)}{4x^3}\,{\rm d}x. $$Let $0 < r < R$. Consider $\gamma_1$ the line segment joining $r$ to $R$, $\gamma_2$ the circular arc oriented counterclockwise joining $R$ to $-R$, $\gamma_3$ the line segment joining $-R$ to $-r$ and $\gamma_4$ the circular arc oriented clockwise joining $-r$ to $r$. Consider $\gamma = \gamma_1 \ast \gamma_2 \ast \gamma_3 \ast \gamma_4$ the concatenation. Sketch:

Consider the function: $$ f(z) = \frac{3e^{iz}-e^{3iz}}{4z^3}. $$The only singularity is $z = 0$ (triple pole). Expanding in Laurent: $$ f(z) = \frac{1}{4z^3}\left(3\sum_{n \geq 0}\frac{i^nz^n}{n!} - \sum_{n \geq 0}\frac{3^ni^nz^n}{n!}\right) = \sum_{n \geq 0} \left(\frac{(3-3^n)i^n}{4\cdot n!}\right)z^{n-3}. $$ Since $f$ is holomorphic in $\gamma$ and inside it, by Cauchy-Goursat we get: $$ \oint_{\gamma} f(z)\,{\rm d}z = \int_{\gamma_1} f(z)\,{\rm d}z + \int_{\gamma_2} f(z)\,{\rm d}z + \int_{\gamma_3}f(z)\,{\rm d}z + \int_{\gamma_4}f(z)\,{\rm d}z = 0.$$ Let's analyze everything sistematically.

Parametrizing $\gamma_1(x) = x$, with $r \leq x \leq R$, we have: $$ \int_{\gamma_1}f(z)\,{\rm d}z = \int_r^R \frac{3e^{ix}-e^{3ix}}{4x^3}\,{\rm d}x. $$For $\gamma_2$, we have that its lenght is $\pi R$, $|e^{iz}| = e^{{\rm Re}(iz)} = e^{-{\rm Im}(z)} < 1$, and similarly $|e^{3iz}| < 1$, since for all $z$ in $\gamma_2$ we have ${\rm Im}(z) > 0$. Hence: $$ \left|\int_{\gamma_2} f(z)\,{\rm d}z\right| \leq \frac{\pi R(3+4)}{4 R^3} = \frac{\pi}{R^2} \stackrel{R\, \to \,+\infty}{\longrightarrow} 0. $$

Parametrizing $\gamma_3^-(x) = -x$, with $r \leq x \leq R$, we have: $$ \int_{\gamma_3}f(z)\,{\rm d}z =- \int_r^R \frac{3e^{-ix}-e^{-3ix}}{4x^3}\,{\rm d}x, $$once the signs in $(-x^3) = -x^3$ and ${\rm d}z = -{\rm d}x$ cancel each other. Notice here that: $$ \int_{\gamma_1}f(z)\,{\rm d}z + \int_{\gamma_3}f(z)\,{\rm d}z = 2i\int_r^R \left(\frac{\sin x}{x}\right)^3\,{\rm d}x.$$

For $\gamma_4$, we have: $$\begin{align} \int_{\gamma_4}f(z)\,{\rm d}z &= \int_{\gamma_4} \sum_{n \geq 0}\left(\frac{(3-3^n)i^n}{4 \cdot n!}\right)z^{n-3}\,{\rm d}z \\ &= \int_{\gamma_4}\frac{1}{2z^3}\,{\rm d}z + \int_{\gamma_4}\frac{3}{4z}\,{\rm d}z + \int_{\gamma_4} \sum_{n \geq 3} \left(\frac{(3-3^n)i^n}{4 \cdot n!}\right)z^{n-3}\,{\rm d}z \end{align}$$

We have: $$ \int_{\gamma_4} \frac{1}{2z^3}\,{\rm d}z = -\frac{1}{4z^2}\Bigg|_{-r}^{r} = 0, \quad \int_{\gamma_4} \frac{3}{4z}\,{\rm d}z = \frac{3}{4}\int_{\gamma_4} -i \frac{{\rm d}z}{-iz} = \frac{3}{4}\int_0^\pi -i\,{\rm d}t = -\frac{3\pi i}{4}, $$ and: $$ \left|\int_{\gamma_4} \sum_{n \geq 3}\frac{(3-3^n)i^n}{4\cdot n!}z^{n-3}\right| \leq \pi r \sum_{n \geq 3}\frac{3^n-3}{4\cdot n!}r^{n-3} = \sum_{n \geq 3}\frac{\pi(3^n-3)}{4 \cdot n!}r^{n-2} \stackrel{r \to 0}{\longrightarrow} 0. $$

Making first $r \to 0$, and then $R \to +\infty$, in $\oint_\gamma f = 0$, we get: $$ 2i\int_0^{+\infty}\left(\frac{\sin x}{x}\right)^3\,{\rm d}x - \frac{3\pi i}{4} = 0 \implies \int_0^{+\infty}\left(\frac{\sin x}{x}\right)^3\,{\rm d}x = \frac{3\pi}{8}, $$as desired.

Solution 5:

Related technique. You can use the Laplace transform technique. Recalling the Laplace transform

$$F(s)= \int_{0}^{\infty} f(x) e^{-sx}dx. $$

Taking $ f(x) = \frac{\sin(x)^3}{x^3} $ gives

$$ F(s)= \frac{\pi \,{s}^{2}}{8}+\frac{3\,\pi}{8}- \frac{3( {s}^{2}-1) }{8}\,\arctan \left( s \right) +\frac{( {s}^{2}-9)}{8}\,\arctan \left( \frac{s}{3} \right) $$ $$+\frac{3s}{8}\, \left( -\ln \left( {s}^{2}+9 \right) +\ln \left( {s} ^{2}+1 \right) \right). $$

Taking the limit as $s\to 0$ gives the desired result $\frac{3\pi}{8}$.

Another Laplace transform approach: Referring to the problem, we can use the following relation

$$ \begin{align} \int_0^\infty F(u)g(u) \, du & = \int_0^\infty f(u)G(u) \, du \\[6pt] L[f(t)] & = F(s) \\[6pt] L[g(t)] & = G(s)\end{align} $$

Let

$$ G(u)=\frac{1}{u^3} \implies g(u)=\frac{u^2}{2!}, $$

and

$$ f(u)= \sin(u)^3 \implies F(u) = {\frac {6}{ \left( {u}^{2}+1 \right) \left( {u}^{2}+9 \right) }}. $$

Now,

$$ \int_0^\infty \frac{\sin^3 x}{x^3} \, dx = \frac{6}{2}\int_0^\infty \frac{u^2}{\left( {u}^{2}+1 \right) \left( {u}^{2}+9 \right)} \, du = \frac{3\pi}{8}$$.