A rough idea to prove $\int_{0}^{\infty}\frac{\sin(x)}{x}\frac{\sin(x/3)}{x/3}dx = \frac{\pi}{2} ?$

Question

I have a rough idea of approach to prove the Borwein integral in $(1)$ via Complex-Analytic techniques, is it valid ?

$(1)$

$$ \int_{0}^{\infty}\frac{\sin(x)}{x}\frac{\sin(x/3)}{x/3}dx = \frac{\pi}{2}.$$

From glancing at it seems one can consider an Indented Contour $\Gamma_{R}$, and define in $(1.2)$

$(1.2)$

$$\Gamma_{R}^{1}(t) = t \, \, \, \, \text{if} \, \, \, -R \leq t \leq -1/R $$

$$\Gamma_{R}^{2}(t) = e^{it}/R \, \, \, \, \text{if} \, \, \, \pi \leq t \leq 2 \pi$$

$$\Gamma_{R}^{3}(t) = e^{it}/R \, \, \, \, \text{if} \, \, \, \pi \leq t \leq 2 \pi$$

$$ \Gamma_{R}^{4}(t) = Re^{it} \, \, \, \, \text{if} \, \, \, 0 \leq t \leq \pi$$

Our original integral $\Psi(x) = \frac{\sin(x)}{x}\frac{\sin(x/3)}{x/3}$ can be defined on $\Gamma_{R}$ as $\Psi(z)=\frac{e^{iz}}{z}\frac{e^{iz}}{z/3}$ so our original integral now becomes

$$I = \oint_{\Gamma_{R}}\frac{e^{iz}}{z}\frac{e^{iz}}{z/3}dz; $$

Then it seems via Cauchy's Theorem one has in $(1.3)$

$(1.3)$ $$\, \, \, \, \, \, \lim_{R \rightarrow \infty} \bigg(\oint_{\Gamma_{R}^{1}} \Psi(x)dx + \oint_{\Gamma_{R}^{2}} \Psi(z)dz + \oint_{\Gamma_{R}^{3}}\Psi(x)dx + \oint_{\Gamma_{R}^{4}} \Psi(z)dz \bigg) = 0 $$

To finish the argument it seems one would have to invoke Jordan's Lemma on the last integral in $(1.3)$, then parametrize the integral over $\Gamma_{R}^{4}$ then in $(1.4)$ we have

$(1.4)$

$$\lim_{R \rightarrow \infty} \oint_{\Gamma_{R}}\frac{e^{iz}}{z}\frac{e^{iz}}{z/3}dz \rightarrow \int_{0}^{\pi}\frac{e^{ire^{i\theta}}}{re^{i\theta}}\frac{e^{ire^{i\theta}}}{re^{i\theta}/3}d\theta = \frac{\pi}{2}$$

Answer 1

As an alternative to finding a suitable integration contour, the Laplace transform often deserves a thought. For any $a,b>0$ we have $$\begin{eqnarray*} I(a,b)&=&\int_{0}^{+\infty}\frac{\sin(ax)\sin(bx)}{x^2}\,dx=\int_{0}^{+\infty}\frac{\cos((a-b)x)-\cos((a+b)x)}{2x^2}\,dx\\&=&J(a+b)-J(a-b)\end{eqnarray*} $$ where $$ J(c)=\int_{0}^{+\infty}\frac{1-\cos(cx)}{2x^2}\,dx = \int_{0}^{+\infty}\frac{\sin^2\left(\frac{c}{2}x\right)}{x^2}\,dx\stackrel{\mathcal{L},\mathcal{L}^{-1}}{=}\int_{0}^{+\infty}\frac{c^2}{2c^2+2s^2}\,ds=\frac{\pi}{4}|c| $$ ensures: $$ I(a,b) = \frac{\pi}{4}\left(|a+b|-|a-b|\right)=\color{red}{\frac{\pi}{2}\min(a,b).}$$

Answer 2

Perhaps you might be interested in seeing a real method used to evaluate this integral.

We begin by enforcing a substitution of $x \mapsto 3x$. This gives \begin{align*} \int_0^\infty \frac{\sin x}{x} \frac{\sin (x/3)}{x/3} \, dx &= \int_0^\infty \frac{\sin (3x) \sin x}{x^2} \, dx\\ &= 3 \int_0^\infty \frac{\sin^2 x}{x^2} \, dx - 4 \int_0^\infty \frac{\sin^4 x}{x^2} \, dx, \tag1 \end{align*} since $\sin (3x) = 3 \sin x - 4 \sin^3 x$.

For the first of these integrals, as is well-known $$\frac{\pi}{2} = \int_0^\infty \frac{\sin x}{x} \, dx,$$ (this is just the Dirichlet integral) enforcing a substitution of $x \mapsto 2x$ gives \begin{align*} \frac{\pi}{2} &= \int_0^\infty \frac{\sin (2x)}{x} \, dx = \int_0^\infty \frac{2 \sin x \cos x}{x} \, dx\ = \int_0^\infty \frac{(\sin^2 x)'}{x} \, dx\\ &= \left [\frac{\sin^2 x}{x} \right ]_0^\infty + \int_0^\infty \frac{\sin^2 x}{x^2} \, dx = \int_0^\infty \frac{\sin^2 x}{x^2} \, dx. \end{align*}

And for the second of the integrals, as we just found above $$\frac{\pi}{2} = \int_0^\infty \frac{\sin^2 x}{x^2}dx.$$ Enforcing a substitution of $x \mapsto 2x$ gives $$\frac{\pi}{2} = \frac{1}{2} \int_0^\infty \frac{\sin^2 (2x)}{x^2} \, dx.$$ Since $$\sin^2 (2x) = 4 \cos^2 x \sin^2 x = 4(1 - \sin^2 x)\sin^2 x = 4 \sin^2 x - 4 \sin^4 x,$$ we have $$\frac{\pi}{2} = 2 \int_0^\infty \frac{\sin^2 x}{x^2} \, dx - 2 \int_0^\infty \frac{\sin^4 x}{x^2} \, dx = 2 \cdot \frac{\pi}{2} - 2 \int_0^\infty \frac{\sin^4 x}{x^2} \, dx,$$ or $$\int_0^\infty \frac{\sin^4 x}{x^2}dx = \frac{\pi}{4}.$$

So returning to (1) we have $$\int_0^\infty \frac{\sin x}{x} \frac{\sin (x/3)}{x/3} \, dx = 3 \cdot \frac{\pi}{2} - 4 \cdot \frac{\pi}{4} = \frac{\pi}{2},$$ as required to show.