How do I show that $\int_0^\infty \frac{\sin(ax) \sin(bx)}{x^{2}} \, \mathrm dx = \pi \min(a,b)/2$

Question

Recently I found a claim saying that $$ \int_0^\infty \left( \frac{\sin ax}{x}\right)\left( \frac{\sin bx}{x}\right) \mathrm{d}x= \pi \min(a,b)/2 $$ from what I can see this seems to be true. I already know that $\int_{0}^\infty \operatorname{sinc}xy\,\mathrm{d}y = \pi/2$, and so independant of $y$. My suspicion is that this is closely related to the integral above.

Can someone give me some suggestions for evaluating the integral above? Also are there any generalizations for the integral? Eg $$ \int_{0}^{\infty} \left( \prod_{k=1}^N \frac{\sin (a_k \cdot x)}{x} \right) \,\mathrm{d}x $$ Where $a_k, \cdots, a_N$ are arbitrary positive constants. It seems related to the Borwein Integral, but there are some subtle differences.

Answer 1

In several steps:

• Trigonometric relation $$\left( \frac{\sin ax}{x}\right)\left( \frac{\sin bx}{x}\right)=\frac{1-\cos(a+b)x}{2x^2}-\frac{1-\cos(a-b)x}{2x^2}$$
• Dirichlet integral $$\int_0^\infty \frac{\sin \alpha t}{t}dt=\frac{\pi}{2}\mathrm{sgn}(\alpha)$$
• Integration by parts $$\int_0^\infty\frac{1-\cos(\alpha)t}{t^2}=\alpha\int_0^\infty \frac{\sin \alpha t}{t}dt=\frac{\pi}{2}|\alpha|$$
• Combine $$\int_0^\infty \left( \frac{\sin ax}{x}\right)\left( \frac{\sin bx}{x}\right) \mathrm{d}x=\frac{\pi}{4}(|a+b|-|a-b|)=\frac{\pi}{2}\min(a,b)$$

Answer 2

One way is to to this by residues. Another way to integrate once by parts to get $$I=\int_0^{\infty}\frac{b\sin ax\cos bx+a\cos ax \sin bx}{x}dx,$$ then to use the formula $2\sin \alpha\cos\beta=\sin(\alpha+\beta)+\sin(\alpha-\beta)$ and the mentioned integral (note that your formula needs to be corrected on the left and on the right) $$\displaystyle \int_0^{\infty}\frac{\sin xy}{x}\,dx=\frac{\pi}{2}\mathrm{sgn}(y).$$ This gives \begin{align}I&=\frac{\pi}{4}\Bigl[b\,\mathrm{sign}(a+b)+b\,\mathrm{sign}(a-b)+a\,\mathrm{sign}(a+b)+a\,\mathrm{sign}(b-a)\Bigr]=\\ &=\frac{\pi}{4}\left(|a+b|-|a-b|\right), \end{align} For $a,b>0$ the last expression is obviously equal to $\pi\min\{a,b\}/2$.

Answer 3

A very easy way to see this is to use Parseval's theorem for Fourier transforms. In general, Parseval's theorem states that, for two functions $f$ and $g$, each having respective FTs $\hat{f}$ and $\hat{g}$, related by

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

etc., then

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

The FT of $\sin{(a x)}/x$ is given by

$$\int_{-\infty}^{\infty} dx \, \frac{\sin{a x}}{x} e^{i k x} = \begin{cases} \pi & |k| \lt a \\ 0 & |k| \gt a\end{cases}$$

Similarly,

$$\int_{-\infty}^{\infty} dx \, \frac{\sin{b x}}{x} e^{i k x} = \begin{cases} \pi & |k| \lt b \\ 0 & |k| \gt b\end{cases}$$

By Parseval, we take the integral of the product of the transforms, which is the product of two rectangles. The product is clearly nonzero over the smaller of the two widths, i.e. $2 \min\{a,b\}$. Thus,

$$\int_{-\infty}^{\infty} dx \, \frac{\sin{a x}}{x} \, \frac{\sin{b x}}{x} = \frac{\pi^2}{2 \pi} 2 \min\{a,b\} $$

Therefore

$$\int_{0}^{\infty} dx \, \frac{\sin{a x}}{x} \, \frac{\sin{b x}}{x} = \frac{\pi}{2} \min\{a,b\}$$

Answer 4

Let $b_{k} >0$ and $a \ge \sum_{k=1}^{n} b_{k}$.

Generalizing the answer HERE, we can use contour integration to quickly show that

$$\int_{0}^{\infty} \frac{\sin(ax)}{x} \prod_{k=1}^{n} \frac{\sin \left( b_{k}x \right)}{x} \, dx= \frac{1}{2} \int_{-\infty}^{\infty} \frac{\sin(ax)}{x} \prod_{k=1}^{n} \frac{\sin \left( b_{k}x \right)}{x} \, dx = \frac{\pi}{2} \prod_{k=1}^{n}b_{k}. \tag{1}$$

Under the conditions stated above, the function $$e^{iaz} \prod_{k=1}^{n} \sin \left( b_{k}x \right) = e^{iaz} \prod_{k=1}^{n} \frac{e^{ib_{k}z}-e^{-ib_{k}z} }{2i}$$ is bounded in the upper half-plane.

So by integrating the function $$f(z) = \frac{e^{iaz}}{z} \prod_{k=1}^{n} \frac{\sin \left( b_{k}z \right)}{z} $$ around an indented contour that consists of the real axis and the semiciricle above it, we get (in the limit), $$\text{PV} \int_{-\infty}^{\infty} \frac{e^{iax}}{x} \prod_{k=1}^{n} \frac{\sin \left( b_{k}x \right)}{x} \, dx - \pi i \, \text{Res} [f(z), 0] = 0,$$ where $$\text{Res}[f(z) ,0] = \lim_{z \to 0}e^{iaz} \prod_{k=1}^{n}\frac{\sin \left( b_{k}z \right)}{z} =1\times \prod_{k=1}^{n} b_{k}. $$

Taking the imaginary parts of both sides of equation leads to the result.

Answer 5

First of all, we use compound angle formula and IBP. $$ \begin{aligned} & \int_{0}^{\infty} \frac{\sin (a x) \sin (b x)}{x^{2}} d x \\ =& \frac{1}{2} \int_{0}^{\infty}[\cos (a-b) x-(\sin (a+b) x] d\left(-\frac{1}{x}\right) \quad \text{(Via compound angle formula)}\\ \stackrel{IBP}{=} &-\left[\frac{1}{2 x}(\cos (a-b) x-\cos (a+b) x)\right]_{0}^{\infty}+\frac{1}{2} \int_{0}^{\infty} \frac{-(a-b) \sin (a-b) x}{x} d x \\\quad &+\frac{1}{2} \int_{0}^{\infty} \frac{(a+b) \sin (a+b) x}{x} d x \end{aligned} $$

Then using the famous formula $$ \int_{0}^{\infty} \frac{\sin k x}{x} d x=(\operatorname{sgn} k) \frac{\pi}{2}, $$

we can now conclude that $$ \begin{aligned} I &=\left\{\begin{array}{ll} \frac{1}{2}\left[\frac{-(a-b) \pi}{2}+\frac{(a+b) \pi}{2}\right] & \text { if } a \geqslant b \\ \frac{1}{2}\left[\frac{(a-b) \pi}{2}+\frac{(a+b) \pi}{2}\right] & \text { if } b>a \end{array}\right.\\ &=\left\{\begin{array}{ll} \frac{b \pi}{2} & \text { if } a \geqslant b \\ \frac{a \pi}{2} & \text { if } b>a \end{array}\right.\\ &=\frac{\min (a, b) \pi}{2} \end{aligned} $$

Answer 6

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

It's sufficient to consider the case $\ds{\bbox[10px,#ffe,border:1px dotted navy] {\ds{a, b \in \mathbb{R}_{\ >\ 0}}}}$.

\begin{align} \mbox{Note that}\ &\int_{0}^{\infty}{\sin\pars{ax} \over x}\,{\sin\pars{bx} \over x}\,\dd x = {1 \over 2}\,b\int_{-\infty}^{\infty}{\sin\pars{ax} \over ax} \,{\sin\pars{\bracks{b/a}ax} \over \pars{b/a}ax}\,a\,\dd x \\[5mm] = &\ {1 \over 2}\,b\int_{-\infty}^{\infty}{\sin\pars{x} \over x} \,{\sin\pars{x/\mu} \over x/\mu}\,\dd x \,,\qquad\mu \equiv {a \over b} > 0 \label{1}\tag{1} \end{align}

---

\begin{align} &\left.\int_{-\infty}^{\infty}{\sin\pars{x} \over x} \,{\sin\pars{x/\mu} \over x/\mu}\,\dd x \,\right\vert_{\ \mu\ >\ 0} = \int_{-\infty}^{\infty} \pars{{1 \over 2}\int_{-1}^{1}\expo{\ic kx}\,\dd k} \pars{{1 \over 2}\int_{-1}^{1}\expo{-\ic qx/\mu}\,\dd q}\dd x \\[5mm] = &\ {1 \over 2}\,\pi \int_{-1}^{1}\int_{-1}^{1}\int_{-\infty}^{\infty}\expo{\ic\pars{k - q/\mu}x} {\dd x \over 2\pi}\,\dd k\,\dd q = {1 \over 2}\,\pi\int_{-1}^{1}\int_{-1}^{1}\delta\pars{k - q/\mu}\,\dd k\,\dd q \\[5mm] = &\ {1 \over 2}\,\pi\int_{-1}^{1}\bracks{-1 < {q \over \mu} < 1}\dd q = \pi\int_{0}^{1}\bracks{q < \mu}\dd q = \pi\braces{\bracks{\mu < 1}\int_{0}^{\mu}\dd q + \bracks{\mu > 1}\int_{0}^{1}\dd q} \\[5mm] = &\ \bracks{a < b}\pi\,{a \over b} + \bracks{a > b}\pi\label{2}\tag{2} \end{align}

---

With \eqref{1} and \eqref{2}: \begin{align} \mbox{} \\ \int_{0}^{\infty}{\sin\pars{ax} \over x}\,{\sin\pars{bx} \over x}\,\dd x & = {1 \over 2}\,\pi\braces{\vphantom{\Large A}\bracks{a < b}a + \bracks{a > b}b} = \bbx{\ds{{1 \over 2}\,\pi\,\min\braces{a,b}}} \\ & \end{align}