# Using Laplace Transforms to solve $\int_{0}^{\infty}\frac{\sin(x)\sin(x/3)}{x(x/3)}\:dx$

So, I've come across the following integral (and it's expansion) many times and in my study so far, Complex Residues have been used to evaluate it. I was hoping to find an alternative approach using Laplace Transforms. I believe the method I've taken is correct, but I'm concerned there may certain theorems/tests I should have applied first

$$I = \int_{0}^{\infty} \frac{\sin\left(\frac{x}{1}\right)\sin\left(\frac{x}{3}\right)}{\left(\frac{x}{1}\right)\left(\frac{x}{3}\right)}\:dx$$

The first step is to make a slight change of variable $$x = 3u$$, which gives us

$$I = \int_{0}^{\infty} \frac{\sin\left(3u\right)\sin\left(u\right)}{u^2} \: du$$

Here I employ the Feynman Trick but with two variables, i.e.

$$I(a,b) = \int_{0}^{\infty} \frac{\sin\left(3ua\right)\sin\left(ub\right)}{u^2}\: du$$

Take the Laplace Transform w.r.t '$$a$$'

$$\mathscr{L}_{a} \left[I(a,b)\right] = \int_{0}^{\infty} \frac{\mathscr{L}_{a}\left[\sin\left(3ua\right)\right]\sin\left(ub\right)}{u^2}\: du = \int_{0}^{\infty} \frac{3u\sin\left(ub\right)}{\left(s^2 + 9u^2\right)u^2}\: du$$

Or

$$\overline{I}(s,b) = \int_{0}^{\infty} \frac{3\sin\left(ub\right)}{\left(s^2 + 9u^2\right)u}$$

Now apply the Laplace Transform w.r.t '$$b$$'. Here $$\omega$$ will be used as the alternate '$$s$$' variable. Hence we arrive at

$$\mathscr{L}_{b}\left[\overline{I}(s,b)\right] = \int_{0}^{\infty} \frac{3\mathscr{L}_{b}\left[\sin\left(ub\right)\right]}{\left(s^2 + 9u^2\right)u}\:du = \int_{0}^{\infty} \frac{3u}{\left(s^2 + 9u^2\right)u\left(\omega^2 + u^2\right)}\:du$$

Or

$$\overline{\overline{I}}\left(s,\omega\right) = \int_{0}^{\infty} \frac{3}{\left(s^2 + 9u^2\right)\left(\omega^2 + u^2\right)}\:du = \frac{3\pi}{2s\omega}\left(\frac{1}{s + 3\omega} \right)$$

In no specific order we now take the Inverse Laplace Transform w.r.t. '$$\omega$$'

$$\overline{I}\left(s,b\right) = \mathscr{L}_{\omega}^{-1}\left[\frac{3\pi}{2s\omega}\left(\frac{1}{s + 3\omega} \right) \right] = \frac{3\pi}{2}\left[\frac{1}{s^2} - \frac{e^{\frac{sb}{3}}}{s^2}\right]$$

We now take the Inverse Laplace Transform w.r.t. '$$s$$'

$$I(a,b) = \frac{3\pi}{2}\mathscr{L}_{s}^{-1}\left[ \frac{1}{s^2} - \frac{e^{\frac{sb}{3}}}{s^2}\right] = \frac{3\pi}{2}\left[a - \left(a - \frac{b}{3} \right)\mathcal{H}\left(a - \frac{b}{3}\right) \right]$$

And so,

$$I = I(1,1) = \frac{3\pi}{2}\left[1 - \left(1 - \frac{1}{3} \right)\mathcal{H}\left(1 - \frac{1}{3}\right) \right] = \frac{\pi}{2}$$

As required.

Is this a stroke of luck? or is it just employing the Dominated Convergence Theorem and Fubini's Theorm (as I believe is valid here).

Not sure what you are asking with your "Is this a stroke of luck?" question...

Just in case, here's a different approach to the integral:

$$I=\int_0^\infty \frac{\sin (3x) \sin(x)}{x^2} dx=\frac{1}{2}\int_0^\infty \frac{\cos (2x)-\cos(4x)}{x^2} dx$$

$$I=\int_0^\infty \frac{\cos (x)-\cos(2x)}{x^2} dx$$

Now we also introduce a parameter, though we only need one:

$$I(a)=\int_0^\infty \frac{\cos (ax)-\cos(2ax)}{x^2} dx$$

$$I'(a)=\int_0^\infty \frac{2\sin (2ax)-\sin(ax)}{x} dx$$

The integral can now be safely separated into two terms, and each has a well known value:

$$\int_0^\infty \frac{\sin(x)}{x} dx=\frac{\pi}{2}$$

So:

$$I'(a)=\int_0^\infty \frac{2\sin (2ax)-\sin(ax)}{x} dx=\frac{\pi}{2}$$

Integrating (the constant of integration is determined by $$I(0)$$), we have:

$$I(a)=\frac{\pi}{2}a$$

$$I(1)=\frac{\pi}{2}$$

The proofs of the $$\text{sinc}$$ integral can be found elsewhere, including this site. Evaluating the integral $$\int_0^\infty \frac{\sin x} x \ dx = \frac \pi 2$$?

• I was unsure if the method I took was valid or a stroke of luck. So I posted the method and then asked if it was a stroke of luck. Where did I lose you?
– user150203
Commented Nov 12, 2018 at 8:36
• @DavidG, there's no luck in Mathematics... Either the method is correct or not, the fact that it brings you the correct answer is a big point in its favor. In my opinion, it's too complicated, and also Laplace transform is just complex analysis in disguise... But I'm pretty sure it's correct. If it wasn't, I wouldn't call it the "stroke of luck" in any case Commented Nov 12, 2018 at 8:39
– user150203
Commented Nov 12, 2018 at 9:13
• @DavidG, I agree with you as well :) Commented Nov 12, 2018 at 9:14

This starts out similarly to Yuriy S's answer, but the execution seems a bit simpler.

Substituting $$x\mapsto3x$$, the integral is equal to \begin{align} \int_0^\infty\frac{\sin(3x)\sin(x)}{x^2}\,\mathrm{d}x &=\int_0^\infty\frac{\cos(2x)-\cos(4x)}{2x^2}\,\mathrm{d}x\tag1\\ &=\int_0^\infty\int_2^4\frac{\sin(ax)}{2x}\,\mathrm{d}a\,\mathrm{d}x\tag2\\ &=\frac12\int_2^4\int_0^\infty\frac{\sin(x)}x\,\mathrm{d}x\,\mathrm{d}a\tag3\\[6pt] &=\frac\pi2\tag4 \end{align} Explanation:
$$(1)$$: $$\cos(3x-x)-\cos(3x+x)=2\sin(3x)\sin(x)$$
$$(2)$$: $$\int_2^4\sin(ax)\,\mathrm{d}a=\frac{\cos(2x)-\cos(4x)}x$$
$$(3)$$: swap order of integration then substitute $$x\mapsto x/a$$
$$(4)$$: $$\int_0^\infty\frac{\sin(x)}x\,\mathrm{d}x=\frac\pi2$$

• Sorry, this is probably being pedantic, but do you have include that you employed Fubini's Theorem as part of the explanation?
– user150203
Commented Nov 12, 2018 at 10:35
• @DavidG: we don't use Fubini directly, but since the infinite integrals converge uniformly (that is, if we replace $\infty$ by $n$, each integral is within $1/n$ of the full integral), we can apply Fubini to the finite integrals and take the limit.
– robjohn
Commented Nov 12, 2018 at 13:34
• Ah thanks for the clarification. I didn't realise Fubini's theorem didn't apply for rectangles of infinite length.
– user150203
Commented Nov 13, 2018 at 5:15
• @DavidG: Fubini doesn't apply since the absolute value of the function is not integrable on $[0,\infty)$. It is integrable on $[0,n]$, so we can apply Fubini there and then limit because of the uniform convergence.
– robjohn
Commented Nov 13, 2018 at 11:21

$$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$$ \begin{align} &\bbox[10px,#ffd]{\int_{0}^{\infty}{\sin\pars{x}\sin\pars{x/3} \over x\pars{x/3}}\,\dd x} = {1 \over 2}\int_{-\infty}^{\infty}{\sin\pars{x} \over x} {\sin\pars{x/3} \over x/3}\,\dd x \\[5mm] = &\ {1 \over 2}\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/3}\dd q}\,\dd x \\[5mm] = &\ {\pi \over 4}\int_{-1}^{1}\int_{-1}^{1}\int_{-\infty}^{\infty} \expo{\ic\pars{k - q/3}x}{\dd x \over 2\pi}\,\dd k\,\dd q = {\pi \over 4}\int_{-1}^{1}\int_{-1}^{1} \delta\pars{k - {q \over 3}}\,\dd k\,\dd q \\[5mm] = &\ {\pi \over 4}\int_{-1}^{1}\bracks{-1 < {q \over 3} < 1}\,\dd q = {\pi \over 4}\int_{-1}^{1}\,\dd q = \bbx{\pi \over 2} \end{align}