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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).

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3 Answers 3

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

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  • $\begingroup$ 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? $\endgroup$
    – user150203
    Commented Nov 12, 2018 at 8:36
  • $\begingroup$ @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 $\endgroup$
    – Yuriy S
    Commented Nov 12, 2018 at 8:39
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    $\begingroup$ There is bad math that leads to correct answers. But I do take your point. Re complicated - I agree. $\endgroup$
    – user150203
    Commented Nov 12, 2018 at 9:13
  • $\begingroup$ @DavidG, I agree with you as well :) $\endgroup$
    – Yuriy S
    Commented Nov 12, 2018 at 9:14
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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 $x\mapsto x/a$
$(4)$: $\int_0^\infty\frac{\sin(x)}x\,\mathrm{d}x=\frac\pi2$

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    $\begingroup$ Sorry, this is probably being pedantic, but do you have include that you employed Fubini's Theorem as part of the explanation? $\endgroup$
    – user150203
    Commented Nov 12, 2018 at 10:35
  • $\begingroup$ @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. $\endgroup$
    – robjohn
    Commented Nov 12, 2018 at 13:34
  • $\begingroup$ Ah thanks for the clarification. I didn't realise Fubini's theorem didn't apply for rectangles of infinite length. $\endgroup$
    – user150203
    Commented Nov 13, 2018 at 5:15
  • $\begingroup$ @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. $\endgroup$
    – robjohn
    Commented Nov 13, 2018 at 11:21
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}

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