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So this integral reminds me of the Dirichlet integral but I am not sure if I can use similar methods to solve it. I want to prove

$$\int_0^{\infty} \frac{\sin^3(x)}{x^2} dx = \frac{3\ln(3)}{4} $$

I tried parameterizing with

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

or

$$ I(a) := \int_0^{\infty} \frac{\sin^3(x)}{x^2}e^{-ax}dx$$

But none of them worked out for me. Not sure what to do. I would really like to use real methods and not complex analysis, since I haven’t learned it yet.

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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[15px,#ffd]{\int_{0}^{\infty}{\sin^{3}\pars{x} \over x^{2}}\,\dd x} = \int_{0}^{\infty}\sin^{3}\pars{x}\ \overbrace{\pars{\int_{0}^{\infty}t\expo{-xt}\,\dd t}} ^{\ds{1 \over x^{2}}}\ \dd x \\ = &\ \int_{0}^{\infty}t\int_{0}^{\infty}\ \overbrace{3\sin\pars{x} - \sin\pars{3x} \over 4}^{\ds{\sin^{3}\pars{x}}}\ \expo{-tx}\,\dd x\,\dd t \\[5mm] = &\ {1 \over 4}\,\Im\int_{0}^{\infty}t\int_{0}^{\infty} \pars{3\expo{\ic x} - \expo{3\ic x}}\expo{-tx}\dd x\,\dd t \\[5mm] = &\ {1 \over 4}\,\Im\int_{0}^{\infty}t\int_{0}^{\infty} \bracks{3\expo{-\pars{t - \ic}x} - \expo{-\pars{t - 3\ic}x}} \dd x\,\dd t \\[5mm] = &\ {1 \over 4}\int_{0}^{\infty} \pars{{3t \over t^{2} + 1} - {3t \over t^{2} + 9}}\dd t \\[5mm] = &\ {1 \over 4}\bracks{{3 \over 2}\,\ln\pars{t^{2} + 1} - {3 \over 2}\,\ln\pars{t^{2} + 9}}_{\ 0}^{\infty} = {1 \over 4}\braces{{3 \over 2}\bracks{-\ln\pars{1 \over 9}}} \\[5mm] = &\ \bbx{{3 \over 4}\,\ln\pars{3}}\ \approx\ 0.8240 \\ &\ \end{align}

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  • $\begingroup$ That was pretty nice $\endgroup$ – MichaelCatliMath Jul 16 at 20:48
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    $\begingroup$ Thanks. $\displaystyle\large ^{@}\left(\bullet \qquad\bullet \atop \smile\right)^{\! @}$. $\endgroup$ – Felix Marin Jul 18 at 17:01
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    $\begingroup$ You have by far the best formatting 😂 $\endgroup$ – Maximilian Janisch Aug 1 at 0:19
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    $\begingroup$ @MaximilianJanisch Thanks. I like it when it's very clear. $\endgroup$ – Felix Marin Aug 1 at 1:22
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METHODOLOGY $1$: Using the Laplace Transform

Let $I$ be given by the integral

$$I=\int_0^\infty \frac{\sin^3(x)}{x^2}\,dx$$

Appealing to This Theorem of the Laplace Transform, we first note that for $f(x)=\sin^3(x)$ and $g(x)=\frac1{x^2}$ we have $$\begin{align}\mathscr{L}\{f\}(x)&=\frac{6}{x^4+10x^2+9}\tag 1\\\\ \mathscr{L}^{-1}\{g\}(x)&=x\tag2 \end{align}$$ whence using $(1)$ and $(2)$ in the theorem shows that $$\begin{align} I&=\int_0^\infty \frac{\sin^3(x)}{x^2}\,dx\\\\ &=\int_0^\infty \mathscr{L}\{f\}(x)\mathscr{L}^{-1}\{g\}(x)\,dx\\\\ &=\int_0^\infty \frac{6x}{x^4+10x+9}\,dx\\\\ &=\frac34\int_0^\infty\left(\frac{x}{x^2+1}-\frac{x}{x^2+9}\right)\,dx\\\\ &=\frac38\left.\left(\log(x^2+1)-\log(x^2+9)\right)\right|_{0}^\infty\\\\ &=\frac34\log(3) \end{align}$$ as was to be shown.


METHODOLOGY $2$: Using Feynman's Trick

Let $F(s)$ be given by the integral

$$F(s)=\int_0^\infty \frac{\sin^3(x)}{x^2}e^{-sx}\,dx$$

Differentiating $F(s)$ twice, we find that

$$F''(s)=\frac{6}{s^4+10s^2+9}$$

Integrating $F''(s)$ once reveals

$$F'(s)=\frac34 \arctan(s)-\frac14\arctan(s/3)+C_1$$

Integrating $F'(s)$ we find that

$$F(s)=\frac34 s\arctan(s)-\frac38 \log(s^2+1)-\frac14 s\arctan(s/3)+\frac38\log(s^2+9)+C_1s+C_2$$

Using $\lim_{s\to\infty}F(s)=0$, we find that $C_1=-\pi/4$ and $C_2=0$. Setting $s=0$ yields the coveted result

$$\begin{align} F(0)&=\int_0^\infty \frac{\sin^3(x)}{x^2}\,dx\\\\ &=\frac34\log(3) \end{align}$$

as expected!

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  • $\begingroup$ Hi Mark ! May I confess that I never felt comfortable with Laplace transforms ? What a shame at my gae, isn't it ? Cheers. $\endgroup$ – Claude Leibovici Apr 30 at 3:49
  • $\begingroup$ Hi Claude. That little theorem of the Laplace Transform can be useful to evaluate integrals. $\endgroup$ – Mark Viola Apr 30 at 3:51
  • $\begingroup$ That’s an awesome solution. For method two, I got the same exact stuff I guess I messed up when doing the calculations for c1 and c2. I appreciate two methods, very cool. I haven’t learned LaPlace either but I plan on it because it seems extremely useful. $\endgroup$ – MichaelCatliMath Apr 30 at 5:08
  • $\begingroup$ Pleased you liked the solutions. I embedded a link to the theorem from Laplace Transforms that I applied herein. It is a really cool tool to evaluate integrals over the positive reals. $\endgroup$ – Mark Viola Apr 30 at 5:56
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    $\begingroup$ @Aryadeva Thank you! Much appreciate your note. $\endgroup$ – Mark Viola Apr 30 at 16:37
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Frullani Integration $$ \begin{align} \int_0^\infty\frac{\sin^3(x)}{x^2}\,\mathrm{d}x &=\int_0^\infty\frac{3\sin(x)-\sin(3x)}{4x^2}\,\mathrm{d}x\tag1\\ &=\lim_{\substack{a\to0^+\\A\to\infty}}\int_a^A\frac{3\sin(x)-\sin(3x)}{4x^2}\,\mathrm{d}x\tag2\\ &=\frac34\lim_{\substack{a\to0^+\\A\to\infty}}\left(\int_a^A\frac{\sin(x)}{x^2}\,\mathrm{d}x -\int_{3a}^{3A}\frac{\sin(x)}{x^2}\,\mathrm{d}x\right)\tag3\\ &=\frac34\left(\lim_{a\to0^+}\int_a^{3a}\frac{\sin(x)}{x^2}\,\mathrm{d}x-\lim_{A\to\infty}\int_A^{3A}\frac{\sin(x)}{x^2}\,\mathrm{d}x\right)\tag4\\ &=\frac34\left(\lim_{a\to0^+}\int_a^{3a}\left(\frac1x+O(x)\right)\mathrm{d}x-\lim_{A\to\infty}\int_A^{3A}O\!\left(\frac1{x^2}\right)\mathrm{d}x\right)\tag5\\[1pt] &=\frac34\log(3)+\lim_{a\to0^+}O\!\left(a^2\right)-\lim_{A\to\infty}O\!\left(\frac1A\right)\tag6\\[3pt] &=\frac34\log(3)\tag7 \end{align} $$ Explanation:
$(1)$: trig identity
$(2)$: write integral as a limit
$(3)$: separate into two integrals and substitute $x\mapsto x/3$ in the right integral
$(4)$: subtract integrals
$(5)$: $\sin(x)=x+O\!\left(x^3\right)$ as $x\to0$ and $\sin(x)=O(1)$ as $x\to\infty$
$(6)$: integrate
$(7)$: evaluate limit


Note that $(1)$ is the classic Frullani Integral when written as $$\newcommand{\sinc}{\operatorname{sinc}} \frac34\int_0^\infty\frac{\sinc(x)-\sinc(3x)}x\,\mathrm{d}x=\frac34\log(3)\tag8 $$ since $\lim\limits_{x\to0}\sinc(x)=1$ and $\lim\limits_{x\to\infty}\sinc(x)=0$.

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    $\begingroup$ Hi Rob. (+1) If you integrate by parts the integral in $(2)$, the result is a pure Frullani integral. $\endgroup$ – Mark Viola Apr 30 at 21:54
  • $\begingroup$ @MarkViola: no need for Integration by Parts if you think of $(1)$ as $$\newcommand{\sinc}{\operatorname{sinc}} \frac34\int_0^\infty\frac{\sinc(x)-\sinc(3x)}{x}\,\mathrm{d}x $$ $\endgroup$ – robjohn Apr 30 at 22:13
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    $\begingroup$ Indeed; that's even better! In fact, "classical" Frullani result does not directly apply to the cosine function (i.e., $\lim_{x\to \infty}\cos(x)$ fails to exist). So we need to generalize the "classical" result a bit by noting that for $ab>0$, $\lim_{L\to\infty}\int_{aL}^{bL}\frac{\cos(x)}{x}\,dx=0$ by the Cauchy criterion for improper Riemann integrals. $\endgroup$ – Mark Viola Apr 30 at 22:29
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Using real methods $$I(x)=\int \frac{\sin^3(x)}{x^2} dx $$ One integration by parts gives $$I(x)=-\frac{\sin ^3(x)}{x}+3\int \frac{ \sin ^2(x) \cos (x)}{x} \,dx$$ Now $$\sin ^2(x) \cos (x)=\cos(x)-\cos^3(x)= \frac 14 \left(\cos(x)-\cos(3x) \right)$$ $$\int \frac{ \sin ^2(x) \cos (x)}{x} \,dx= \frac 14 \left(\int\frac{ \cos (x)}{x} \,dx -\int\frac{ \cos (3x)}{3x} \,d(3x)\right)$$ $$I(x)=-\frac{\sin ^3(x)}{x}+\frac 3 4\left(\text{Ci}(x)-\text{Ci}(3 x) \right)$$ When $x \to \infty$, $I(x) \to 0$ All of that makes that we have to deal with the limit of $I(x)$ when $x \to0$. A Taylor series gives the expected result.

Edit

Using Taylor series, or, much better, Padé approximants, we can compute with a reasonable accuracy $$\int_a^{\infty} \frac{\sin^3(x)}{x^2} dx=\frac{3\log(3)}4+a^2\frac{-\frac{1}{2}+\frac{10283 }{198840}a^2-\frac{295703 }{83512800}a^4 } {1+\frac{3643 }{24855}a^2+\frac{317893 }{41756400}a^4 }$$ which is quite good for $0 \leq a \leq 2$.

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    $\begingroup$ Hi Claude … (+1) for the nice expansion! $\endgroup$ – Mark Viola Apr 30 at 21:55
  • $\begingroup$ Please what was it the significance of $\text{Ci}$? $\endgroup$ – Sebastiano Apr 30 at 22:03
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    $\begingroup$ @Sebastiano. Cosine integral. $\endgroup$ – Claude Leibovici May 1 at 1:49
  • $\begingroup$ Thank you very much. I never have treated this concept when I was in the university. I think that it is: mathworld.wolfram.com/CosineIntegral.html $\endgroup$ – Sebastiano May 1 at 10:50
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If you write $\frac{1}{x^2}$ as $\int_0^\infty ye^{-xy}dy$ and use $2i\sin x=e^{ix}-e^{-ix}$, the integral becomes$$\begin{align}&\frac{i}{8}\int_{[0,\,\infty)^2}y(e^{-x(y-3i)}-3e^{-x(y-i)}+3e^{-x(y+i)}-e^{-x(y+3i)})dxdy\\&=\frac{i}{8}\int_0^\infty y\left(\frac{1}{y-3i}-\frac{3}{y-i}+\frac{3}{y+i}-\frac{1}{y+3i}\right)dy\\&=\frac34\int_0^\infty\left(\frac{y}{y^2+1}-\frac{y}{y^2+9}\right)dy\\&=\frac38\left[\ln\frac{y^2+1}{y^2+9}\right]_0^\infty=\frac34\ln3.\end{align}$$

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  • $\begingroup$ (+1) Nice answer! I hope that you don't mind that I edited a minor typo. $\endgroup$ – Mark Viola Apr 30 at 21:53
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\begin{align} \int_0^{\infty} \frac{\sin^3(x)}{x^2} dx &= \frac14\int_0^{\infty} (3\sin x- \sin 3x)d(-\frac1x)dx\\ &=\frac34 \int_0^{\infty} \frac{\cos x- \cos 3x}{x}dx\\ &= \frac34 \int_0^\infty dx\int_1^3\sin ux du\\ & =\frac34\int_1^3 du \lim_{t\to0}\int_0^\infty{e^{-t x}\sin u x}\, dx\\ &=\frac34\int_1^3 du \lim_{t\to0}\ \frac u {t^2+u^2} =\frac34\int_1^3 \frac 1u du = \frac34\ln3 \end{align}

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  • $\begingroup$ The interchange of integrals was not legitimate, which is the reason for the "contrived" limiting operation. Using Frullani, you could immediately conclude that $\int_0^\infty \frac{\cos(x)-\cos(3x)}{x}\,dx=\log(3)$. So, why not just do that? $\endgroup$ – Mark Viola Apr 30 at 4:16
  • $\begingroup$ In your edited version, why not just introduce the term $e^{-tx}$ in the Frullani integral and use Feynman's Trick? Equivalently, write $\frac1x=\int_0^\infty e^{-tx}\,dt$. Then, interchange the integrals. It's rigorous and efficient. Aside, the limit in your development must be taken from the right. $\endgroup$ – Mark Viola Apr 30 at 4:33
  • $\begingroup$ @MarkViola - it is not an outright exchange $\endgroup$ – Quanto Apr 30 at 4:33
  • $\begingroup$ Exchanging operations is not always legitimate. In your original development, the interchange was not legitimate and rendered an undefined integral, which you altered with a limiting operation. $\endgroup$ – Mark Viola Apr 30 at 4:35
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    $\begingroup$ Please read the comments. You have an undefined limit. It must be taken from the right. Aside, you could have (1) used Frullani and avoided superfluous work or (2) written $\frac1x=\int_0^\infty e^{-tx}\,dt$ and avoided any need for a questionable interchange of a limit and integrals. $\endgroup$ – Mark Viola Apr 30 at 4:46
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Using this integration
(1)...$\int_{0}^{\infty}\frac{\sin(ax)\sin(bx)}{x}dx=\frac{1}{2}\log({\frac{a+b}{a-b}})$ So :

$$\int_{0}^{\infty}\frac{\sin^{3}(ax)}{x^2}dx=\int_{0}^{\infty}(-\frac{1}{x})^{'}\sin^{3}(ax)dx=[-\frac{\sin^{3}(ax)}{x}]_{0}^{\infty}+\int_{0}^{\infty}\frac{3a\sin^2(ax)\cos(ax)}{x}dx=\frac{3}{2}a\int_{0}^{\infty}\frac{\sin(2ax)\sin(ax)}{x}dx=\frac{3}{2}a\frac{1}{2}\log(\frac{2a+a}{2a-a})=\frac{3}{4}a\log(3)$$ So we put $$a=1$$ we find $$\int_{0}^{\infty}\frac{\sin^{3}(x)}{x^2}dx=\frac{3}{4}\log(3)$$

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