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

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.

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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!

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

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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 $\lim\limits_{x\to\infty}\sinc(x)=0$.

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 $0 \leq a \leq 2$.