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

Question

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.

Answer 1

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!

Answer 2

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

Answer 3

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 $0 \leq a \leq 2$.

Answer 4

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$$\begin{align}&\frac{i}{8}\int_{[0,\,\infty)^2}y(e^{-x(y-3i)}-3e^{-x(y-i)}+3e^{-x(y+i)}-e^{-x(y+3i)})dxdy\\&=\frac{i}{8}\int_0^\infty y\left(\frac{1}{y-3i}-\frac{3}{y-i}+\frac{3}{y+i}-\frac{1}{y+3i}\right)dy\\&=\frac34\int_0^\infty\left(\frac{y}{y^2+1}-\frac{y}{y^2+9}\right)dy\\&=\frac38\left[\ln\frac{y^2+1}{y^2+9}\right]_0^\infty=\frac34\ln3.\end{align}$$

Answer 5

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}

Answer 6

\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}

Answer 7

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 $$\int_{0}^{\infty}\frac{\sin^{3}(x)}{x^2}dx=\frac{3}{4}\log(3)$$