Integration involving $\sin x/x$ Let $f$ be a differentiable function satisfying:
$$\int_0^{f(x)}f^{-1}(t)dt-\int_0^x(\cos t-f(t))dt=0$$
and $f(\pi)=0$, Considering $g(x)=f(x)\forall x\in\mathbb R_0=\mathbb R+{0}$.
If $$\int_0^{\infty}(g(x))^3dx=A$$ and $$\int_0^{\infty}\frac{1-g(x)}{x^2}dx=\frac{kA}{k+1}$$ then k is?

First I did :
$$\int_0^{f(x)}f^{-1}(t)dt-\int_0^x(\cos t-f(t))dt=0\\\int_0^{f(x)}f^{-1}(t)dt+\int_0^xf(t)dt=\int_0^x\cos tdt\\xf(x)=\sin x$$
So $f(x)=\dfrac{\sin x}x$
But how can someone calculate?
$$\int_0^{\infty}\frac{\sin^3x}{x^3}dx$$
NB, limit to highschool level
 A: We have $4\sin(x)^3=3\sin(x)-\sin(3x)$. Let $\varepsilon>0$.  Then let
$$4A_{\varepsilon}=3\int_{\varepsilon}^{+\infty}\frac{\sin(x)}{x^3}dx-\int_{\varepsilon}^{+\infty}\frac{\sin(3x)}{x^3}$$
We have by putting $3x=u$ in the last integral
$$4A_{\varepsilon}=3\int_{\varepsilon}^{+\infty}\frac{\sin(x)}{x^3}dx-9\int_{3\varepsilon}^{+\infty}\frac{\sin(u)}{u^3}du$$
Hence
$$4A_{\varepsilon}=3\int_{\varepsilon}^{+\infty}\frac{\sin(x)-x}{x^3}dx+3\int_{\varepsilon}^{+\infty}\frac{dx}{x^2}-9\int_{3\varepsilon}^{+\infty}\frac{\sin(x)-x}{x^3}-9\int_{3\varepsilon}^{+\infty}\frac{dx}{x^2}$$
and:
$$4A_{\varepsilon}=-3\int_{\varepsilon}\frac{1-g(x)}{x^2}dx+9\int_{3\varepsilon}\frac{1-g(x)}{x^2}dx$$
Now if we let $\varepsilon\to 0$, we get
$$4\int_0^{+\infty}(\frac{\sin(x)}{x})^3dx=6\int_{0}\frac{1-g(x)}{x^2}dx$$
and we are done.
A: You can use this way to evaluate. Let
$$ h(a)=\int_0^\infty e^{-ax}\frac{\sin^3x}{x^3}dx, a\ge 0 $$
Then $h(\infty)=h'(\infty)=h''(\infty)=0$ and
\begin{eqnarray}
h'''(a)&=&-\int_0^\infty e^{-ax}\sin^3xdx\\
&=&-\frac{3}{4}\int_0^\infty e^{-ax}\sin xdx+\frac{1}{4}\int_0^\infty e^{-ax}\sin(3x)dx\\
&=&-\frac{3}{4}\frac{1}{a^2+1}+\frac{1}{4}\frac{3}{a^2+9},
\end{eqnarray}
where
$$ \int_0^\infty e^{-ax}\sin(bx)dx=\frac{b}{a^2+b^2}. $$
Then integrating three times, you can get
$$ \int_0^\infty\frac{\sin^3x}{x^3}dx=\frac{3\pi}{8}.$$
A: 
But how can someone calculate $\quad\quad\quad\displaystyle\int_0^{\infty}\frac{\sin^3x}{x^3}dx$

Assuming that you know $$\int_0^{\infty}\frac{\sin x}{x}=\frac{\pi}{2}$$

Use$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\displaystyle\sin^3 x=\frac{3}{4}\sin x - \frac{1}{4}\sin 3x$

And then Integrate by parts
For example
$$\int\frac{\sin^3xdx}{x^3}=\int\frac{3\sin xdx}{4x^3}-\int\frac{\sin xdx}{4x^3}$$
$$\int\frac{\sin xdx}{x^3}=-\frac{\sin x}{2x^2}-\int\frac{\cos x}{2x^2}dx$$
$$\int\frac{\cos x}{x^2}dx=-\frac{\cos x}{x}+\int\frac{\sin x}{x}dx$$
This method seems reasonable as $\text{Mathematica } 10.0.1.0 $ yields follwing Indefinite Integral
$$\int\frac{\sin^3x}{x^3}dx=\frac{-3 x^2 \text{Si}(x)+9 x^2 \text{Si}(3 x)-4 \sin ^2(x) (\sin (x)+3 x \cos (x))}{8 x^2}+c$$
Pat your back if you get

$$\large\int_0^{\infty}\frac{\sin^3x}{x^3}dx=\frac{3\pi}{8}$$

