Integration involving $\sin x/x$

Question

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?

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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

Answer 1

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.

Answer 2

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

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

Answer 3

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