If $\int_ {0}^ \infty (\frac{ \sin x}{x})^3 = A$, then $\int_ {0}^ \infty \frac{ x- \sin x}{x^3} = k A $. Find the value of k
The main problem is how to get started as in the given integral , there is a term of $(\sin x)^3$, where as in the integral to be computed , there is no cube. Please help :)
Edit :I'm still in highschool so don't know how to compute this integral and that is why the first one was given
As $\sin^3 x=(3\sin x-\sin 3x)/4$, we have \begin{align} A&=\int_{0}^{\infty}\frac{(3x-\sin 3x)-3(x-\sin x)}{4x^3}\,dx \\ &=\frac{1}{4}\int_{0}^{\infty}\frac{3x-\sin 3x}{x^3}\,dx-\frac{3}{4}\int_{0}^{\infty}\frac{x-\sin x}{x^3}\,dx \\ &=\frac{9}{4}\int_{0}^{\infty}\frac{y-\sin y}{y^3}\,dy-\frac{3}{4}\int_{0}^{\infty}\frac{y-\sin y}{y^3}\,dy \\ &=\frac{3}{2}\int_{0}^{\infty}\frac{x-\sin x}{x^3}\,dx. \end{align} Thus $k=2/3$.
We know that $-\sin^3(x)=\frac{\sin(3x)-3\sin(x)}{4}$ thus $-\int_0^{\infty} \frac{\sin^3(x)}{x^3}dx=\frac{1}{4}\int_0^{\infty} \frac{3(x-\sin(x))-(3x-\sin(3x))}{x^3}dx$ putting in the second bracket $3x=u$ and again letting $u=x$ without any loss of generality we have $$-4A=3\int_0^{\infty}\frac{x-\sin(x)}{x^3}dx-\int_0^{\infty} \frac{u-\sin(u)}{\frac{u^3}{27}}\frac{du}{3}$$ let the unknown integral have value $k$ thus we have $-4A=-6k$ hence we have $\int_0^{\infty} \frac{x-\sin(x)}{x^3}dx=\frac{2A}{3}$
