representing $$I_1=\int_0^\infty{\frac{\sin x}{x^3}}dx$$ in terms of $$I_2=\int_0^\infty{\frac{\sin^3x}{x^3}}dx$$
I used $\sin(3x)=3\sin(x)-4\sin^3(x)$ and put the value of $\sin^3(x)$ in $I_2$ got $$I_2 = \int_0^\infty{\frac{3\sin(x)-\sin(3x)}{4x^3}}dx$$ $$I_2 = \int_0^\infty{\frac{3\sin(x)}{4x^3}}dx - \int_0^\infty{\frac{\sin(3x)}{4x^3}}dx$$ Now I replaced $x$ with $x/3$ in second integral: $$I_2 = \int_0^\infty{\frac{3\sin(x)}{4x^3}}dx - \int_0^\infty{\frac{9\sin(x)}{4x^3}}dx$$ This gives that $$I_2=\frac{-3}{2}\int_0^\infty{\frac{\sin x}{x^3}}dx$$
But according to Wolfram alpha LHS converges but RHS diverges. Can anyone point out where am I doing wrong?
$I_{1}$ does not exist (in Lebesgue sense nor improper sense: $\lim_{a\rightarrow0+}\int_{a}^{\infty}\frac{\sin x}{x^{3}}dx$). The reason is that when $x$ is small, $\sin x\approx x$. However, $\lim_{a\rightarrow0+}\int_{a}^{b}\frac{dx}{x^{2}}=+\infty$.