Integrate $\int^\infty_0 \dfrac{\sin x}{x^3}\,dx$

Question

By using this method we can evaluate $\displaystyle\int^\infty_0\dfrac{x-\sin x}{x^3}\,dz=\dfrac{\pi}{4}$ and I intended to solve $\displaystyle\int^\infty_0 \dfrac{\sin x}{x^3}\,dx=\displaystyle\int^\infty_0\dfrac{1}{x^2}\,dx-\dfrac{\pi}{4}$. But how can I integrate $\displaystyle\int^\infty_0\dfrac{1}{x^2}\,dx$ when the integral does not converge?

Answer 1

You must keep in mind that an equality like $$ \int_A (f(x)+g(x))\,dx = \int_A f(x)\, dx + \int_A g(x)\,dx $$

only holds if all three integrals exist. In your example, as it was pointed out in the comments, this is not the case.