# convergence test for improper integrals

in an old book on calculus in the chapter about improper integrals there was this theorem that says when $$\int_{a}^{\infty }f(x)dx$$ can be written in the form $$\int_{a}^{\infty} \frac{\phi (x)}{x^k}dx$$ you can test for convergence by verifying that $$|\phi (x)| and $$k>1$$ what i don't understand is how do we write $$f(x)$$ in that form what is the choice of $$\phi(x)$$and$$x^k$$ depend on for example
$$\int_{0}^{\infty}e^{-x^2}dx$$ can be written as $$\frac{x^2 e^{-x^2}}{x^2}$$ and therefor converges but why did we choose$$x^2$$not $$x$$ for example ?

Because one of the assumptions is that $$k>1$$. Besides, although you can indeed write $$e^{-x^2}$$ as $$\frac{xe^{-x^2}}x$$, since $$\int_a^\infty\frac{\mathrm dx}x$$ diverges, you deduce nothing from that.