0
$\begingroup$

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)|<M $ 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 ?

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.