0
$\begingroup$

How can I determine if the following improper integral is converging or diverging

$$ \int_1^{+\infty} \Bigg( \frac{x^2}{4x^4+3x^3+2x^2+x} \Bigg) dx $$ $$ \int_0^{1} \Bigg( \frac{x^2}{4x^4+3x^3+2x^2+x} \Bigg) dx $$

So I see the two improper integral are the same, with differnet bounds. My understanding is to integrate it and then compute the bounds. If it approaches infinity or negative infinity it much diverge. If it approaches a finite number it converges. However i'm having trouble integrating this.

EDIT: I mistakengly put a p as an exponent

$\endgroup$
  • 1
    $\begingroup$ Your integrand is comparable to (and in fact smaller than) $1/x^2$ as $x \to \infty$. Then you can apply the p-test and set $2p>1$. $\endgroup$ – Andrew Li Mar 22 at 0:46
  • $\begingroup$ As a first step, divide the numerator and denominator by $x$ $\endgroup$ – Mark Viola Mar 22 at 0:59
1
$\begingroup$

Hint

$$\frac{x^2}{4x^4+3x^3+2x^2+x}<{x^2\over 4x^4}={1\over 4x^2}$$

$\endgroup$
1
$\begingroup$

The integrand common to these integrals is $O(x)$ for small $x$, finite and positive for $x>0$, and $O(x^{-2})$ for $x<0$, so each integral converges. In particular, if $\epsilon>0$ then $\int_0^\epsilon x^p dx$ converges for $p>-1$, while $\int_\epsilon^\infty x^p dx$ converges for $p<-1$.

$\endgroup$

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.