2
$\begingroup$

Is it correct to just consider the asymptotic behaviour of the integrand in an improper integral to determine whether or not it converges?

For example,

$\frac{1}{(x+3)^2}\sim_{\infty}\frac{1}{x^2}$. Since $\int_1^{\infty}\frac{1}{x^2} dx$ converges, can I conclude that $\int_1^{\infty}\frac{1}{(x+3)^2} dx$ does as well?

$\endgroup$
  • 1
    $\begingroup$ That's correct. $\endgroup$ – user63181 Mar 19 '13 at 7:51
2
$\begingroup$

For that example, simply pose $t = x+3$, your integral becomes $$\int_{4}^{\infty} \frac{1}{t^2} \mathrm{d}t$$ which converges. That's it.

But be careful, the asymptotic behaviour of $\frac{1}{(x-3)^2}$ at $x\to\infty$ is also $\frac{1}{x^2}$... But the integral

$$ \int_{1}^{\infty} \frac{1}{(x-3)^2}$$ does not converge because there's an issue at $x=3$.

$\endgroup$

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.