2

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?

Hypercube
  • 576
  • 1
  • 5
  • 14

1 Answers1

2

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 $x=3$.

mwoua
  • 865