# Convergence of improper integrals and asymptotic behaviour

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?

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That's correct. –  Sami Ben Romdhane Mar 19 '13 at 7:51

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 $x=3$.