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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

1 Answer 1

up vote 2 down vote accepted

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 $x=3$.

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