# Convergence of an improper integral - I

What's the value of:$$\int_a^\infty x^{-2}dx, a>0$$ And why it converge?

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It doesn't converge. –  Gautam Shenoy Nov 18 '12 at 16:50
I'm sorry, I wrote the wrong integral –  Pizzirani Leonardo Nov 18 '12 at 16:53

## 2 Answers

The integral doesn't converge. $$\int x^{-2} dx = \dfrac{x^{-2+1}}{-2+1} + C = - \dfrac1x+C$$ Hence, for $a>0$, we have that $$\int_a^{\infty} x^{-2} dx= \left [-\dfrac1x \right]_{x=a}^{x=\infty} = \left. \dfrac1x \right \vert_{a} = \dfrac1a$$

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You can compute this integral explicitly:

$$\int_{0}^{\infty} \frac{1}{x^2}dx = \lim_{\epsilon \rightarrow 0} \lim_{N \rightarrow \infty} \int_{\epsilon}^N \frac{1}{x^2}dx = \lim_{\epsilon \rightarrow 0} \lim_{N \rightarrow \infty} -\frac{1}{x}\bigg\vert_{\epsilon}^N = \lim_{\epsilon \rightarrow 0} \lim_{N \rightarrow \infty} \left(-\frac{1}{N} + \frac{1}{\epsilon}\right) = \infty.$$

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