What's the value of:$ \int_a^\infty x^{-2}dx, a>0 $ And why it converge?
Convergence of an improper integral - I
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integration
convergence-divergence
improper-integrals
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0I'm sorry, I wrote the wrong integral – 2012-11-18
2 Answers
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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:
$\int_{0}^{\infty} \frac{1}{x^2}dx = \lim_{\epsilon \rightarrow 0} \lim_{N \rightarrow \infty} \int_{\epsilon}^N \frac{1}{x^2}dx = \lim_{\epsilon \rightarrow 0} \lim_{N \rightarrow \infty} -\frac{1}{x}\bigg\vert_{\epsilon}^N = \lim_{\epsilon \rightarrow 0} \lim_{N \rightarrow \infty} \left(-\frac{1}{N} + \frac{1}{\epsilon}\right) = \infty.$