Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
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 2

up vote 2 down vote accepted

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 $$\int_a^{\infty} x^{-2} dx= \left [-\dfrac1x \right]_{x=a}^{x=\infty} = \left. \dfrac1x \right \vert_{a} = \dfrac1a$$

share|improve this answer

You can compute this integral explicitly:

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.