[tex] \int_{-1}^1 \frac{dx}{x} = \lim_{a \rightarrow 0} \int_{-1}^a \frac{dx}{x} + \int_a^1 \frac{dx}{x} = 0 [/tex].(adsbygoogle = window.adsbygoogle || []).push({});

Since 2a also goes to 0 for 'a' going to 0, then we have [tex] \int_{-1}^1 \frac{dx}{x} = \lim_{a \rightarrow 0} \int_{-1}^{2a} \frac{dx}{x} + \int_a^1 \frac{dx}{x} = ln2[/tex].

It seems like the increase in area beneath the curve is because 2a goes to 0 slower than 'a' does. So the area from -1 to 2a is less than the area from 'a' to 1 as 'a' goes to 0, and so when you add the two areas there is a positive value left over, equaling ln2.

My question is though, do we say that the improper integral [tex] \int_{-1}^1 \frac{dx}{x} [/tex] is convergent? I thought that we say it converges if [tex] \lim_{\epsilon \rightarrow \infty} \int_{-1}^{x_{\epsilon}} \frac{dx}{x} + \int_{y_{\epsilon}}^1 \frac{dx}{x} [/tex] where [tex] x_{\epsilon}; y_{\epsilon} [/tex] are both sequences tending to 0, always yields the same value regardless ofhowthey go to 0.

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# Improper integral confusion

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