I'm confused as to why $$\int_{-\infty}^\infty\frac{1}{x}dx$$ diverges. If $\frac{1}{x}$ is an odd function shouldn't the area to the left of origin be the opposite of the area to the right of origin, resulting in a net area of 0?
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1Both "areas" are infinite. $1/x$ is not integrable in any neighborhood of zero because of its singularity there. You can define something called the principal value of the integral, which is zero. – Mar 13 '14 at 23:53
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For some basic information about writing math at this site see e.g. here, here, here and here. – frabala Mar 14 '14 at 00:25
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@user135330 Even if you replace the $x$ by and $x^2$ the integral still diverges. Do you see why? – imranfat Mar 14 '14 at 15:04
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@imranfat Yes because 1/x^2 goes to infinity at x=0 – user135330 Mar 15 '14 at 00:22
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@user135330 yep :) – imranfat Mar 15 '14 at 22:23
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As an improper integral, $\int_{-\infty}^{\infty} \frac{1}{x} dx = \lim_{N \to \infty} \int_{-N}^{ - \frac{-1}{N}}\frac{1}{x} dx + \lim_{M \to \infty} \int_{\frac{1}{M}}^{M}\frac{1}{x} dx$
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1Technically, even that is a simplification, since the limits of each of your two integrals should vary independently, so there should be four limits in all. – Mar 14 '14 at 00:51
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Could it be something like this $ \int_{-\infty}^{\infty} \frac{1}{x} dx = \int_{-\infty}^{-1} \frac{1}{x} dx + \int_{-1}^{0} \frac{1}{x} dx + \int_{0}^{1} \frac{1}{x} dx + \int_{1}^{\infty} \frac{1}{x} dx$ ? – john May 30 '17 at 02:39