Is 1/x Integrable on Intervals Containing 0?

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Homework Statement



Prove that 1/x is NOT integrable on any intervals containing 0

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The Attempt at a Solution



would it be sufficient to say that the anti-derivative, ln x, blows up at 0? would this answer be rigorous enough for an analysis course?

or do i have to use some kind of partition argument?
 
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You mean you can't say that since 1/x AND ln(x) do not exist at 0, it cannot be integrated over 0?
 
I believe you have to show that you can find a partition such that the upper sum is greater than any alpha ( > 0).

Basically, you show that the integral over zero can get as large as you want.
 
sorry, i have to show that its not improper riemann integrable
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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