Does this prove that $y={1\over x}$ has no midpoint?

Question

I was looking at the area under $y={1\over x}$, and randomly I wondered if there was some point at which the areas on either side were equal.

I set about finding the answer using this method: $$\\ \lim_{a\to0^+} {\int_a^p{dx\over x}}=\lim_{b\to\infty} {\int_p^b{dx\over x}}\\ \lim_{a\to0^+}\ln p - \ln a=\lim_{b\to\infty}\ln b - \ln p\\ 2\ln p=\lim_{a\to0^+}\ln a+\lim_{b\to\infty} \ln b\\ 2\ln p = -\infty+\infty\\ $$ Upon seeing the last line, I concluded that given the indeterminate form, $p\in(0,\infty)$, which I interpreted to mean "$p$ does not exist"

Is this rigorous enough, or did I skip some steps?

Answer 1

We have that $\lim_{a\to0^+} {\int_a^p{dx\over x}}$ diverges and $\lim_{b\to\infty} {\int_p^b{dx\over x}}$ diverges too. So, since these two areas tend to $\infty$, you cannot compare those two. Hence you cannot find a point where these two areas are equal.

From the indeterminate that you found you cannot conclude the non-existence of $p$ cause an indeterminate form of $\infty -\infty$ with some manipulations can lead to a number or not.

Answer 2

Your statement that $p$ does not exist is correct.

You know that the total integral is $\infty$ and you are looking for a point which divides infinity in two equal parts.

Now if the point $p$ is a real point,then the first half of the integral will be bounded and can not be infinity.

Thus even without integration we notice that such a point does not exist.

The same argument goes for every integral which diverges to infinity.