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Why $\log|x|$ is a locally integrable function and $1/|x|$ is not? Id know how their graphs look like but I don't know what is the exact difference causing local integrability of the first one.

luka5z
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    Note first, that the problematic part is only around $0$. $\log |x|$ is dominated by any positive power of $|x|$, in particular $-\log |x|=\log \frac{1}{|x|}\le C \frac{1}{|x|^{1/2}}$. But the latter is integrable at $0$ (prove by looking at the primitive of $x^{-1/2}$, which is $x^{1/2}$ up to constants). – J.R. Jan 28 '14 at 16:18
  • Thanks. I just read some articles that claims that the first functions has mild singularity at 0 so it is locally integrable, and the second function has strong singularity and it isn't. I can't spot the difference in their graphs which makes one integrable and the second one not... – luka5z Jan 28 '14 at 16:22
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    You cannot see it in the graphs so easily, the thing is that $\log x$ goes to $-\infty$ really slowly as $x\rightarrow 0$, while $-1/x$ does it really fast. You can see this by plugging in say $x=10^{-100}$. Then $\log x=-100$, while $-1/x=-10^{100}$ (a number with $100$ zeros!). – J.R. Jan 28 '14 at 16:24
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    The function $1/|x|^\alpha$ is locally integrable at $0$ if and only if $\alpha<1$. $1/|x|$ is the first example (the "slowest singularity") where it fails. The proof is by going close to $0$ up to some $\epsilon$ and computing the integral using a primitive (antiderivative) and then taking the limit as $\epsilon\rightarrow 0$. – J.R. Jan 28 '14 at 16:26
  • And how to prove that $\log|x|$ is locally integrable? Calculate antiderivative and the see what happens at $0$? Use d'Hospital? – luka5z Jan 28 '14 at 16:29
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    I just did that in the first comment. Compare with $1/|x|^{1/2}$. – J.R. Jan 28 '14 at 16:30
  • Oh, sorry, I forgot. Thank you a lot. – luka5z Jan 28 '14 at 16:31
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    Ok, so maybe I better post all this as an answer, so this question can be considered done. – J.R. Jan 28 '14 at 16:32
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    Never mind, somebody else took it over. – J.R. Jan 28 '14 at 16:32
  • sorry... didn't mean to scoop you. I think we were probably typing simultaneously, me in the answers, you in the comments. – BaronVT Jan 28 '14 at 16:43

2 Answers2

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First: there are two obstacles to global integrability here: behavior at $0$ and behavior at $\infty$. The function $1/|x|$ fails in both regards, that is

$$ \int_0^b 1/|x| \,dx = \infty $$

for any $b$, and

$$ \int_a^\infty 1/|x|\,dx = \infty $$

for any $a$.

Since we're only concerned with local integrability, we only examine the first situation (indeed, $\log$ will fail global integrability). Further, since it is clear that $\int_a^b \,dx$ will be finite for any $a>0$, $b<\infty$ for either of these functions, we only have to look at the local integrability around $a = 0$.

Typically one evaluates improper integrals in the following way:

$$ \lim_{\epsilon \to 0} \int_\epsilon^b 1/|x|\,dx = \lim_{\epsilon \to 0}\log x|_\epsilon^b= \log(b) - \lim_{\epsilon \to 0}\log\epsilon = \log b - (-\infty) = \infty $$

so this failed to be integrable in a neighborhood of $0$.

On the other hand, for $\log$, $$ \lim_{\epsilon \to 0} \int_\epsilon^b \log x\,dx = \lim_{\epsilon \to 0}\left[x\log x - x\right]_\epsilon^b= b\log b - b - \lim_{\epsilon \to 0}\left(\epsilon\log\epsilon - \epsilon\right) = b\log b - b - \lim_{\epsilon \to 0} \epsilon \log \epsilon$$

finally, you just need to prove (you can use L'Hopital's rule) $$ \lim_{\epsilon \to 0} \epsilon \log \epsilon = 0 $$ to show that this integral converges.

BaronVT
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To show that $\log|x|$ is locally integrable in $\mathbf R^d$ for $d \geqslant 2$, we can use a change of variables to spherical coordinates to deal with the singularity at $0$. Let $B$ be the ball of radius $1$ centered at $0$ in $\mathbf R^d$, and $\sigma$ the usual surface measure on the sphere $S^{d-1}\subset\mathbf R^d$. Then we have \begin{align*} \int_B\log|x|\,\mathrm dx &= \int_{S^{d-1}}\int_0^1\log (r)\, r^{d-1}\,\mathrm dr\,\mathrm d\sigma \\ &= \sigma\big(S^{d-1}\big)\int_0^1\log(r)\,r^{d-1}\,\mathrm dr. \end{align*} Because $d \geqslant 2$, we have $d-1 \geqslant 1$, hence $r^{d-1} \leqslant r$ for all $r\in [0,1]$. Hence for all $r\in (0,1]$, we have the upper bound $$0\leqslant -\log(r)\,r^{d-1} \leqslant -\log(r)\,r,$$ so we can conclude that the integral $\log |x|$ converges absolutely on $B$, and quickly conclude that $\log |x|$ is locally integrable on $\mathbf R^d$ for $d\geqslant 2$.

Alex Ortiz
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