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Given the following integral:

$$ \int_0^2 \frac{1}{\ln(x)} dx $$

Does it converge?

Iv'e gone this far: $$ \int_0^2 \frac{1}{\ln(x)} dx = \int_0^1 \frac{1}{\ln(x)} dx + \int_1^2 \frac{1}{\ln(x)} dx $$

Now I'm having trouble calculating each. I can only tell that: $ \int \frac{1}{\ln(x)} dx \gt \int\frac1x dx $

Bérénice
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Squirrel
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  • $\frac{1}{\log(x)}$ is not greater than $1/x$ for $x<1$ since $\log(x)<0$ for $x<1$. – Mark Viola Jun 20 '16 at 15:55
  • Hi! It's been a while. I hope you're staying safe and healthy during the pandemic. I've reached out to contact you a few times, but am unsure whether you've received the notes? If you would, please let me know how I can improve my answer. I really want to give you the best answer I can. And feel free to up vote an answer as you see fit. ;-) – Mark Viola Feb 28 '21 at 19:40

3 Answers3

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PRIMER:

In THIS ANSWER, I showed using only the limit definition of the exponential function along with Bernoulli's Inequality that the logarithm function satisfies the inequalities

$$\frac{x-1}{x}\le \log(x)\le x-1 \tag 1$$

for $x>0$.

Note from $(1)$, that we have for $x>0$, $x\ne1$

$$\frac{x}{x-1} \ge \frac{1}{\log(x)}\ge \frac{1}{x-1}$$

Therefore, the singularity at $x=1$ renders the integral divergent.

However, the Cauchy Principal Value

$$\text{PV}\int_0^2 \frac{1}{\log(x)}\,dx=\lim_{\epsilon \to 0^+}\left(\int_0^{1-\epsilon}\frac{1}{\log(x)}\,dx+\int_{1+\epsilon}^2 \frac{1}{\log(x)}\,dx\right)$$

does converge.

Mark Viola
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  • @Squirrel Hi! It's been a while. I hope you're staying safe and healthy during the pandemic. I've reached out to contact you a few times, but am unsure whether you've received the notes? If you would, please let me know how I can improve my answer. I really want to give you the best answer I can. And feel free to up vote an answer as you see fit. ;-) – Mark Viola Feb 28 '21 at 19:40
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Just continuing from Mark's answer, in terms of Gregory coefficients: $$\begin{eqnarray*}\text{PV}\int_{0}^{2}\frac{dx}{\log(x)} &=& \lim_{\varepsilon\to 0^+}\int_{\varepsilon}^{1}\left(\frac{1}{\log(1+x)}+\frac{1}{\log(1-x)}\right)\,dx\\&=& 2\sum_{n\geq 0}\frac{G_{2n+1}}{2n+1} \\&=&\int_{0}^{+\infty}\frac{\log(x+2)-\log(x)}{\pi^2+\log^2(x)}\,dx\end{eqnarray*}$$ but both the series representation and the integral one are quite slow-converging to $$\text{LogIntegral}(2)=1.04516378\ldots$$

Jack D'Aurizio
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If $1\ne x$ then $\log x < x-1$. You can tell that's true just because the tangent line to $y=\ln x$ at $x=1$ is $y=x-1$ and the graph is concave downward at that point. Therefore $$ \int_1^2 \frac {dx} {\log x} > \int_1^2 \frac {dx} {x-1} = +\infty. $$ In a similar way, one can deduce that $$ \int_0^1 \frac{dx}{\log x} = -\infty. $$

In this sort of situation you can find things like $$ \lim_{\varepsilon\,\downarrow\,0} \left( \int_0^{1-\varepsilon} \frac{dx}{\log x} + \int_{1+\varepsilon}^2 \frac{dx}{\log x} \right) \ne \lim_{\varepsilon\,\downarrow\,0} \left( \int_0^{1-\varepsilon} \frac{dx}{\log x} + \int_{1+2\varepsilon}^2 \frac{dx}{\log x} \right). $$ The limit on the left above is the Cauchy principal value. I think that exists. Maybe I'll be $\text{back }\ldots$

Michael Hardy
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