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When providing a counterexample for $0<f(x)=O(x^{1+\varepsilon})\;\forall\epsilon>0\Longrightarrow \int_{1}^{+\infty}\frac{dx}{f(x)}=+\infty$
I realized that the error of the approximation $$ \mathcal{J}=\int_{1}^{+\infty}\frac{dx}{x\log^2(x+1)}\approx 2$$ is less than $7\cdot 10^{-3}$.

Q: Is such approximate identity just a numerical coincidence, or is there some reason for expecting in advance that the RHS is very close to $2$?

Of course from $$ \mathcal{J}=\int_{0}^{+\infty}\frac{2\,dt}{\left(t+\log\left(2\cosh t\right)\right)^2}$$ it is not hard to believe that $\mathcal{J}\approx\int_{0}^{+\infty}\frac{2\,dt}{(t+\log 2)^2}=\frac{2}{\log 2}$ or that $$ \mathcal{J}\approx\int_{0}^{+\infty}\frac{2\,dt}{\left(t+\log(2)-1+\sqrt{1+t^2}\right)^2}$$ where the RHS has an explicit form involving a lot of $(\log 2)$s, namely $$ \frac{2\log\log 2+\frac{2}{\log 2}-\log^2 2+2\log 2-3}{(1-\log 2)^3}=2.014532719\ldots$$ Small addendum: by exploiting $\frac{1}{\log^2(x+1)}=\int_{0}^{+\infty}s(x+1)^{-s}\,ds$ we also have $$ \mathcal{J}= \int_{0}^{+\infty}{}_2 F_1\left(s,s;s+1;-1\right)\,ds$$ where the integrand function is way less elementary but much more well-behaved.

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    $\begingroup$ Are you sure it IS close to 2? When I integrate from $1$ to $1,000,000$ I get $1.921177267644058$ $\endgroup$ – sku Mar 9 '18 at 18:55
  • $\begingroup$ What is even more strange is this: $$\int_1^{10^9} f(x)\ dx \approx 1.9453$$ $$\int_1^{10^{12}} f(x)\ dx \approx 1.95737$$ $$\int_1^{10^{17}} f(x)\ dx \approx 1.9453$$ $$\int_1^{10^{21}} f(x)\ dx \approx 1.96212$$ $$\int_1^{10^{23}} f(x)\ dx \approx 2.01638$$ $$\int_1^{10^{40}} f(x)\ dx \approx 0.841742$$ and so on. $\endgroup$ – Von Neumann Mar 9 '18 at 19:00
  • $\begingroup$ @sku: for obvious reasons, the original integral is extremely slow-convergent. A numerical evaluation is simpler by using the second form of $\mathcal{J}$, and we actually have $$ \mathcal{J}=1.99355968\ldots $$ $\endgroup$ – Jack D'Aurizio Mar 9 '18 at 19:17
  • $\begingroup$ @VonNeumann: which procedure have you used to compute such approximations? I guess numerical errors are accumulating there, beyond the $10^{20}$ threshold. $\endgroup$ – Jack D'Aurizio Mar 9 '18 at 19:19
  • $\begingroup$ The explicit approximation we get by using the second form and replacing $\log\cosh(t)$ with $\sqrt{1+t^2}-1$ is $$ \frac{2\log\log 2+\frac{2}{\log 2}-\log^2 2+2\log 2-3}{(1-\log 2)^3}=2.014532719\ldots$$ $\endgroup$ – Jack D'Aurizio Mar 9 '18 at 19:23
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Thoughts:

  1. I think it is just a mathematical coincidence. A few months ago, I asked this question about integrating $x^{-x}$ from $0$ to $+\infty$ and you gave a rather neat answer showing that it is just below the threshold of $2$ - note that Wolfram has the exact value of $1.9954$ ($5$.d.p.). I suspect that it is the same with your integral, and many other variants.

  2. Integrating by parts with $f(x)=\log^{-2}(x+1)$ and $g'(x)=x^{-1}$ gives $$\mathcal{J}=\left[\frac{\log x}{\log^2(x+1)}\right]_1^{+\infty}+2\int_1^{+\infty}\frac{\log x}{(x+1)\log^3(x+1)}\,dx$$ so it is a case of showing that $$\mathcal{K}=\int_1^{+\infty}\frac{\log x}{(x+1)\log^3(x+1)}\,dx=\int_2^{+\infty}\frac{\log(u-1)}{u\log^3u}\,du\approx1$$ But unfortunately, there is no closed form of the indefinite integral of $\mathcal{K}$.

  3. We can attempt to find a Maclaurin series for the integrand. It is not that tedious actually since the denominator fits nicely with the standard series of $\log(x+1)$. The first three terms are $$\frac1{x^3}+\frac1{x^2}+\frac1{12x}$$ but note that this diverges due to the final term. However, integrating this from $1$ to $500$ gives a value of around $2.02$.

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