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Can someone explain to me how to solve the integral \begin{align*} \int_0^1 \sqrt{\log\left(\frac{1}{2x}+1\right)}\mathrm{d} x \end{align*}

I know it is finite by simulation in R

fun <- function(x) sqrt(log(1+1/(2*x)))
integrate(fun, 0, 1)
0.9304908 with absolute error < 5.4e-07

by I also would like to solve it with pen and paper :) Thanks for your help!

wonderer
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  • First thing I tried: integral is equal to $\int_0^1 \sqrt{\log(2x+1) - \log(2x)}$ – gist076923 Mar 01 '23 at 20:56
  • Also, where does this integral appear in your studies? – gist076923 Mar 01 '23 at 20:56
  • Yeah, I tried the same as you! I bumped into that integral when dealing with the log-entropy metric of sets. – wonderer Mar 01 '23 at 21:19
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    Equivalent to $(1/2)\int_{3/2}^{\infty}\sqrt{\ln u}/(u-1)^2 du$. $\sqrt{\ln u}/(u-1)^2$ doesn't look likely to have a closed-form antiderivative. – eyeballfrog Mar 01 '23 at 21:31
  • Mmmm, I see. Let me slightly change the question then. How can I prove that it is finite? – wonderer Mar 01 '23 at 22:29
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    It is finite. You know that $\log (1+a)<a$ for $a>0$. Therefore, $\log (1+1/2x)<1/2x$. Since $\int_0^1 x^{-1/2}dx$ is convergent, your integral is convergent by comparison. – GReyes Mar 02 '23 at 00:54

1 Answers1

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I am not sure that we could find a closed form result but we can use nice bounds using for example the Padé approximants of the upper and lower bounds of $\log(1+x)$ (have a look at Table $1$ here).

Using the second one, after changing $x$ to $\frac 1{2x}$, we have $$\sqrt{\frac{3(4 x+1)}{24x^2+12 x +1}}<\sqrt{\log \left(1+\frac{1}{2 x}\right)}<\sqrt{\frac{12 x+1}{8 x (3 x+1)}}$$ Both lhs and rhs lead to explicit antiderivatives (not very simple) and then definite integrals. Integrated between $0$ and $1$,the decimal representations of the results are $0.91495$ and $1.02867$ while the solution is $0.93049$ as you found.

Claude Leibovici
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