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The following integral was proposed in a paper by O. Furdui, namely

$$\int_0^1 \log^2(\sqrt{1+x}-\sqrt{1-x}) \ dx$$

and then the generalization

$$\int_0^1 \log^2(\sqrt[k]{1+x}-\sqrt[k]{1-x}) \ dx$$

As regards the first integral, my approach was to combine the integration by parts and the variable change, that is $\sqrt{1+x}-\sqrt{1-x} \mapsto x$, and then we get another integral that can be decomposed into $2$ integrals where the hardest part gets reduced to computing $$\int_0^{\sqrt{2}/2}\frac{\arcsin(x)}{x} \ dx$$
that is pretty straightforward by variable change combined with the integration by parts.

And here is a supplementary question

$$\int_0^1 \log^3(\sqrt{1+x}-\sqrt{1-x}) \ dx$$

I'd be also interested in other approaching ways if possible.

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  • $\begingroup$ Would this by any chance be helpful ? Or double differentiation under the integral sign with regard to a parametric exponent ? $($Probably not, but these are the only things that come to mind$\ldots)$ $\endgroup$ – Lucian Oct 31 '14 at 23:44
  • $\begingroup$ @Lucian That method does work... if you are willing to keep up with messy algebra. $\endgroup$ – Ali Caglayan Nov 1 '14 at 0:53
  • $\begingroup$ I think that $$\int_0^1 \log^k\left(\sqrt[k]{1+x}-\sqrt[k]{1-x}\right) dx$$is a cute generalisation. $\endgroup$ – Ali Caglayan Nov 1 '14 at 0:55
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Let $\sqrt{1+x}-\sqrt{1-x}\mapsto2\sin{x}$. We get, for the first integral, \begin{align} \color{#6F00FF}{\int^1_0\ln^2\left(\sqrt{1+x}-\sqrt{1-x}\right)\ {\rm d}x} =&\int^\frac{\pi}{4}_02\cos{2x}\ln^2(2\sin{x})\ {\rm d}x\\ =&\sin{2x}\ln^2(2\sin{x})\Bigg{|}^\frac{\pi}{4}_0-\color{#FF4F00}{\int^\frac{\pi}{4}_02\left(1+\cos{2x}\right)\ln(2\sin{x})\ {\rm d}x}\\ =&\frac{1}{4}\ln^2{2}+2\sum^\infty_{n=1}\frac{1}{n}\int^\frac{\pi}{4}_0\cos(2nx)\ {\rm d}x-\sin{2x}\ln(2\sin{x})\Bigg{|}^\frac{\pi}{4}_0\\ &+\int^\frac{\pi}{4}_02\cos^2{x}\ {\rm d}x\\ =&\frac{1}{4}\ln^2{2}+\sum^\infty_{n=1}\frac{\sin(n\pi/2)}{n^2}-\frac{1}{2}\ln{2}+\left[x+\frac{1}{2}\sin{2x}\right]^\frac{\pi}{4}_0\\ =&\color{#6F00FF}{\mathbf{G}+\frac{1}{4}\ln^2{2}-\frac{1}{2}\ln{2}+\frac{\pi}{4}+\frac{1}{2}} \end{align} For the third integral, \begin{align} \int^1_0\ln^3\left(\sqrt{1+x}-\sqrt{1-x}\right)\ {\rm d}x =&\int^\frac{\pi}{4}_02\cos{2x}\ln^3(2\sin{x})\ {\rm d}x\\ =&\sin{2x}\ln^3(2\sin{x})\Bigg{|}^\frac{\pi}{4}_0-\int^\frac{\pi}{4}_03(1+\cos{2x})\ln^2(2\sin{x})\ {\rm d}x\\ =&\frac{1}{8}\ln^3{2}-{\rm Re}\int^\frac{\pi}{4}_03\ln^2(1-e^{i2x})\ {\rm d}x-\int^\frac{\pi}{2}_\frac{\pi}{4}3x^2\ {\rm d}x\\ &-\frac{3}{2}\sin{2x}\ln^2(2\sin{x})\Bigg{|}^\frac{\pi}{4}_0+\color{#FF4F00}{\int^\frac{\pi}{4}_06\cos^2{x}\ln(2\sin{x})\ {\rm d}x}\\ =&-\frac{3\mathbf{G}}{2}+\frac{1}{8}\ln^3{2}-\frac{7\pi^3}{64}-\frac{3}{8}\ln^2{2}+\frac{3}{4}\ln{2}-\frac{3\pi}{8}-\frac{3}{4}\\ &-\Re\int^\frac{\pi}{4}_03\ln^2(1-e^{i2x})\ {\rm d}x \end{align} The remaining integral can be evaluated like so \begin{align} -{\rm Re}\int^\frac{\pi}{4}_03\ln^2(1-e^{i2x})\ {\rm d}x =&-\frac{3}{2}{\rm Im}\int^i_1\frac{\ln^2(1-z)}{z}{\rm d}z\\ =&{\rm Im}\left\{-\frac{3}{2}\ln{i}\ln^2(1-i)-3\int^i_1\frac{\ln{z}\ln(1-z)}{1-z}{\rm d}z\right\}\\ =&\frac{3\pi^3}{64}-\frac{3\pi}{16}\ln^2{2}-{\rm Im}\Bigg{\{}3{\rm Li}_2(1-i)\ln(1-i)-3{\rm Li}_3(1-i)\Bigg{\}} \end{align} Plucking $z=i$ in the dilogarithm reflection formula, $$\underbrace{{\rm Li}_2(i)}_{\displaystyle\small{-\frac{\pi^2}{48}+i\mathbf{G}}}+{\rm Li}_2(1-i)=\frac{\pi^2}{6}\underbrace{-\ln{i}\ln(1-i)}_{\displaystyle\small{-\frac{\pi^2}{8}-i\frac{\pi}{4}\ln{2}}}$$ which implies $${\rm Li}_2(1-i)=\frac{\pi^2}{16}-i\left(\mathbf{G}+\frac{\pi}{4}\ln{2}\right)$$ Thus we have $$-{\rm Re}\int^\frac{\pi}{4}_03\ln^2(1-e^{i2x})\ {\rm d}x=3\Im{\rm Li}_3(1-i)+\frac{3\mathbf{G}}{2}\ln{2}+\frac{3\pi^3}{32}+\frac{3\pi}{16}\ln^2{2}$$ Finally, \begin{align} \color{#6F00FF}{\int^1_0\ln^3\left(\sqrt{1+x}-\sqrt{1-x}\right)\ {\rm d}x =}&\color{#6F00FF}{3\Im{\rm Li}_3(1-i)+\frac{3\mathbf{G}}{2}\ln{2}-\frac{3\mathbf{G}}{2}-\frac{\pi^3}{64}+\frac{3\pi}{16}\ln^2{2}}\\ &\color{#6F00FF}{+\frac{1}{8}\ln^3{2}-\frac{3}{8}\ln^2{2}+\frac{3}{4}\ln{2}-\frac{3\pi}{8}-\frac{3}{4}} \end{align}

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    $\begingroup$ Very nice, I love to have many approaches in place. (+1) $\endgroup$ – user 1357113 Nov 10 '14 at 12:55
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    $\begingroup$ +1) Nice answer. And I note with no small curiosity that this same imaginary part of a trilogarithm term has once again reared its ugly head to stand in the way of what would otherwise be a pleasant closed form solution. Perhaps we should just assign it a Greek letter and move on? =) $\endgroup$ – David H Nov 10 '14 at 13:08
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    $\begingroup$ Wow. And an intersting trilogarithm imaginary part again! +1 by the way. $\endgroup$ – user153012 Nov 10 '14 at 13:08
  • $\begingroup$ Thanks for all your compliments. I guess the trilogarithm will have to stay that way, as I prefer it to its hypergeometric form (The latter is way uglier in my opinion). $\endgroup$ – M.N.C.E. Nov 10 '14 at 13:14
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    $\begingroup$ @M.N.C.E. Fantastic solution ! (+1) :-) I think it appears as Problem 253 in the latest issue of La Gaceta de la RSME, proposed by O. Furdui :) .. consider sending it ? :D $\endgroup$ – r9m Nov 10 '14 at 13:46
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This response may be unnecessary after M.N.C.E.'s nice answer, but for what it's worth here's an alternative approach.

Let's start by transforming the integral via the following Euler substitution:

$$\sqrt{1-x^2}=1-xt.$$

Solving for the variable $t$ as a function of $x$ and vice versa, we find,

$$t=\frac{1-\sqrt{1-x^2}}{x},$$

and

$$x=\frac{2t}{1+t^2}.$$

The advantage of this substitution over the one suggested in the question statement is that it quickly converts the integrand to a product of a rational function and the logarithm (squared) of a rational function. A single integration by parts can then reduce the power of the logarithm to $1$:

$$\begin{align} \mathcal{I} &=\int_{0}^{1}\ln^2{\left(\sqrt{1+x}-\sqrt{1-x}\right)}\,\mathrm{d}x\\ &=\int_{0}^{1}\ln^2{\left(\frac{|1+t|}{\sqrt{1+t^2}}-\frac{|1-t|}{\sqrt{1+t^2}}\right)}\cdot\frac{2(1-t^2)}{(1+t^2)^2}\,\mathrm{d}t\\ &=2\int_{0}^{1}\frac{(1-t^2)}{(1+t^2)^2}\ln^2{\left(\frac{1+t-1+t}{\sqrt{1+t^2}}\right)}\,\mathrm{d}t\\ &=2\int_{0}^{1}\frac{(1-t^2)}{(1+t^2)^2}\ln^2{\left(\frac{2t}{\sqrt{1+t^2}}\right)}\,\mathrm{d}t\\ &=\frac12\int_{0}^{1}\frac{(1-t^2)}{(1+t^2)^2}\ln^2{\left(\frac{4t^2}{1+t^2}\right)}\,\mathrm{d}t\\ &=\left[\frac{t}{1+t^2}\ln^2{\left(\frac{4t^2}{1+t^2}\right)}\right]_{0}^{1}-\frac12\int_{0}^{1}\frac{t}{1+t^2}\cdot\frac{4\ln{\left(\frac{4t^2}{1+t^2}\right)}}{t(1+t^2)}\,\mathrm{d}t\\ &=\frac{\ln^2{(2)}}{4}-2\int_{0}^{1}\frac{\ln{\left(\frac{4t^2}{1+t^2}\right)}}{(1+t^2)^2}\,\mathrm{d}t.\\ \end{align}$$

The last integral of course can be systematically evaluated in terms of dilogarithms.

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  • $\begingroup$ Good, that might be another way! (+1) Also the substitution in statement is very powerful and makes things very easy. $\endgroup$ – user 1357113 Nov 10 '14 at 13:32
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I'd be also interested in other approaching ways, if possible.

Too long for a comment:

  • On one hand, x varies in between $0$ and $1$, implying a trigonometric substitution.

  • On the other hand, expressions of the form $\sqrt{1\pm x}$ automatically bring to mind the famous

$~\quad~$ trigonometric formulas $\dfrac{1+\cos2t}2=\cos^{2}t$ and $\dfrac{1-\cos2t}2=\sin^2t$.

  • On the third hand, we notice that exactly two out of the three possible types of conic sections are present in our integrand: hyperbolas, represented by the natural logarithm, whose derivative the hyperbola is; and parabolas, represented by the two square roots. Now, recalling that in almost all of Agatha Christie's mystery novels the culprit is precisely the character one least suspects, we are thereby inclined to suspect that a substitution relating to the third and mysteriously absent conic section, the circle, might be in order.

  • On the fourth hand, if life has taught me anything over the past few months, it's that it is not the actual, literal mathematical expressions which appear in an integrand that one should pay attention to $($since they are nothing more than mere masks and false appearances, meant to conceal their true identity, and deceive the audience, like the pretty, half-dressed show-girls in an illusionist's act, diverting the public's attention away from the latter's sleight of hand, and preventing them from discovering his tricks$)$, but rather the value-domain which they occupy, and the way in which they relate to each other. In this particular case, it is really hard not to

$~\quad~$ notice that $\Big(\sqrt{1+x}\Big)^2+\Big(\sqrt{1-x}\Big)^2=2$. Does any particular trigonometric formula come
$~\quad~$ to mind ?

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For the generalized integral \begin{align} \int_{0}^{1} \ln^{2}\left( \sqrt[k]{1+x} - \sqrt[k]{1-x} \right) \, dx \end{align} one possible result, although not readily seen to be reducible in a quick manor, can be shown to be \begin{align} & \int_{0}^{1} \ln^{2}(\sqrt[k]{1+x} - \sqrt[k]{1-x} ) \, dx \nonumber\\ & \hspace{5mm} = \frac{2(1 - \ln 2)^{2}}{k^{2}} + \frac{2}{k} \, \sum_{r, s = 1}^{\infty} \frac{(-1)^{s}} {r \, s(s+1)} \, F_{1}\left( s+2; - \frac{r}{k}, \frac{r}{k}; s+2; 1, -1 \right) \nonumber\\ & \hspace{5mm} + 2 k \, \sum_{r=1}^{\infty} \frac{(-1)^{r-1} \, H_{r}}{(r+1)(r+k+1)} \, {}_{2}F_{1} \left( 1, \frac{r+1}{k} ; \frac{r+2k+1}{k} ; -1 \right) \end{align} where $F_{1}$ is Appell's hypergeometric function, ${}_{2}F_{1}$ is Gauss' hypergeometric function and $H_{n}$ is the Harmonic number.

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