How to Evaluate the Integral? $\int_{0}^{1}\frac{\ln\left( \frac{x+1}{2x^2} \right)}{\sqrt{x^2+2x}}dx=\frac{\pi^2}{2}$

Question

I am trying to find a closed form for $$ \int_{0}^{1}\ln\left(\frac{x + 1}{2x^{2}}\right) {{\rm d}x \over \,\sqrt{\,{x^{2} + 2x}\,}\,}. $$

I have done trig substitution and it results in $$ \int_{0}^{1}\ln\left(\frac{x + 1}{2x^{2}}\right) {{\rm d}x \over \,\sqrt{\,{x^{2} + 2x}\,}\,} = \int_{0}^{\pi/3}\sec\left(\theta\right) \ln\left(\frac{\sec\left(\theta\right)} {2\left[\sec\left(\theta\right) - 1\right]^{\,2}} \right){\rm d}\theta $$ which doesn't help.

By part integration with $\displaystyle u = \ln\left(\frac{x + 1}{2x^{2}} \right)$, $\displaystyle\,\,{\rm d}v=\frac{\displaystyle\,\,{\rm d}x}{\,\sqrt{\,{x^{2} + 2x}\,}\,}$ also makes it more complicated.

I appreciate any help on this problem.

Answer 1

FDP and phi-rate in their answers, each has shown the integral equals $$I=\frac{17\pi^2}{24}-\frac{\ln^2(2+\sqrt3)}{2}+\frac{1}{2}\operatorname{Li}_2(-7+4\sqrt3)-4\operatorname{Li}_2(2-\sqrt3)$$

So $I = \pi^2/2 \DeclareMathOperator{\Li}{Li}$ is equivalent to $$\tag{*} \text{Li}_2(-\chi^2) - 8 \text{Li}_2(\chi) = \log ^2 \chi-\frac{5 \pi ^2}{12} \quad \chi=2-\sqrt{3}$$

Most such numerical identities of polylogs are non-trivial, proving $(*)$ is actually the hard part. I outline two proofs of $(*)$, but only the first proof will be self-contained.

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First proof: begin with the five-term functional equation (due to Kummer) $$\text{Li}_2\left(\frac{x (1-y)^2}{(1-x)^2 y}\right)=\text{Li}_2\left(\frac{1-y}{1-x}\right)+\text{Li}_2\left(-\frac{x (1-y)}{1-x}\right)+\text{Li}_2\left(\frac{x (1-y)}{(1-x) y}\right)+\text{Li}_2\left(-\frac{1-y}{(1-x) y}\right)+\frac{\log ^2(y)}{2}$$ let $u$ be such that $(1-u)/(1+u) = iu$, then $iu^2 = \chi$. Make $x=-u,y=u$, yields $$\begin{aligned} \Li_2 (u^2) &= \Li_2(iu^2)+\Li_2(-i)+\Li_2(-iu)+\Li_2(iu) + \frac{1}{2}\log^2 u \\ \iff\Li_2 (-i\chi) &= \Li_2(\chi)+\Li_2(-i)+\frac{1}{2}\Li_2(i\chi) + \frac{1}{2}\log^2 u\end{aligned}$$

taking sum with its complex conjugate, using $\Li_2(x) + \Li_2(-x) = \Li_2(x^2)/2$ proves $(*)$. QED.

One could prove a tri-log version $$\tag{**} \text{Li}_3\left(-\chi^2\right)-16 \text{Li}_3\left(\chi\right) = -\frac{23 \zeta (3)}{2}-\frac{2}{3} \log ^3\left(\sqrt{3}+2\right)+\frac{5}{6} \pi ^2 \log \left(\sqrt{3}+2\right)$$ via a nine-term functional equation of $\Li_3$.

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Second proof: The number $\Li_n(\alpha)$ for certain special algebraic $\alpha$ is contained in a certain $\mathbb{Q}$-vector space known as colored multiple zeta values (CMZVs) of level 12. Such $\alpha$ includes $\chi,-\chi^2,\chi^4,15 \sqrt{3}-26$ etc.

For small $n$, the structure of this vector space is completely understood, a spanning set is recorded in my Mathematica package here. As a consequence, any relation between them can be demonstrated, such as $(*),(**)$. Further examples obtained from this approach are:

$$51 \text{Li}_2(\chi)+\text{Li}_2\left(15 \sqrt{3}-26\right)-\frac{9}{4} \text{Li}_2(\chi^4) = \frac{61 \pi ^2}{24}-6 \log ^2\left(\sqrt{3}+2\right)$$

$$-1224 \text{Li}_3(\chi)-8 \text{Li}_3\left(15 \sqrt{3}-26\right)+\frac{27}{2} \text{Li}_3(\chi^4) = -851 \zeta (3)-48 \log ^3\left(\sqrt{3}+2\right)+61 \pi ^2 \log \left(\sqrt{3}+2\right)$$

I don't know how to prove them from functional equations, but such proofs should exist. For another $\Li_2$ example, see here.

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Some remarks:

• We also have, with $G$ Catalan constant $$\int_{0}^{1}\frac{\log\left( \frac{x+1}{2x^2} \right)\log (1+x)}{\sqrt{x^2+2x} }dx = \frac{4 \pi G}{3}-\frac{1}{2} \pi ^2 \log (2)$$

• The original purpose of the package mentioned above is to calculate integrals. It could also do this one directly. Since it only accepts integrand from 0 to 1, without radicals. We need to first perform the substitution $1+x = \frac{3+u^2}{3-u^2}$, $$\int_{0}^{1}\frac{\ln\left( \frac{x+1}{2x^2} \right)}{\sqrt{x^2+2x}}dx = \int_0^1 \frac{2 \log \left(\frac{1}{8} \left(\frac{9}{u^4}-1\right)\right)}{\sqrt{3} \left(1-\frac{u^2}{3}\right)} du$$ and

  MZIntegrate[(2 Log[1/8 (-1 + 9/u^4)])/(Sqrt[3] (1 - u^2/3)), {u, 0, 1}]
  

  immediately gives the result $\pi^2/2$.

Answer 2

You can "simplify" the problem using a first integration by parts to get rid of the logarithm $$u=\log \left(\frac{x+1}{2 x^2}\right)\quad \implies \quad du=-\frac{x+2}{x^2+x}$$ $$dv=\frac{1}{\sqrt{x^2+2 x}}\quad \implies \quad v=2 \tanh ^{-1}\left(\sqrt{\frac{x}{x+2}}\right)$$

Using the bounds $u\,v=0$ and we are left with $$I=2\int_0^1\frac{(x+2) }{x^2+x}\tanh ^{-1}\left(\sqrt{\frac{x}{x+2}}\right)dx$$

Now $$\sqrt{\frac{x}{x+2}}=t \implies x=\frac{2 t^2}{1-t^2}\implies dx=\frac{4 t}{\left(1-t^2\right)^2}$$ $$I=8\int_0^{\frac{1}{\sqrt{3}}}\frac{\tanh ^{-1}(t)}{t-t^5}\,dt$$ Now, using partial fraction decomposition $$\frac{1}{t-t^5}=\frac{1}{t(1-t^2)(1+t^2)}=-\frac{t}{2 \left(t^2+1\right)}-\frac{1}{4 (t-1)}-\frac{1}{4 (t+1)}+\frac{1}{t}$$ and now would arrive a bunch of polylogarithm functions.

The simplest is

$$\int \frac{\tanh ^{-1}(t)}{t}\,dt=\frac{1}{2} (\text{Li}_2(t)-\text{Li}_2(-t))$$

Fortunately, between the given bounds everything simplify a lot.

I let you the pleasure of computing the pieces.

Edit

If we write

$$\frac{\tanh ^{-1}(t)}{t-t^5}=\frac{\tanh ^{-1}(t)}{t}+\sum_{n=1}^\infty t^{4n-1}\,\tanh ^{-1}(t)$$ we have $$I=8\int_0^{\frac{1}{\sqrt{3}}}\frac{\tanh ^{-1}(t)}{t-t^5}\,dt=4 \left(\text{Li}_2\left(\frac{1}{\sqrt{3}}\right)-\text{Li}_2\left (-\frac{1}{\sqrt{3}}\right)\right)+$$ $$\sum_{n=1}^\infty\frac{9^{-n} \log \left(2+\sqrt{3}\right)-B_{\frac{1}{3}}\left(2 n+\frac{1}{2},0\right)}{n}$$ that is to say $$I=4 \left(\text{Li}_2\left(\frac{1}{\sqrt{3}}\right)-\text{Li}_2\left (-\frac{1}{\sqrt{3}}\right)\right)+\log \left(\frac{9}{8}\right) \log \left(2+\sqrt{3}\right)-\sum_{n=1}^\infty \frac{B_{\frac{1}{3}}\left(2 n+\frac{1}{2},0\right)}{n}$$ Numerically, the sum of the first and second terms is $4.96991$ and the summation is only $0.03511$

Edit

There is something very strange : two different $CAS$ give as result $$I =\frac{17 \pi ^2}{24}-\frac{1}{2} \left(8 \text{Li}_2\left(2-\sqrt{3}\right)-\text{Li}_2\left(-7+4 \sqrt{3}\right)+\log ^2\left(2-\sqrt{3}\right)\right)$$ without any further simplification while Wolfram Alpha gives $1.$ when it is written as $$\int_0^1 \frac2{ \pi ^2} \frac{\log \left(\frac{x+1}{2 x^2}\right)}{\sqrt{x^2+2 x}}\,dx$$ Without the factor, it just return a decimal value.

Using RootApproximant[%] also behaves differently with or without the factor inside the integrand. $$\frac{1}{2} \left(8 \text{Li}_2\left(2-\sqrt{3}\right)-\text{Li}_2\left(-7+4 \sqrt{3}\right)+\log ^2\left(2-\sqrt{3}\right)\right)=2.056167583560283045590519$$ Passed to the $ISC$, it is returned as $\frac{5 \pi ^2}{24}$

Answer 3

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Previous try

It should be $\pi^2/2$ but showing this requires invoking the Cauchy residue theorem, as is often the case when $\pi$ creeps in integrals.
The demonstration proceeds in two steps.

I - Contour integration

1. First refer to C. Leibovici's answer:

$$ \int \frac{1}{\sqrt{x^2+2 x}} \, dx = 2 \tanh ^{-1}\left(\sqrt{\frac{x}{x+2}}\right)$$

then integrate by parts using this result, which yields:

$$I = \int_{0}^{1/\sqrt{3}} \frac{8 \tanh ^{-1}(v)}{v \left(1-v^4\right)} \, dv$$

2. Now notice that the (simple) poles of the integrand $f(v)$ are:

$$ \{-1,0,-i,i,1\} $$

3. Choose a rectangular integration contour such as: $$ \{\frac{-1}{\sqrt{3}}, 0^{+}, \frac{1}{\sqrt{3}}, c\, i + \frac{1}{\sqrt{3}}, c \, i -\frac{1}{\sqrt{3}}\},$$ with: $\, \, c \rightarrow \, +\infty $
   $0^{+}$ intends to symbolize a little dent around the 0 pole, this is to make it clean but anyway the residue there is null.

4. Compute the residue at $i$, this is easy and I'll leave it to you. Hint: notice that $ \tanh^{-1}(i) = -i \frac{\pi}{4} $, which is the true reason why you get in $\pi$ along the way.

5. Check that at infinity along the upper side of the rectangle contour, $f(v)$ is negligible, which is fairly easy too.

6. The hard bit is to show that the real part of the contour integration along one or the other of the vertical sides of the rectangle is null just at the points chosen. With other points, it is not null, and it does not cancel (it adds up) when the contour integration along the two vertical sides is performed:

$$ \Re{ \int_{\pm\frac{1}{\sqrt{3}}}^{\pm\frac{1}{\sqrt{3}} + i \infty} f(v) \, dv} = 0 \,\,\, (\mathscr{C})$$

The easy bit is that, in any case, the imaginary part does obviously cancel, owing to the fact that $f(v)$ is even.

The hard bit is not demonstrated yet, but does not look intractable.
It is anyway an improvement to replace a conjecture on a multiple of $\pi^2$ by a conjecture about a null value, with a hint at optimization (for which there are lots of methods around).

7. Applying Cauchy's residue theorem you now get the result:

$$ \int_R f(z) \, dz = 2i\pi \, Res(f, i) = 2i\pi (-(i \frac{\pi}{2})) = \pi^2$$

8. Considering the fact that the integral vanishes along the three sides above the real axis when you let $c$ grow to infinity, you now are left with the integration along the real segment $[-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}]$, whose value is 2I, owing to the parity of $f$. Hence the result.

II - Towards the proof of conjecture ($\mathscr{C}$)

Brute-force integration of $f(v)$ along a vertical rectangular side (let us take the right one) $[\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} + i y]$:

$$f(\frac{1}{\sqrt{3}} + i \,y) = \frac{8 \tanh ^{-1}\left(\frac{1}{\sqrt{3}}+i y\right)}{\left(1-\left(\frac{1}{\sqrt{3}}+i y\right)^4\right) \left(\frac{1}{\sqrt{3}}+i y\right)}$$

This gives the imaginary part (owing to contour integration, $f$ must be multiplied by $i$, the derivative of the parametrized contour side):

$$g(y) = \frac{18 \left(90 \sqrt{3} y^4 \tan ^{-1}(\frac{y}{\frac{1}{\sqrt{3}}-1})-2 \sqrt{3} \left(-45 y^4+30 y^2+8\right) \tan ^{-1}(\frac{y}{1+\frac{1}{\sqrt{3}}})+4 \sqrt{3} \left(15 y^2+4\right) \tan ^{-1}\left(\frac{1}{2} \left(\sqrt{3}+3\right) y\right)+3 \left(-9 y^4+30 y^2+4\right) y \log \left(1-\frac{4 \sqrt{3}}{3 y^2+2 \sqrt{3}+4}\right)\right)}{\left(3 y^2+1\right) \left(336 y^2+27 \left(y^2+2\right) \left(3 y^2-2\right) y^4+64\right)}$$

Integration between $y = 0$ and $y= N$ (let us denote this integral $J(N)$) then using an asymptotic series when N tends to infinity reduces $\mathscr{(C)}$ to $\mathscr{(C')}$:

$$J(N) \rightarrow 0 \,\,\, \mathscr{(C')}$$

(numerically this much is proved).

Plot of $J(N)$ as $N \rightarrow +\infty$

When proved this will prove ($\mathscr{C}$) and complete the demonstration. I'll leave this for further edits.

This attack procedure may well work if one finds a good absolutely dominating function for $\Im{f(\frac{1}{\sqrt{3}} + i \,y)}$ for which integration and asymptotic approximation may work without too much toil.

III - Edit: going further and final comments

Integrating $g(y)$ in II is a bit cumbersome, to say the least, but can be done.
Doing this goes through the following steps:

1. Decompose the denominator into simple poles to find the values for $\alpha$ above.

2. Integrate:

$$I_k(N) = \int_0^N \frac{u^k \tan^{-1}(u)}{u - \alpha} \, du, \,\, k=0, 2, 4, \, \alpha \in \mathbb{R} $$

3. Integrate:

$$ L_k(N) = \int_0^N y^k \frac{\log(1-\frac{4 \sqrt{3}}{3 y^2+2 \sqrt{3}+4})}{y - \alpha} \, dy, k=1, 3, 5, \, \alpha \in \mathbb{R} $$

4. Expand each of the above results by N and sum the different polynomial contributions

5. Simplify the resulting expression and let N tend to infinity.

On these steps:

Point 1

The poles of the denominator are :

$$\left\{-\frac{i}{\sqrt{3}}, \frac{i}{\sqrt{3}}, -i \sqrt{\frac{4}{3}-\frac{2}{\sqrt{3}}}, i \sqrt{\frac{4}{3}-\frac{2}{\sqrt{3}}}, \frac{1}{3} \left(-3-i \sqrt{3}\right),\frac{1}{3} \left(3-i \sqrt{3}\right), \frac{1}{3} \left(-3+i \sqrt{3}\right), \frac{1}{3} \left(3+i \sqrt{3}\right), -i \sqrt{\frac{2}{3} \left(\sqrt{3}+2\right)}, i \sqrt{\frac{2}{3} \left(\sqrt{3}+2\right)}\right\}$$

So the denominator easily simplifies into a sum of 10 simple poles. Each must be decomposed into 3 arctangent-type and 1 logarithm-type numerator. Then the polynomial factors are expanded to end up with $I_k(N)$ and $L_k(N)$ expressions. In all there must be 60 $I_k$-ype integrals and 30 $L_k$-type integrals to consider.

Point 2

Actually the case $k = 0$ is well-known: it is a generalized inverse tangent integral of second order (see 1), for which an asymptotic expansion is known:

$$Ti_2(N, -\alpha) = \frac{1}{2} \pi \ln{N} - \frac{1}{4} \pi \ln{(1 + \alpha^2)} - Ti_2(\alpha) + \tan^{-1}(\alpha) \ln{|\alpha|} + O(\frac{1}{N})$$

The other cases can be obtained from this starting point by classic integration techniques (integration by parts and auxiliary parameter derivation), so there is reason to consider that an asymptotic expansion of the arctangent part of $J(N)$ is algorithmically doable.

Also, maybe the good strategy would be to find an asymptotic expansion for $I_k(N)$ from the above formula before tackling the 60 $I_k$-type integrals. This would simplify the task by an order of magnitude.

Further, $L_k(N)$ is a sum of classic integrals that can be found in Gradshtein or Prudnikov.

There is reason to believe that proving or refuting conjecture $\mathscr{(C')}$ is an algorithmically doable task. It is just quite cumbersome and will take a few dozen classic operations; yet not anything that exceeds the power of CAS systems, so there must be a conclusion in the end of the daunting calculus: true or false.

We expect the asymptotic logarithms and arctangents to cancel out easily; however we will be stranded with the second-order inverse tangent expressions taken at the 10 poles. These are closely associated values, so we expect them to cancel out; but to reach this conclusion we will need formulas relating $Ti_2$ values of comparable complexity as the above solution using the five-term Kummerian $Li_2$ relation (see 1, appendix). This may be doable again but, at the end of the day, looks like a much-complexified version of the same solution.

It looks fascinating that for such integrals, there seems to be no way out other than by manipulating polylogs and associated functions.

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1 Leonard Lewin, Polylogarithms and associated functions, 1981, North Holland, chapter 3 and expansion p. 288 (11).

Answer 4

Hyperbolic substitution is very convenient for this integral. We can substitute $x+1=\cosh t$

$$I=\int_0^1 \ln{\left(\frac{x+1}{2x^2}\right)}\frac{dx}{\sqrt{x^2+2x}}=\int_0^{\ln{(2+\sqrt 3)}}\ln\left(\frac{\cosh t}{2(\cosh t-1)^2}\right)dt$$$$=\int_0^{\ln{(2+\sqrt 3)}}\ln\left(\frac{e^{-t}(1+e^{-2t})}{(1-e^{-t})^4}\right)dt$$ When we split the integral we get $$I=-\frac{\ln^2(2+\sqrt 3)}{2}+\int_0^{\ln{(2+\sqrt 3)}}\ln (1+e^{-2t})dt-4\int_0^{\ln{(2+\sqrt 3)}}\ln(1-e^{-t})dt$$ For the first integral we can apply the substitution $u=-e^{-2t}$, and for the second integral we can use $v=e^{-t}$ This will allow us to use the dilogarithm. $$I=-\frac{\ln^2(2+\sqrt 3)}{2}+\frac{1}{2}\int_{-1}^{-7+4\sqrt3}\frac{-\ln (1-u)}{u}du-4\int_1^{2-\sqrt 3}\frac{-\ln(1-v)}{v}dv$$ Using $\int \frac{-\ln(1-x)}{x}dx=\operatorname{Li}_2(x)+C$ we get $$I=\frac{17\pi^2}{24}-\frac{\ln^2(2+\sqrt3)}{2}+\frac{1}{2}\operatorname{Li}_2(-7+4\sqrt3)-4\operatorname{Li}_2(2-\sqrt3)$$ ATTEMPT 2

Refer to C. Leibovici's answer, apply partial fractions and integration by parts $$I=8\int_0^{\frac{1}{\sqrt3}}\frac{\operatorname{artanh} x}{x(1-x^4)}dx=8\int_0^{\frac{1}{\sqrt3}}\frac{\operatorname{artanh}x}{x}dx+2\int_0^{\frac{1}{\sqrt3}}\frac{4x^3\operatorname{artanh}x}{1-x^4}dx$$$$=8X_2\left(\frac{1}{\sqrt3}\right)+\ln(2-\sqrt3)\ln\left(\frac{8}{9}\right)+2\int_0^\frac{1}{\sqrt3} \frac{\ln(1-x^4)}{1-x^2}dx$$ $X_2(t)$ is the Legendre chi function

Using $x^4-1=(x+1)(x-1)(x+i)(x-i)$ in conjunction with logarithm properties and partial fractions, we can decompose the integral into 8 integrals of the form $\int \frac{\ln(x+a)}{x+b}dx$. 2 of them, $\int \frac{\ln(1+x)}{1+x}dx$ and $\int \frac{\ln(1+x)}{1+x}dx$ can be evaluated without dilogarithms. The other 6 each give us 2 dilogarithmic terms. This gives us 12 dilogarithms to simplify in total, and it might work

Answer 5

\begin{align}J&=\int_{0}^{1}\frac{\ln\left( \frac{x+1}{2x^2} \right)}{\sqrt{x^2+2x}}dx\\ &\overset{z=\frac{x}{2+x}}=2\int_0^{\frac{1}{\sqrt{3}}}\frac{\ln\left(\frac{1-z^4}{8z^4}\right)}{1-z^2}dz\\ &\overset{\text{IBP}}=-\underbrace{\left[\ln\left(\frac{1-z}{1+z}\right)\ln\left(\frac{1-z^4}{8z^4}\right)\right]_0^{\frac{1}{\sqrt{3}}}}_{=0}-4\int_0^{\frac{1}{\sqrt{3}}} \frac{\ln\left(\frac{1-z}{1+z}\right)}{z(1-z^2)(1+z^2)}dz\\ &\overset{w=\frac{1-z}{1+z}}=-\int_{\frac{\sqrt{3}-1}{\sqrt{3}+1}}^1 \frac{(1+w)^3\ln w}{w(1-w)(1+w^2)}dw\\ &=\frac{1}{2}\ln^2\left(2+\sqrt{3}\right)-4\int_{\frac{\sqrt{3}-1}{\sqrt{3}+1}}^1 \frac{\ln w}{1-w}dw-2\underbrace{\int_{\frac{\sqrt{3}-1}{\sqrt{3}+1}}^1 \frac{w\ln w}{1+w^2}dw}_{\text{IBP}}\\ &=\frac{1}{2}\ln^2\left(2+\sqrt{3}\right)-4\int_{\frac{\sqrt{3}-1}{\sqrt{3}+1}}^1 \frac{\ln w}{1-w}dw-\Big[\ln(1+w^2)\ln w\Big]_{\frac{\sqrt{3}-1}{\sqrt{3}+1}}^1+\underbrace{\int_{\frac{\sqrt{3}-1}{\sqrt{3}+1}}^1 \frac{\ln(1+w^2)}{w}dw}_{y=w^2}\\ &=\frac{3}{2}\ln^2\left(2-\sqrt{3}\right)+2\ln 2\ln\left(2-\sqrt{3}\right)-4\int_{2-\sqrt{3}}^1 \frac{\ln w}{1-w}dw+\frac{1}{2}\int_{(2-\sqrt{3})^2}^1 \frac{\ln(1+y)}{y}dy\\ &=\frac{3}{2}\ln^2\left(2-\sqrt{3}\right)+2\ln 2\ln\left(2-\sqrt{3}\right)+4\text{Li}_2\left(\sqrt{3}-1\right)+\frac{1}{2}\text{Li}_2\Big(-(2-\sqrt{3})^2\Big)\underbrace{-\frac{1}{2}\text{Li}_2\left(-1\right)}_{=\frac{\pi^2}{24}}\\ &=\frac{3}{2}\ln^2\left(2-\sqrt{3}\right)+2\ln 2\ln\left(2-\sqrt{3}\right)+4\text{Li}_2\left(\sqrt{3}-1\right)+\frac{1}{2}\text{Li}_2\Big(-(2-\sqrt{3})^2\Big)+\frac{\pi^2}{24} \end{align}

To be continued...

First addendum:: Using, \begin{align}0<x<1,\text{Li}_2\left(x\right)+\text{Li}_2\left(1-x\right)=\frac{\pi^2}{6}-\ln x\ln(1-x)\end{align}

One obtains, \begin{align}&=\frac{3}{2}\ln^2\left(2-\sqrt{3}\right)+2\ln 2\ln\left(2-\sqrt{3}\right)+4\text{Li}_2\left(\sqrt{3}-1\right)+\frac{1}{2}\text{Li}_2\Big(-(2-\sqrt{3})^2\Big)+\frac{\pi^2}{24}\\ &=\boxed{\frac{17\pi^2}{24}-\frac{1}{2}\ln^2\left(2-\sqrt{3}\right)-4\text{Li}_2\left(2-\sqrt{3}\right)+\frac{1}{2}\text{Li}_2\Big(-(2-\sqrt{3})^2\Big)}\\ \end{align}

Answer 6

Let $$x=y^2,\quad y^2+1=t,\tag1$$ then $$I=\int\limits_0^1 \dfrac{\ln\left(\dfrac{x+1}{2x^2}\right)}{\sqrt{x^2+2x}}dx =\int\limits_0^1 \dfrac{\ln\left(\dfrac{y^2+1}{2y^4}\right)}{\sqrt{y^4+2y^2}}\, 2y\,\text dy =\int\limits_1^2 \dfrac{\ln\left(\dfrac{t}{2(t-1)^2}\right)}{\sqrt{t^2-1}}\,\text dt,$$ $$I=I_1-2I_2-I_3,$$ where \begin{cases} I_1=\int\limits_1^2 \dfrac{\ln t\,\text dt}{\sqrt{t^2-1}} = \dfrac{\pi^2}{24} -\dfrac12 \operatorname{Li_2}\left(\dfrac{2 - \sqrt3}4\right) - \ln^2 2-\dfrac14\arccos^2 2\\ I_2=\int\limits_1^2 \dfrac{\ln(t-1)\,\text dt}{\sqrt{t^2-1}} =-\dfrac{\pi^2}3+2\operatorname{Li}_2(2-\sqrt3) +6\operatorname{arcsinh}^2\dfrac1{\sqrt2} -4 \operatorname{arcsinh}\dfrac 1{\sqrt2}\,\log(1+\sqrt3)\\ I_3=\int\limits_1^2 \dfrac{\ln 2\text dt}{\sqrt{t^2-1}} = \dfrac12\ln2 \ln\left(4\sqrt{3}+7\right) = \ln2 \ln\left(2+\sqrt{3}\right)\\ \arccos 2 = i\ln(2+\sqrt3), \end{cases}

Therefore, $$I=\dfrac{17\pi^2}{24}-\dfrac12 \operatorname{Li_2}\left(\dfrac{2 - \sqrt3}4\right)-4\operatorname{Li}_2(2-\sqrt3)-\ln^2 2 -\ln\left(2+\sqrt{3}\right)\ln 2$$ $$+\dfrac1{4}\ln^2(2+\sqrt3) -12 \operatorname{arcsinh}^2 \dfrac1{\sqrt2}+8\operatorname{arcsinh}\dfrac 1{\sqrt2}\,\ln(1+\sqrt3),$$ $$I\approx4.9348022005446793094$$ (see also Wolfram Alpha calculations), in accordance with numeric calculations.

Answer 7

Let the given integral be $I$. After the substitution $x+1=\sec t $, integrtation by parts and some tricks in the integrand, we have $$I=\int_0^{\frac{\pi}{3}}\frac{\sin t+\tan t}{1-\cos t}\ln\left(\frac{\cos t}{1-\sin t}\right)dt.$$ After the $z=\tan\left(\frac{t}{2}\right)$ substitution, $$I=\int_0^{\frac{1}{\large\sqrt{3}}}\frac{4\ln\left(\frac{1+z}{1-z}\right)}{z(1-z^2)(1+z^2)}dz.$$ I think this expression is obtained by Claude Leibovici and FDP.

user12030145 mentioned about this contour integral. Let $C_1=\{z=-\frac{1}{\sqrt{3}}+iy| y:\infty\rightarrow 0\}$, $C_2=\{z=\frac{1}{\sqrt{3}}+iy| 0<y<\infty\}$ and $f(z)=\frac{4\ln\left(\frac{1+z}{1-z}\right)}{z(1-z^2)(1+z^2)}$. Then, branch cut of the logarithm is outside and the only singularity inside and on the contour $C_1+[-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}]+C_2$ is $z=i$ with residue $Res_{z=i}f(z)=\pi^2$. $z=0$ is fake singularity. Hence, since $f(z)$ is even function, the integral on $C_1+C_2$ is equal to the integral on the line $\{\frac{1}{\sqrt{3}}+iy|y\in\Bbb{R}\}$ and by residue theorem we have $$2I+\int_{\frac{1}{\sqrt{3}}-i\infty}^{\frac{1}{\sqrt{3}}+i\infty}f(z)dz=\pi^2$$ So, the problem reduces to showing that $\int_{\frac{1}{\sqrt{3}}-i\infty}^{\frac{1}{\sqrt{3}}+i\infty}f(z)dz=0.$ Nothing new so far, since I think user12030145 conjectured about such a thing. I think the real part of this integral is zero because of the Maclaurin series of $f(z)$, so it has to be zero.

Answer 8

We want to evaluate $$ I=\int^{1}_{0}\frac{\log\left((x+1)/x^2\right)}{\sqrt{x^2+2x}}dx. $$ We can easily write $I$ in the form $$ I=2\int^{1}_{1/2}\frac{\log(\sqrt{t}/(1-t))}{t\sqrt{1-t^2}}dt=4\int^{1}_{1/\sqrt{2}}\log\left(\frac{t}{1-t^2}\right)\frac{dt}{t\sqrt{1-t^4}}= $$ $$ =\tanh^{(-1)}\left(\frac{\sqrt{3}}{2}\right)\log 2-\int^{1}_{1/2}\frac{t+1}{t-1}t^{-1}\tanh^{(-1)}\left(\sqrt{1-t^2}\right)dt. $$ Set $t=i\tan t'$, then $I$ becomes $$ I=\tanh^{(-1)}\left(\frac{\sqrt{3}}{2}\right)\log 2- $$ $$ -\int^{-i\infty}_{-i\tanh^{(-1)}(1/2)}\left(-2i-\cot t'+\tan t'\right)\tanh^{(-1)}(\sec t')dt'= $$ $$ \frac{1}{2}i\pi \log(2+\sqrt{3})-\int^{-i\infty}_{-i\tanh^{(-1)}\left(\frac{1}{2}\right)}\left(-2it'-\log\left(\frac{1}{2}\sin(2t')\right)\right)\csc(t')dt'= $$ $$ \frac{1}{2}i\pi \log(2+\sqrt{3})+2i\int^{-i\infty}_{-i\tanh^{(-1)}\left(\frac{1}{2}\right)}t'\csc(t')dt'+ $$ $$ +\int^{-i\infty}_{-i\tanh^{(-1)}\left(\frac{1}{2}\right)}\log\left(\frac{1}{2}\sin(2t')\right)\csc(t')dt'. $$ But $$ \int t\csc tdt=-2t\tanh^{(-1)}(e^{it})-2iLi_2(e^{it})+\frac{1}{2}iLi_2(e^{2it}) $$ and $$ \int\log\left(\sin(t)\right)\csc(t)dt= $$ $$ =[\log(\sin t)+\log(1-i\tan(t/2))+\log(1+i\tan(t/2))- $$ $$ -1/2\log(\tan(t/2))]\log(\tan(t/2))+Li_2(-i\tan(t/2))+Li_2(i\tan(t/2)). $$ Hence using the above we get the evaluation. (Note that $\int \frac{1}{\sqrt{t^2+2t}}dt$ can evaluated easily).