How to evaluate $I=\int_0^1\arctan^3\left(\sqrt{\frac{x}{4-x}}\right)\frac{\ln x}{x}dx$ in closed form?

Question

I stumbled across this question on MSE:

• The title integral on the question is fairly simple, but this supplementary integral seems much harder.
• After substituting $x = 4\sin^{2}\left(t\right)$ and integrating by parts, we get \begin{align*} I & =\int_{0}^{1}\arctan^{3}\left(\sqrt{\frac{x}{4 -x}}\right)\frac{\ln\left(x\right)}{x}\,{\rm d}x \\[2mm] & = 4\int_0^{\pi/6}x^3\cot(x)\ln\left(2\sin\left( x\right)\right)\,{\rm d}x \\[2mm] & = -6\int_0^{\pi/6}x^{2}\ln^{2}\left(2\sin\left(x\right)\right)\,{\rm d}x \end{align*}
• There does not exist an obvious continuation after substituting the Fourier series for $\ln\left(2\sin\left(x\right)\right)$, mostly due to the square on the logarithm.
• The upper bound being $\pi/6$ instead of $\pi/2$ or $\pi/4$ makes the obvious substitution $u=\tan\left(x\right)$ questionable.
• Note that $\arctan\left(\sqrt{\frac{x}{4-x}}\right)=\arcsin\left(\frac{\sqrt x}2\right)$

A perhaps related integral which loosely relates with this post $$\int_0^1\arctan^2\left(\sqrt\frac{x}{4-x}\right)\frac{\ln (x)}{x}dx=-\frac{17}{72}\zeta(4)$$

Answer 1

$\underline{\space{\bf\text{0.}}\space}$ Start by: $$ \begin{align} I =& -6\int\limits_{0}^{\pi/6} x^2 \log^2\left(2\sin(x)\right) \,dx \end{align} $$

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$\underline{\space{\bf\text{1.}}\space}$ Change to $ \displaystyle \cos(x)=\sin\left(\frac{\pi}{2}-x\right) $ in order to use the identity: $$ \begin{align} &2\cos(x) = e^{{\small+}ix} + e^{{\small-}ix} \implies \\[1mm] &\log\left(1+e^{{\small-}i\,2x}\right) = \log\left(2\cos(x)\right) + ix \\[1mm] &\log^2\left(1+e^{{\small-}i\,2x}\right) = \log^2\left(2\cos(x)\right) - x^2 + i\,2x\log\left(2\cos(x)\right) \\ \end{align} $$ And because of the integration range, consider only the REAL part: $$ \begin{align} I =& -6\int\limits_{\pi/3}^{\pi/2} \left(\frac{\pi}{2}-x\right)^2 \log^2\left(2\cos(x)\right) \,dx \\[1mm] =& -6\,\,\Re{\int\limits_{\pi/3}^{\pi/2}} \left(\frac{\pi}{2}-x\right)^2 \left[x^2+\log^2\left(1+e^{{\small-}i\,2x}\right)\right] dx \\[1mm] =& \color{blue}{-6\int\limits_{\pi/3}^{\pi/2} \left(\frac{\pi}{2}-x\right)^2 x^2 \,dx} -6\,\,\Re{\int\limits_{\pi/3}^{\pi/2}} \left(\frac{\pi}{2}-x\right)^2 \log^2\left(1+e^{{\small-}i\,2x}\right) \,dx \\[1mm] =& \color{blue}{-\frac{17\,\pi^5}{12960}} -6\,\,\Re{\int\limits_{\pi/3}^{\pi/2}} \left(\frac{\pi}{2}-x\right)^2 \log^2\left(1+e^{{\small-}i\,2x}\right) \,dx \\[1mm] \end{align} $$

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$\underline{\space{\bf\text{2.}}\space}$ Substitute: $ \displaystyle -e^{{\small-}i\,2x} \mapsto x \implies x \mapsto \frac{i}{2}\log(-x) $
Also use: $ \displaystyle \log(-x)=\log(x)-i\pi\, $
And let: $ \displaystyle \color{red}{a} =-e^{{\small-}i\,2\pi/3} =\frac{1}{2}+i\,\frac{\sqrt{3}}{2} $ $$ \begin{align} I =& -\frac{17\,\pi^5}{12960} -6\,\,\Re{\int\limits_{a}^{1}} \left[\frac{\pi}{2}-\frac{i}{2}\left(\log(x)-i\pi\right)\right]^2 \log^2(1-x) \,\frac{i}{2x} \,dx \\[1mm] =& -\frac{17\,\pi^5}{12960} \color{red}{+}\frac{3}{4}\,\,\color{red}{\Re}{\int\limits_{a}^{1}} \,\color{red}{i}\,\frac{\log^2(x)\log^2(1-x)}{x} \,dx \\[1mm] =& -\frac{17\,\pi^5}{12960} \color{red}{-}\frac{3}{4}\,\,\color{red}{\Im}{\int\limits_{a}^{1}} \,\frac{\log^2(x)\log^2(1-x)}{x} \,dx \end{align} $$

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$\color{red}{\underline{\space{\bf\text{3.}}\space\text{(the main trick)}\colon}}$ The beauty of the last integral is the switching between $\,(x)\,$ and $\,(1-x)\,$: $$ \begin{align} \color{blue}{+\int\,\frac{\log^2(x)\log^2(1-x)}{x}\,dx} = \color{blue}{-\int\,\frac{\log^2(1-x)\log^2(x)}{1-x}\,dx} \end{align} $$ Hence, using this property and substituting $\, \displaystyle x \mapsto 1/2+x \,$,
considering the integral range and the IMAGINARY part: $$ \begin{align} \int\limits_{a}^{1}\,\frac{\log^2(x)\log^2(1-x)}{x}\,dx =& \int\limits_{i\sqrt{3}/2}^{1/2}\,\frac{\log^2(1/2+x)\log^2(1/2-x)}{1/2+x}\,dx \\[1mm] \int\limits_{a}^{1}\,\frac{\log^2(x)\log^2(1-x)}{1-x}\,dx =& \int\limits_{i\sqrt{3}/2}^{1/2}\,\frac{\log^2(1/2+x)\log^2(1/2-x)}{1/2-x}\,dx \end{align} $$ Thus, $$ \begin{align} I =& -\frac{17\,\pi^5}{12960} -\frac{3}{4}\,\,\Im{\int\limits_{i\sqrt{3}/2}^{1/2}} \,{\log^2(1/2-x)\log^2(1/2+x)} \left[\frac{1/2}{1/2-x}+\frac{1/2}{1/2+x}\right] dx \\[2mm] =& -\frac{17\,\pi^5}{12960} -\frac{3}{8}\,\,\Im{\int\limits_{i\sqrt{3}/2}^{1/2}} \,\frac{\log^2(1/2-x)}{1/2-x} \,\frac{\log^2(1/2+x)}{1/2+x} \,dx \quad\color{red}{\left\{\small{x}\mapsto{x/2}\right\}} \\[2mm] =& \color{blue}{-\frac{17\,\pi^5}{12960} -\frac{3}{4}\,\,\Im{\int\limits_{i\sqrt{3}}^{1}} \,\frac{\log^2\left((1-x)/2\right)}{1-x} \,\frac{\log^2\left((1+x)/2\right)}{1+x} \,dx} \end{align} $$

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$\underline{\space{\bf\text{4.}}\space}$ Similar to this question, use the identity: $$ \begin{align} 12\log^2(x)\,\log^2(y) =& \,\left[\log(x)+\log(y)\right]^4 + \left[\log(x)-\log(y)\right]^4 -2\left[\log(x)\right]^4 -2\left[\log(y)\right]^4 \\[1mm] =& \,\log^4(x\,y) + \log^4(x/y) -2\log^4(x) -2\log^4(y) \end{align} $$ To split the last integral into four integrals: $$ \begin{align} \color{brown}{I_{_1}(x)} =&\, \color{brown}{\int\frac{\log^4\left((1-x)\,(1+x)\right)}{1-x^2}\,dx} \\[1mm] \color{magenta}{I_{_2}(x)} =&\, \color{magenta}{\int\frac{\log^4\left((1-x)/(1+x)\right)}{1-x^2}\,dx} \\[1mm] \color{olive}{I_{_3}(x)} =&\, \color{olive}{\int\frac{\log^4\left(1-x\right)}{1-x^2}\,dx} \\[1mm] \color{orange}{I_{_4}(x)} =&\, \color{orange}{\int\frac{\log^4\left(1+x\right)}{1-x^2}\,dx} \end{align} $$ And with some WA help (specially for $I_{_1}$): $$ \begin{align} \color{brown}{I_{_1}(x)} =&\, \color{brown}{ \frac{1}{x}\sqrt{\frac{x^2}{1-x^2}}\,\times } \\ &\color{brown}{ \,\,[\,\,384\,\log^0\left({\small({1-x^2})/{4}}\right)\, {_{\small6}F_{\small5}}{\left({\small\frac12,\frac12,\frac12,\frac12,\frac12,\frac12};\,{\small\frac32,\frac32,\frac32,\frac32,\frac32};\,{\small\frac{1}{1-x^2}}\right)} } \\ & \color{brown}{ +\,192\,\log^1\left({\small({1-x^2})/{4}}\right)\, {_{\small5}F_{\small4}}{\left({\small\frac12,\frac12,\frac12,\frac12,\frac12};\,{\small\frac32,\frac32,\frac32,\frac32};\,{\small\frac{1}{1-x^2}}\right)} } \\ & \color{brown}{ +\,\,\,48\,\,\log^2\left({\small({1-x^2})/{4}}\right)\, {_{\small4}F_{\small3}}{\left({\small\frac12,\frac12,\frac12,\frac12};\,{\small\frac32,\frac32,\frac32};\,{\small\frac{1}{1-x^2}}\right)} } \\ & \color{brown}{ +\,\,\,\,8\,\,\,\,\log^3\left({\small({1-x^2})/{4}}\right)\, {_{\small3}F_{\small2}}{\left({\small\frac12,\frac12,\frac12};\,{\small\frac32,\frac32};\,{\small\frac{1}{1-x^2}}\right)} } \\ & \color{brown}{ +\,\,\,\,1\,\,\,\,\log^4\left({\small({1-x^2})/{4}}\right)\, {_{\small2}F_{\small1}}{\left({\small\frac12,\frac12};\,{\small\frac32};\,{\small\frac{1}{1-x^2}}\right)}] } \\[3mm] \color{magenta}{I_{_2}(x)} =&\, \color{magenta}{ -{\small\frac{1}{10}}\log^5\left({\small\frac{1-x}{1+x}}\right) } \\[3mm] \color{olive}{I_{_3}(x)} =&\, \color{olive}{ -{\small\frac{1}{10}}\log^5\left({\small\frac{1-x}{2}}\right) } \\ &\, \color{olive}{ -\,{\small\frac{1}{2}}\,\log^4\left({\small\frac{1-x}{2}}\right) \text{ Li}_{\small1}{\left({\small\frac{1-x}{2}}\right)} } \\ &\, \color{olive}{ +\,\,2\,\,\log^3\left({\small\frac{1-x}{2}}\right) \text{ Li}_{\small2}{\left({\small\frac{1-x}{2}}\right)} } \\ &\, \color{olive}{ -\,\,6\,\,\log^2\left({\small\frac{1-x}{2}}\right) \text{ Li}_{\small3}{\left({\small\frac{1-x}{2}}\right)} } \\ &\, \color{olive}{ +\,12\,\log^1\left({\small\frac{1-x}{2}}\right) \text{ Li}_{\small4}{\left({\small\frac{1-x}{2}}\right)} } \\ &\, \color{olive}{ -\,12\,\log^0\left({\small\frac{1-x}{2}}\right) \text{ Li}_{\small5}{\left({\small\frac{1-x}{2}}\right)} } \\[3mm] \color{orange}{I_{_4}(x)} =&\, \color{orange}{ +{\small\frac{1}{10}}\log^5\left({\small\frac{1+x}{2}}\right) } \\ &\, \color{orange}{ +\,{\small\frac{1}{2}}\,\log^4\left({\small\frac{1+x}{2}}\right) \text{ Li}_{\small1}{\left({\small\frac{1+x}{2}}\right)} } \\ &\, \color{orange}{ -\,\,2\,\,\log^3\left({\small\frac{1+x}{2}}\right) \text{ Li}_{\small2}{\left({\small\frac{1+x}{2}}\right)} } \\ &\, \color{orange}{ +\,\,6\,\,\log^2\left({\small\frac{1+x}{2}}\right) \text{ Li}_{\small3}{\left({\small\frac{1+x}{2}}\right)} } \\ &\, \color{orange}{ -\,12\,\log^1\left({\small\frac{1+x}{2}}\right) \text{ Li}_{\small4}{\left({\small\frac{1+x}{2}}\right)} } \\ &\, \color{orange}{ +\,12\,\log^0\left({\small\frac{1+x}{2}}\right) \text{ Li}_{\small5}{\left({\small\frac{1+x}{2}}\right)} } \end{align} $$

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$\underline{\space{\bf\text{5.}}\space}$ Define: $ \displaystyle f(x) = \color{brown}{I_{_1}{(x)}} + \color{magenta}{I_{_2}{(x)}} - \color{olive}{2\,I_{_3}{(x)}} - \color{orange}{2\,I_{_4}{(x)}} $
And calculate the limits at the integral boundaries: $$ \begin{align} \lim_{x\to 1}\,f(x)\, =&\, -4\,\pi^2\zeta(3)+48\,\zeta(5)-i\,\frac{19\,\pi^5}{30} \\[1mm] \lim_{x\to i\sqrt{3}}\,f(x)\, =&\, +i\,\frac{\pi^5}{81}-i\,192 \,{_{\small6}F_{\small5}}{\left({\small\frac12,\frac12,\frac12,\frac12,\frac12,\frac12};\,{\small\frac32,\frac32,\frac32,\frac32,\frac32};\,{\small\frac14}\right)} \end{align} $$

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$\underline{\space{\bf\text{6.}}\space}$ Simplify: $$ \begin{align} &\quad\,\, I = -\frac{17\,\pi^5}{12960} -\frac{3}{4}\,\frac{1}{12}\,\,\Im\left[\,f\left(1\right)-f\left(i\sqrt{3}\right)\,\right] \\[1mm] &\boxed{\quad \color{red}{I =\frac{253\,\pi^5}{6480} - 12\,{_{\small6}F_{\small5}}{\left({\small\frac12,\frac12,\frac12,\frac12,\frac12,\frac12};\,{\small\frac32,\frac32,\frac32,\frac32,\frac32};\,{\small\frac14}\right)}} \quad \\} \end{align} $$

Answer 2

If we let $\sqrt{\frac{x}{4-x}}=t$

$$I=2\int_0^{\frac 1{\sqrt 3}}\frac{\tan ^{-1}(t)^3}{t \left(t^2+1\right)}\,\log \left(\frac{4 t^2}{t^2+1}\right)\,dt $$

$$\tan ^{-1}(t)^3=\sum_{n=0}^\infty (-1)^n \frac{a_n}{b_n}\,t^{2n+3}$$ where the $a_n$ and $b_n$ correspond to sequences $A002429$ and $A231121$ in $OEIS$ (they have explicit closed forms).

We then face integrals $$J_n=\int_0^{\frac 1{\sqrt 3}} \frac{t^{2(n+1)} }{t^2+1}\,\log \left(4 t^2\right)\,dt- \int_0^{\frac 1{\sqrt 3}} \frac{t^{2(n+1)} }{t^2+1}\,\log \left( t^2+1\right)\,dt$$

The required antiderivatives express in terms of hypergeometric functions

$$\int_0^{\frac 1{\sqrt 3}} \frac{t^{2(n+1)} }{t^2+1}\,\log \left(4 t^2\right)\,dt=-\frac{1}{2} 3^{-n-\frac{1}{2}} \Phi \left(-3,2,-n-\frac{1}{2}\right)-$$ $$\frac{1}{8} (-1)^n \left(8 \pi \log (2)+4 i \log \left(\frac{4}{3}\right) B_{-3}\left(-n-\frac{1}{2},0\right)\right)$$

$$\int_0^{\frac 1{\sqrt 3}} \frac{t^{2(n+1)} }{t^2+1}\,\log \left( t^2+1\right)\,dt=\frac{ \gamma}{2} 3^{-n-\frac{5}{2}} \Phi \left(-\frac{1}{3},1,n+\frac{5}{2}\right)-$$ $$\frac{1}{2} 3^{-n-\frac{5}{2}} \Gamma \left(n+\frac{5}{2}\right)\, _2\tilde{F}_1^{(\{0,0,0\},\{0,1\},0)}\left(1,n+\frac{5}{2};n+\frac{7}{2};-\frac{1}{3}\right)$$

Computing the partial sums

$$\left( \begin{array}{cc} p & I_{(p)} \\ 0 & -0.0645119 \\ 1 & -0.0573551 \\ 2 & -0.0584397 \\ 3 & -0.0582429 \\ 4 & -0.0582829 \\ 5 & -0.0582741 \\ 6 & -0.0582762 \\ 7 & -0.0582757 \\ 8 & -0.0582758 \\ 9 & -0.0582758 \\ \end{array} \right)$$

The $ISC$ does not anything looking like $-0.0582757833331166168483583\cdots$

Answer 3

A physicist's approach:

Let's consider the general case

$$I(n) =\int_{0}^{1}\arctan^{n}\left(\sqrt{\frac{x}{4 -x}}\right)\frac{\ln\left(x\right)}{x}\,{\rm d}x$$

By going through the same procedures as in the original post, we get

$$I(n)=-2n\int_0^{\pi/6}x^{n-1}\ln^{2}\left(2\sin\left(x\right)\right)\,dx$$

Now we use the fact that in the integration range $ \left (0,\frac{\pi}{6} \right )$ $\sin(x)$ differs little from $\frac{3x}{\pi}$:

I.e.

$$\sin(x)\approx \frac{3x}{\pi}$$

Therefore

$$I(n)\approx-2n\int_0^{\pi/6}x^{n-1}\ln^{2}\frac{6x}{\pi}\,dx=-4\frac{\left (\frac{\pi}{6}\right )^n }{n^2}$$

From the result, it is easy to see according to which law the integral approaches zero as a function of $n$.