A Challenging Integral: $\int_{0}^{1} \frac{x \arcsin(x/2) \log(x)}{x^2-1} \ dx=\frac{5 \pi^3}{1296}$

I wish to evaluate the integral $$I=\int_{0}^{1} \frac{x \arcsin(x/2) \log(x)}{x^2-1} \ dx.$$ I used Mathematica's Rationalize command to see that it is equal to $5 \pi^3/1296,$ but I do not know how to prove it analytically.

Context

The integral appears as part of an alternative expression for the triple integral $$J=\int_{0}^{1} \int_{0}^{1}\int_{0}^{1} \frac{x^2y}{\sqrt{4-x^2}{\sqrt{4-x^2y^2} \sqrt{4-x^2y^2z^2}}} \ dz \ dy \ dx.$$ It is easy to evaluate $J$ directly and show it is $\frac{\pi^3}{1296}.$ However, if we reverse the order of integration and integrate with respect to $y$ first, we get

$$J=- \int_{0}^{1} \int_{0}^{1} \frac{\log \left(\frac{\sqrt{4-x^2 z^2}+\sqrt{4-x^2} z}{2 z+2}\right)}{\sqrt{4-x^2} z} \ dz \ dx.$$ The integral can be expanded into the triple integral $$- \int_{0}^{1} \int_{0}^{1} \int_{0}^{z}\frac{x^2}{z \left(x^2 \left(t^2 \sqrt{4-x^2}+\sqrt{4-t^2 x^2}\right)-4 \left(\sqrt{4-t^2 x^2}+\sqrt{4-x^2}\right)\right)} \ dt \ dz \ dx.$$ Reversing the order of integration and integrating with respect to $x$ first, we can deduce \begin{align*} J &= \int_{0}^{1} \int_{t}^{1}\frac{\frac{\pi }{6} t-\sin ^{-1}\left(\frac{t}{2}\right)}{\left(t-t^3\right) z} \ dz \ dt \\ &= -\int_{0}^{1}\frac{\frac{\pi}{6} t \log(t) -\sin ^{-1}\left(\frac{t}{2}\right) \log(t)}{t-t^3} \ dt \\ &=\frac{\pi}{6} \int_{0}^{1}\frac{\log(t)}{t^2-1} \ dt + \int_{0}^{1} \frac{\sin ^{-1}\left(\frac{t}{2}\right) \log(t)}{t-t^3} \ dt \\ &= \frac{\pi^3}{48} + \int_{0}^{1} \frac{\log (t) \sin ^{-1}\left(\frac{t}{2}\right)}{t} \ dt - \int_{0}^{1} \frac{t \log (t) \sin ^{-1}\left(\frac{t}{2}\right)}{t^2-1} \ dt, \end{align*} in which we used the well-known result $$\int_{0}^{1} \frac{\log(t)}{t^2-1} \ dt = \frac{\pi^2}{8}$$ and partial fractions on the second integral term in the second to last equality.

Recalling $J=\pi^3/1296,$ rearranging gives $$ \begin{align*} -\frac{13 \pi ^3}{648} &=\int_{0}^{1} \frac{\log (t) \sin ^{-1}\left(\frac{t}{2}\right)}{t} \ dt - \int_{0}^{1} \frac{t \log (t) \sin ^{-1}\left(\frac{t}{2}\right)}{t^2-1} \ dt \\ &=\int_{0}^{1} \frac{\log (t) \sin ^{-1}\left(\frac{t}{2}\right)}{t} \ dt - I. \end{align*}$$

Question

Upon using integrating by parts, it turns out $$\int_{0}^{1} \frac{\log (t) \sin ^{-1}\left(\frac{t}{2}\right)}{t} \ dt= -\int_{0}^{1} \frac{\log^2(t)}{2\sqrt{4-t^2}} \ dt.$$ Further substituting $t=2 \sin(\theta)$ and using the answers in either Interesting log sine integrals $\int_0^{\pi/3} \log^2 \left(2\sin \frac{x}{2} \right)dx= \frac{7\pi^3}{108}$ or Finding $\int^{1}_{0}\frac{\ln^2(x)}{\sqrt{4-x^2}}dx$, we can deduce
$$\int_{0}^{1} \frac{\log (t) \sin ^{-1}\left(\frac{t}{2}\right)}{t} \ dt = -\frac{7 \pi^3}{432}.$$ Hence, assuming this, we can get $$I= \frac{5 \pi^3}{1296}.$$

Can we evaluate $I$ without knowing the value of the integral $\int_{0}^{1} \frac{\log (t) \sin ^{-1}\left(\frac{t}{2}\right)}{t} \ dt$ ? Can $I$ be evaluated with real methods ?

Solution 1:

Step 1. Substituting $x = 2\sin(\theta/2)$ and noting that

\begin{align*} \arcsin(x/2)\log(x) &= (\theta/2) \log(2\sin(\theta/2)) \\ &= \frac{1}{i} \log(e^{i\theta/2}) \log\bigl(i(e^{-i\theta/2} - e^{i\theta/2})\bigr) \\ &= \frac{1}{2}\operatorname{Im}\left[ \log^2\bigl(i(1 - e^{i\theta})\bigr) \right] \end{align*}

for $0 < \theta < \pi$, the integral is recast as

\begin{align*} I &= \frac{1}{2} \int_{0}^{\frac{\pi}{3}} \frac{\sin\theta}{1-2\cos\theta} \operatorname{Im}\left[ \log^2 \left( i(1 - e^{i\theta}) \right) \right] \, \mathrm{d}\theta. \end{align*}

Here, $\log(\cdot)$ is the principal logarithm. To utilize this, for $0<\epsilon<\frac{\pi}{3}$, we introduce the following auxiliary integral

$$ J_{\epsilon} = \frac{1}{2} \int_{0}^{\frac{\pi}{3}-\epsilon} \frac{\sin\theta}{1-2\cos\theta} \left( \log (1 - e^{i\theta}) + \frac{i\pi}{2} \right)^2 \, \mathrm{d}\theta. $$

$J_{\epsilon}$ is related to $I$ by the relation $I = \lim_{\epsilon \to 0^+}\operatorname{Im}(J_{\epsilon})$.

Step 2. Unfortunately, $J_{\epsilon}$ diverges as $\epsilon \to 0^+$ since its real part does. The key idea in the next step is to decompose $J_{\epsilon}$ that allows to separate the diverging real part from $J_{\epsilon}$.

Consider the curve $\gamma_{\varepsilon}$ in $\mathbb{C}$ which is defined by $\gamma_{\varepsilon}(\theta) = 1 - e^{i\theta}$ for $0 \leq \theta \leq \frac{\pi}{3}-\epsilon$, and substitute $z = \gamma_{\varepsilon}(\theta)$. Then

\begin{align*} J_{\epsilon} &= \frac{1}{2} \int_{\gamma_{\epsilon}} \frac{\frac{1}{2i}\bigl((1-z) - \frac{1}{1-z}\bigr)}{1-\bigl((1-z) + \frac{1}{1-z}\bigr)} \left(\log z + \frac{i\pi}{2} \right)^2 \, \frac{\mathrm{d}z}{i(z-1)} \\ &= \frac{1}{4} \int_{\gamma_{\varepsilon}} \left( \frac{3z^2}{z^3+1} - \frac{2z}{z^2-1} \right) \left(\log z + \frac{i\pi}{2} \right)^2 \, \mathrm{d}z \\ &= A_{\epsilon} - B_{\epsilon}, \end{align*}

where $A_{\epsilon}$ and $B_{\epsilon}$ are defined by

$$ A_{\epsilon} = \frac{1}{4} \int_{\gamma_{\varepsilon}} \frac{3z^2}{z^3+1} \left(\log z + \frac{i\pi}{2} \right)^2 \, \mathrm{d}z \quad\text{and}\quad B_{\epsilon} = \frac{1}{4} \int_{\gamma_{\varepsilon}} \frac{2z}{z^2-1} \left(\log z + \frac{i\pi}{2} \right)^2 \, \mathrm{d}z. $$

To further simplify these integrals, we observe that

$$ \log z = \frac{1}{2}\log(z^2) = \frac{1}{3}\log (-z^3) - \frac{i\pi}{3} \qquad \text{for } z \in \gamma_{\epsilon}. $$

Here, the last step is a consequence of the branch cut of $\log(\cdot)$ together with the fact that the contours corresponding to $\gamma_{\varepsilon}$, $\gamma_{\varepsilon}^2$, and $-\gamma_{\varepsilon}^3$ are as follows:

So, by applying further substitutions $-z^3 \mapsto z$ and $z^2 \mapsto z$ to $A_{\epsilon}$ and $B_{\epsilon}$, respectively, we get

\begin{align*} A_{\epsilon} &= \frac{1}{4} \int_{-\gamma_{\epsilon}^3} \frac{1}{z-1} \left( \frac{1}{3}\log z + \frac{i\pi}{6} \right)^2 \, \mathrm{d}z, \\ B_{\epsilon} &= \frac{1}{4} \int_{\gamma_{\epsilon}^2} \frac{1}{z-1} \left( \frac{1}{2}\log z + \frac{i\pi}{2} \right)^2 \, \mathrm{d}z. \end{align*}

Step 3. For $A_{\varepsilon}$, the integrand has a simple pole at $z = 1$. Moreover, the terminal point of $-\gamma_{\epsilon}^3$ has an asymptotic formula of the form

$$ -(1-e^{i(\pi/3 - \epsilon)})^3 = 1 - 3e^{i\pi/6}\epsilon + \mathcal{O}(\epsilon^2) \qquad \text{as $\epsilon \to 0^+$.} $$

So by replacing $-\gamma_{\epsilon}^3$ by the union of

• $z = x$ for $0 \leq x \leq 1-3\epsilon$, which is the line segment from $0$ to $1-3\epsilon$;
• $z = 1 - 3\epsilon e^{i\theta}$ for $0 \leq \theta \leq \frac{\pi}{6}$, which is the circular arc from $1-3\epsilon$ to $1-3\epsilon e^{i\pi/6}$;
• the line segment from $1-3\epsilon e^{i\pi/6}$ to the terminal point of $-\gamma_{\epsilon}^3$;

it follows that

\begin{align*} &A_{\epsilon} \\ &= \frac{1}{4} \int_{0}^{1-3\epsilon} \frac{1}{x-1} \left( \frac{1}{3}\log x + \frac{i\pi}{6} \right)^2 \, \mathrm{d}x + \frac{i}{4} \int_{0}^{\frac{\pi}{6}} \left( \frac{1}{3}\log(1 - 3\epsilon e^{i\varphi}) + \frac{i\pi}{6} \right)^2 \, \mathrm{d}\varphi + \mathcal{O}(\epsilon) \\ &= \frac{i\pi}{36} \int_{0}^{1-3\epsilon} \frac{\log x}{x-1} \, \mathrm{d}x + \frac{i}{4} \cdot \frac{\pi}{6} \left( \frac{i\pi}{6} \right)^2 + \text{[some real number]} + \mathcal{O}(\epsilon). \end{align*}

As $\epsilon \to 0^+$, its imaginary part converges to

$$ \lim_{\epsilon \to 0^+} \operatorname{Im} (A_{\epsilon}) = \frac{\pi}{36} \int_{0}^{1} \frac{\log x}{x-1} \, \mathrm{d}x - \frac{\pi^3}{864} = \frac{\pi^3}{288}. $$

Step 4. For $B_{\epsilon}$, its limit is simply $B_0$. So,

$$ \lim_{\epsilon \to 0^+} B_{\epsilon} = B_0 = \frac{1}{4} \int_{0}^{e^{-2\pi i/3}} \frac{1}{z-1} \left( \frac{1}{2}\log z + \frac{i\pi}{2} \right)^2 \, \mathrm{d}z $$

Performing integration by parts twice, we end up with

$$ \lim_{\epsilon \to 0^+} B_{\epsilon} = -\frac{\pi^2}{144}\log\bigl(1 - e^{-2\pi i/3} \bigr) + \frac{i\pi}{24} \operatorname{Li}_2\bigl( e^{-2\pi i/3} \bigr) - \frac{1}{8} \operatorname{Li}_3\bigl( e^{-2\pi i/3} \bigr)$$

Although the exact value of $\operatorname{Li}_2\left( e^{-2\pi i/3} \right)$ and $\operatorname{Li}_3\left( e^{-2\pi i/3} \right)$ are not known, it only suffices to know the values of $\operatorname{Re}\left[ \operatorname{Li}_2\left( e^{-2\pi i/3} \right) \right]$ and $\operatorname{Im}\left[ \operatorname{Li}_3\left( e^{-2\pi i/3} \right) \right]$. Moreover, these values can be computed using the Fourier series of the Bernoulli polynomials. The upshot of this observation is that

$$ \operatorname{Re}\bigl[ \operatorname{Li}_2\bigl( e^{-2\pi i/3} \bigr) \bigr] = -\frac{\pi^2}{18}, \qquad \operatorname{Im}\bigl[ \operatorname{Li}_3\bigl( e^{-2\pi i/3} \bigr) \bigr] = -\frac{2\pi^3}{81}. $$

Putting altogether, we end up with

$$ \lim_{\epsilon \to 0^+} \operatorname{Im}( B_{\epsilon} ) = -\frac{\pi^3}{2592}, $$

Conclusion. Combining all the efforts and using the relation

$$ I = \lim_{\epsilon \to 0^+} \operatorname{Im}(J_{\epsilon}) = \lim_{\epsilon \to 0^+} \operatorname{Im}(A_{\epsilon} - B_{\epsilon}), $$

the desired equality is proved.