Who's up to the challenge? 64

$\int_0^{2 \ln\phi} \ln\left(2\sinh\left(\dfrac{x}{2}\right)\right) dx = \int_0^{2\ln\phi} \ln\left(\dfrac{e^x-1}{e^{x/2}}\right)dx$ $= \int_0^{2\ln\phi} \ln (e^x-1) dx - \int_0^{2\ln\phi} \dfrac{x}{2}dx$

Now let $u = e^x$ . Then we can rewrite the integral: $=\int_{1}^{\phi^2} \dfrac{\ln(u-1)}{u} du - \ln^2(\phi)$ Let's rewrite it once again using integration by parts: $= -\ln^2(\phi) + \ln(u)\ln(u-1) \large \rvert_{1}^{\phi^2} \normalsize -\int_{1}^{\phi^2} \dfrac{\ln(u)}{u-1} du$ Make the substitution $y=\dfrac{1}{u}$ : $= \ln^2(\phi) + \int_1^{1/\phi^2} \dfrac{\ln(y)}{y^2-y}dy$ $= \ln^2(\phi)+ \int_1^{1/\phi^2} \dfrac{\ln(y)}{y-1} dy - \int_1^{1/\phi^2} \dfrac{\ln(y)}{y} dy$ $= \ln^2(\phi) - \text{Li}_2(1-y) \large \rvert_{1}^{1/\phi^2} \normalsize -\dfrac{\ln^2(y)}{2} \large \rvert_{1}^{1/\phi^2}$ $= -\ln^2(\phi) - \text{Li}_2\left(\dfrac{1}{\phi}\right)$

To find the value of this expression, I am going to use three lemmas involving the dilogarithm:

$\begin{array}{ll} (1) & \text{Li}_2(z^2) = 2\text{Li}_2(z) + 2\text{Li}_2(-z)\\ (2) & \text{Li}_2(z)+\text{Li}_2(1-z) = \dfrac{\pi^2}{6} -\ln(z)\ln(1-z)\\ (3) & \text{Li}_2(z) + \text{Li}_2\left(\dfrac{z}{z-1}\right) = -\dfrac{1}{2} \ln^2(1-z)\end{array}$

These can all be proved easily by differentiating, so I won't waste time with that here.

If we substitute $z = \dfrac{1}{\phi }$ into the first equation and $z= \dfrac{1}{\phi^2}$ into the others, we get the following system of equations:

$\begin{array}{ll} (1) & \text{Li}_2\left(\dfrac{1}{\phi^2}\right) = 2\text{Li}_2\left(\dfrac{1}{\phi}\right)+ 2\text{Li}_2\left(\dfrac{-1}{\phi}\right)\\ (2) & \text{Li}_2\left(\dfrac{1}{\phi^2}\right) + \text{Li}_2\left(\dfrac{1}{\phi}\right)= \dfrac{\pi^2}{6} -2\ln^2(\phi)\\ (3) & \text{Li}_2\left(\dfrac{1}{\phi^2}\right) + \text{Li}_2\left(\dfrac{-1}{\phi}\right) = -\dfrac{1}{2}\ln^2(\phi) \end{array}$

From this system we get $\text{Li}_2\left(\dfrac{1}{\phi}\right) = \dfrac{\pi^2}{10} - \ln^2(\phi)$ . Therefore the value of the integral is $\boxed{-\dfrac{\pi^2}{10}}$ .