Dive

Note that $\begin{aligned} \int_0^1 \ln t \, \ln(1-t)\,dt & = \; -\sum_{n=1}^\infty \tfrac{1}{n}\int_0^1 t^n \ln t\,dt \; = \; \sum_{n=1}^\infty \frac{1}{n(n+1)^2} \; = \; \sum_{n=1}^\infty \left( \frac{1}{n} - \frac{1}{n+1} - \frac{1}{(n+1)^2}\right) \\ & = \; 1 - \sum_{n=2}^\infty \frac{1}{n^2} \; = \; 2 - \tfrac16\pi^2 \end{aligned}$ and hence $\begin{aligned} F(a) & = \; \int_0^a \ln x \, \ln(a-x)\,dx \; = \; a\int_0^1 \ln(at)\, \ln(a(1-t))\,dt \; = \; a\int_0^1[\ln a + \ln t][\ln a + \ln(1-t)]\,dt \\ & = \; a\left[(\ln a)^2 + (\ln a)\int_0^1 \ln t\,dt + (\ln a)\int_0^1 \ln(1-t)\,dt + \int_0^1 \ln t\,\ln(1-t)\,dt\right] \; = \; a\left[(\ln a)^2 - 2\ln a + 2 - \tfrac16\pi^2\right] \end{aligned}$ which makes $F'(a) \; = \; (\ln a)^2 - \tfrac16\pi^2$ and so the turning point of $F$ occurs when $a = e^{\frac{\pi}{\sqrt{6}}}$ , making the answer $\boxed{6}$ .