I was messing around with the Dirichlet Beta Function and was able to get a formula:
$$\int_0^1 \frac{\ln(\ln(p))\ln(p)^{x-1}}{1+p^2}dp = \Gamma(x)(-1)^x(\beta'(x)-\beta(x)(i\pi +\psi(x)) $$ where $\beta(x)$ is the Dirichlet Beta Function and $\beta'(x)$ is its derivative. $\Gamma(x)$ is the Gamma Function and $\psi(x)$ is the Digamma Function and $i$ is the imaginary unit. Plugging in $1$ into the equation, you get the (what I thought to be cool) integral:
$$\int_0^1 \frac{\ln(\ln(p))}{1+p^2}dp = 2\ln(2)+3\ln(\pi)-4\ln(\Gamma(\frac{1}{4}))+i\frac{\pi^2}{4}$$
Looking at a graph and the function itself, it's not even defined from $0$ to $1$ on the real plane, so how is there even a way this integral has a real part? The only thing I can think of is an analytic continuation for the Gamma Function, but I don't really know anything about analytic continuation so it's just a hunch.