Solving the integral $\int\sqrt{\ln(x)}\,dx $

Let $x=e^{-u}$ and you get

$$\int_a^b\sqrt{\ln x}\ dx=-\int_{-\ln a}^{-\ln b}\sqrt{-u}e^{-u}\ du=i\left(\gamma(3/2,-\ln a)-\gamma(3/2,-\ln b)\right)$$

where $\gamma(a,x)$ is the lower incomplete gamma function.

Considering $$I=\int\sqrt{\log( x)}\ dx$$ change variable $$\sqrt{\log( x)}=t\implies x=e^{t^2}\implies dx=2te^{t^2}\,dt$$ which makes $$I=\int 2t^2e^{t^2}\,dt$$ Now, integration by parts $$u=t\implies u'=dt$$ $$v'=2te^{t^2}\,dt\implies v=e^{t^2}$$ $$I=te^{t^2}-\int e^{t^2}\,dt=te^{t^2} -\frac{\sqrt{\pi }}{2} \text{erfi}(t)$$ where appears the imaginary error function.

Back to $x$, $$I=x \sqrt{\log (x)}-\frac{\sqrt{\pi }}{2} \text{erfi}\left(\sqrt{\log (x)}\right)$$ which is real if $x\geq 1$.