What are other methods to evaluate $\int_0^1 \sqrt{-\ln x} \ \mathrm dx$

Here's slightly different approach

We have $$\mathcal I=\int_0^1 \sqrt{-\ln x} \mathrm dx=\int_0^1 \left[\ln\frac1 x\right]^{1/2} \mathrm dx$$ By using IBP $$\left[\ln\frac1 x\right]^{1/2}=u\iff\mathrm du= -\frac 1{2x}\left[\ln\frac1 x\right]^{-1/2}\mathrm dx$$

$$\mathrm dx=\mathrm dv\iff x=v$$

$$\begin{align} \mathcal I &=x\left[\ln\frac1 x\right]^{1/2}\Bigg\rvert_0^1+ \frac 1{2}\int_0^1\left[\ln\frac1 x\right]^{-1/2}\mathrm dx\\ &=\frac 1{2}\int_0^1\left[\ln\frac1 x\right]^{-1/2}\mathrm dx\\ \end{align}$$ Now by substututing $$t^2=\ln\frac1x\implies e^{-t^2}=x\iff \mathrm dx=-2te^{-t^2}\mathrm dt$$ We get $$\mathcal I=\int _0^\infty e^{-t^2}\mathrm dt=\frac{\sqrt\pi}{2}$$ Hence,

$$\int_0^1 \sqrt{-\ln x} \mathrm dx=\frac{\sqrt\pi}{2}$$

As far as I know, there are essentially two ways for proving that $\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$.

The first way is to use integration by parts, leading to $\Gamma(z+1)=z\Gamma(z)$, in order to relate $\Gamma\left(\frac{1}{2}\right)$ to the Wallis product. The latter can be computed by exploiting the Weierstrass product for the sine function: $$\frac{\sin z}{z}=\prod_{n=1}^{+\infty}\left(1-\frac{z^2}{\pi^2n^2}\right)$$ by evaluating it in $z=\frac{\pi}{2}$. The duplication formula for the $\Gamma$ function follows from this approach.

The second way is to use some substitutions in order to relate $\Gamma\left(\frac{1}{2}\right)$ to the gaussian integral $$\int_{-\infty}^{+\infty}e^{-x^2}\,dx$$ that can be evaluated through Fubini's theorem and polar coordinates.

Similar to what Iuʇǝƃɹɐʇoɹ did, let $u=\sqrt{-\ln x}$. Then $x=e^{-u^2}$ and hence $$ \int_0^1\sqrt{-\ln x}dx=-\int_0^\infty ud(e^{-u^2})=-ue^{-u^2}|_0^\infty+\int_0^\infty e^{-u^2}du=\frac{\sqrt{\pi}}{2}. $$