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

$$\int_0^1 \sqrt{-\ln x} dx$$
I'm looking for alternative methods to what I already know (method I have used below) to evaluate this Integral.
$$y=-\ln x$$

$$\bbox[8pt, border:1pt solid crimson]{e^y=e^{-\ln x}=e^{\ln\frac{1}{x}}=\frac{1}{x}}$$

$$\color{#008080}{dx= -\frac{dy}{e^y}}$$

$$\int_0^1 \sqrt{-\ln x} dx=\int_{\infty}^0 \sqrt{y} \left(-\frac{dy}{e^y}\right)=\int_0^{\infty} e^{-y} y^{\frac{1}{2}}dy=\left(\frac{1}{2}\right)!$$

$$\bbox[8pt, border:1pt solid crimson]{\left(\frac{1}{2}\right)!=\frac{1}{2} \sqrt{\pi}}$$

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

Answer

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.