$$\int_0^1 \sqrt{-\ln x} dx$$
I'm looking for alternative methods to what I already know (method I have used below) to evaluate this Integral.
$$y=-\ln x$$
$$\bbox[8pt, border:1pt solid crimson]{e^y=e^{-\ln x}=e^{\ln\frac{1}{x}}=\frac{1}{x}}$$
$$\color{#008080}{dx= -\frac{dy}{e^y}}$$
$$\int_0^1 \sqrt{-\ln x} dx=\int_{\infty}^0 \sqrt{y} \left(-\frac{dy}{e^y}\right)=\int_0^{\infty} e^{-y} y^{\frac{1}{2}}dy=\left(\frac{1}{2}\right)!$$
$$\bbox[8pt, border:1pt solid crimson]{\left(\frac{1}{2}\right)!=\frac{1}{2} \sqrt{\pi}}$$
$$\Large{\color{crimson}{\int_0^1 \sqrt{-\ln x} dx=\frac{1}{2} \sqrt{\pi}}}$$
Best Answer
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,