One more Integral

Let $I$ be the integral, then we have:

$\begin{aligned} I & = \int_0^1 \frac{dx}{\sqrt{-\ln x}} \quad \quad \small \color{#3D99F6}{\text{Let } t = - \ln x, \space x = e^{-t}, \space dx = - e^{-t} dt} \\ & = \int_0^\infty t^{-\frac{1}{2}}e^{-t} dt \\ & = \int_0^\infty t^{\color{#3D99F6}{\frac{1}{2}}-1}e^{-t} dt \\ & = \Gamma \left(\color{#3D99F6}{\frac{1}{2}}\right) \quad \quad \quad \quad \space \space \small \color{#3D99F6}{\Gamma(x) \text{ is Gamma function}} \\ & = \sqrt{\pi} \approx \boxed{1.772} \end{aligned}$

Let $u = \sqrt{-\ln(x)}$ . Then $du = \dfrac{-\dfrac{1}{x}}{2\sqrt{-\ln(x)}} dx \Longrightarrow \dfrac{dx}{\sqrt{-\ln(x)}} = -2x du$ .

But as $u^{2} = -\ln(x) \Longrightarrow x = e^{-u^{2}}$ , and since $u$ goes from $\infty$ to $0$ as $x$ goes from $0$ to $1$ , the given integral can be written as

$-2\displaystyle\int_{\infty}^{0} e^{-u^{2}} du = 2 \int_{0}^{\infty} e^{-u^{2}} du = 2 * \dfrac{\sqrt{\pi}}{2} = \sqrt{\pi} = \boxed{1.772}$ to 3 decimal places.

Note that this last integral, by symmetry, is one-half the value of the Gaussian integral .