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The integral is $$\int_0^1\frac{dx}{\sqrt{-\ln x}}.$$ Not sure if it helps, but it is in the same problem section as $$\int_0^\infty e^{-x^2}dx.$$

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Let $u = -\ln x$. Then $u \to \infty$ as $x \to 0$ and $u \to 0$ as $x \to 1$. Further, since $x = e^{-u}$, $dx = -e^{-u}\, du$. It follows that

$$\int_0^1 \frac{1}{(-\ln x)^{1/2}}\, dx = \int_0^\infty u^{-1/2}e^{-u}\, du.$$

Using the substitution $u = v^2$, we have

$$\int_0^\infty u^{-1/2}e^{-u}\, du = \int_0^\infty v^{-1} e^{-v^2} \cdot 2v\, dv = 2\int_0^\infty e^{-v^2}\, dv = 2\cdot\frac{\sqrt{\pi}}{2} = \sqrt{\pi}.$$

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Using the substitution $y = \sqrt{-\ln x}$, $$ \int_0^1\frac{1}{\sqrt{-\ln x}}dx = 2\int_0^\infty e^{-y^2}dy = \sqrt{\pi}. $$

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This is Euler's first integral formula for the $\Gamma$ function. $\big($Alternately, let $x=e^{-t}$ to arrive at a more familiar expression, then $t=u^2$ to arrive at the expression of the Gaussian integral, whose various evaluations can also be found here$\big)$.

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