SF11

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\int_0^1 \frac{1}{\sqrt{-\ln x}}\,dx

Let -\ln x = t\,; then x = e^{-t}, dx = -e^{-t}\,dt

Limits: as x \to 0^+, t = -\ln x \to \infty; x=1\implies t = \,-\ln 1 = 0

Thus,

\int_0^1 \frac{1}{\sqrt{-\ln x}}\,dx =- \int_\infty^0 \frac{e^{-t}}{\sqrt{t}}\,dt = \int_0^\infty e^{-t}\,t^{-1/2}\,dt = \Gamma(1/2) = \sqrt{\pi}
(see SF8 for the last result)

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