MHB Evaluate Integral: $\int_{0}^{1}\dfrac{dx}{\sqrt{-ln x}}$

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The integral $\int_{0}^{1}\dfrac{dx}{\sqrt{-\ln x}}$ can be evaluated using the substitution $u = -\ln(x)$, transforming the limits as $u$ goes from 0 to $\infty$. This leads to the integral $\int_0^{\infty} e^{-u} u^{-\frac{1}{2}} du$, which is recognized as the Gamma function $\Gamma\left(\frac{1}{2}\right)$. The result of this integral is $\sqrt{\pi}$. An alternative substitution of $-\log x = t^2$ also yields the same result. The final value of the integral is $\sqrt{\pi}$.
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Please evaluate the following integral:
$\int_{0}^{1}\dfrac{dx}{\sqrt{-ln x}}$
 
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Re: evaluate integral-01

Albert said:
Please evaluate the following integral:
$\int_{0}^{1}\dfrac{dx}{\sqrt{-ln x}}$

[math]\displaystyle \begin{align*} \int_0^1{\frac{dx}{\sqrt{-\ln{(x)}}}} &= \int_0^1{\frac{-x\,dx}{-x\,\sqrt{-\ln{(x)}}}} \end{align*}[/math]

Now making the substitution [math]\displaystyle \begin{align*} u = -\ln{(x)} \implies du = -\frac{dx}{x} \end{align*}[/math] and noting that [math]\displaystyle \begin{align*} u(1) = 0 \end{align*}[/math] and as [math]\displaystyle \begin{align*} x \to 0^+ , u \to +\infty \end{align*}[/math], the integral becomes

[math]\displaystyle \begin{align*} \int_0^1{\frac{x\,dx}{x\,\sqrt{-\ln{(x)}}}} &= \int_{\infty}^0{\frac{-e^{-u}\,du}{\sqrt{u}}} \\ &= \int_0^{\infty}{e^{-u}\,u^{-\frac{1}{2}}\,du} \\ &= \int_0^{\infty}{u^{\frac{1}{2} - 1 } \, e^{-u}\,du} \\ &= \Gamma{ \left( \frac{1}{2} \right) } \\ &= \sqrt{\pi} \end{align*}[/math]
 
Re: evaluate integral-01

Another way: using the substitution $-\log x= t^2$ (or equivalently $x=e^{-t^2}$) and the Euler's integral.

$$\int_0^1\frac{dx}{\sqrt{-\log x}}=\int_{+\infty}^0\frac{-2te^{-t^2}dt}{\sqrt{t^2}}=2\int_0^{+\infty}e^{-t^2}dt=2\frac{\sqrt{\pi}}{2}=\sqrt{\pi}$$
 
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