# Evaluate Integral - 01

#### Albert

##### Well-known member
$\int_{0}^{1}\dfrac{dx}{\sqrt{-ln x}}$

#### Prove It

##### Well-known member
MHB Math Helper
Re: evaluate integral-01

$\int_{0}^{1}\dfrac{dx}{\sqrt{-ln x}}$
\displaystyle \displaystyle \begin{align*} \int_0^1{\frac{dx}{\sqrt{-\ln{(x)}}}} &= \int_0^1{\frac{-x\,dx}{-x\,\sqrt{-\ln{(x)}}}} \end{align*}

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

\displaystyle \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*}

#### Fernando Revilla

##### Well-known member
MHB Math Helper
Re: evaluate integral-01

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

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