TITLE: Integral $\int_0^1\frac{dx}{\sqrt{\log \frac{1}{x}}}=\sqrt \pi$
QUESTION [6 upvotes]: Hi I am trying to prove this result below
$$
\mathcal{J}:=\int_0^1\frac{dx}{\sqrt{\log \frac{1}{x}}}=\sqrt \pi.
$$
The result is quite interesting however I realized I am not familiar with working square roots of log functions like this.  I am more stumped than usual because there isn't much to work with here.  The indefinite integral is given by
$$
\int \frac{dx}{\sqrt{\log \frac{1}{x}}}=-\sqrt \pi\, \text{erf}\left(\sqrt {\log \frac{1}{x}}\right),
$$
although I cannot prove this.  Possibly a proof to this will lead to the result of the definite integral.  
It seems the integral is somehow related to a Gaussian integral possibly, I notice the error function and the result $\sqrt \pi$.  A solution would be greatly appreciated and I hope could also be of use to the math community here.  
Thanks.  In case anybody likes this integral and is interested in a similar one, here is another one for you:
$$
\int_0^1 \sqrt{\log \frac{1}{x}} \,dx=\frac{\sqrt \pi}{2}
$$

REPLY [7 votes]: This is Euler's first historical integral expression for the $\Gamma$ function. No substitution necessary. The integral is simply $\Big(-\frac12\Big)!=\Gamma\Big(1-\frac12\Big)=\Gamma\Big(\frac12\Big)=\sqrt\pi.~$ QED.