Integral $\int_0^1\frac{dx}{\sqrt{\log \frac{1}{x}}}=\sqrt \pi$ 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:
$$
\int_0^1 \sqrt{\log \frac{1}{x}} \,dx=\frac{\sqrt \pi}{2}
$$
 A: 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.
A: Note that
$$\int_0^1\left(\log\frac1x \right)^pdx=\int_0^{\infty}u^pe^{-u}du=\Gamma(p+1)$$
by making the simple substitution $x=e^{-u}$.
Your integrals follow immediately from this.
A: Integrating by parts, and borrowing @Pranav's value for the first integral,
$$\int_0^1 \sqrt{\log \frac{1}{x}} \,dx= x \sqrt{\log \frac{1}{x}}\big{|}_0^1- \int_0^1 x \frac{dx}{(-2x)\sqrt{\log \frac{1}{x}}}\\
=\frac12 \frac{}{}\int_0^1  \frac{dx}{\sqrt{\log \frac{1}{x}}}\\
=\frac{\sqrt{\pi}}{2}$$
A: Rewrite $\log(1/x)=-\log(x)$ and use the substitution $-\log x=t^2 \Rightarrow dx=-2te^{-t^2}dt$ to obtain:
$$I=\int_0^1 \frac{dx}{\sqrt{\log \frac{1}{x}}}=\int_0^{\infty} 2e^{-t^2}\,dt=\boxed{\sqrt{\pi}}$$

For the second one, do the same thing to obtain the integral:
$$\int_0^{\infty} 2t^2e^{-t^2}\,dt$$
Since
$$\int_0^{\infty} e^{-(at)^2}\,dt=\frac{\sqrt{\pi}}{2a}$$
Differentiate both the sides wrt $a$ to obtain:
$$\int_0^{\infty} 2at^2e^{-(at)^2}\,dt=\frac{\sqrt{\pi}}{2a^2}$$
Substitute $a=1$ to obtain the answer.
