Calculating the integral $\int_0^1 \frac{dx}{\sqrt{-\ln(x)}}$ using Euler integrals

Question

I'm trying to calculate the integral $\int_0^1 \frac{dx}{\sqrt{-\ln(x)}}$ using Euler integrals ($\Gamma(x)$ and $B$(x,y)$).

I basically have to find a way to make that integral resemble one of the Euler integrals so that I can easily find a value.

I reasoned that since the integration is from $0$ to $1$, I either need to use the $B$ integral or I need to use the $\Gamma$ integral with a substitution that gives the proper boundaries.

Not really knowing where to go from there I tried maybe getting an $e$ in there by writing $e^{\ln(\ln(x))}$ instead of $\ln(x)$, but that gets me nowhere.

I tried using a $B$ integral by using the same variable twice instead of 2 different ones in its definition but I don't think I'll ever get a logarithm and substitution doesn't seem to work either.

I'm having trouble with other integrals, mainly the ones with logarithms, I don't really know where to begin when having to apply Euler integrals.

Answer 1

Change variables to $u=-\log{x}$. Then the limits become $\infty$ and $0$, and $x=e^{-u}$, $dx = -e^{-u} du$, so the integral is, swapping the limits, $$ \int_0^{\infty} u^{-1/2} e^{-u} \, du = \Gamma(1/2) $$

Your original was in fact the form that Euler originally considered for the $\Gamma$-function, by the way: the form we know and love is due to Legendre.

Answer 2

Sub $x=e^{-y}$; then the integral is equal to

$$\int_0^{\infty} dy \, y^{-1/2} e^{-y} = 2 \int_0^{\infty} du \, e^{-u^2} = \sqrt{\pi}$$

The first integral is $\Gamma(1/2)$.

Answer 3

$n!=\displaystyle\int_0^1\big(-\ln x\big)^n~dx~$ was Euler's first historical integral expression for the $\Gamma$ function, so all you have to do is to notice that $n=-\dfrac12$. :-$)$