Calculate :
Is it a definite integral actually ?
If we replace x by $\displaystyle e^t $
then $\displaystyle dx = e^t dt$
the integral $\displaystyle = \int e^t \sqrt{t} ~dt $
I am afraid that i dont have confidence to solve it !
If the u. limit is 1 and l. limit is 0 , then this integral
$\displaystyle \int_0^1 \sqrt{-\ln{x}}~dx ~~$ can be written as
$\displaystyle \int_0^{\infty} e^{-t} \sqrt{t} ~dt = \Gamma(3/2) = \frac{\sqrt{\pi}}{2}$
According to Mathematica, the answer is
$\displaystyle x\sqrt{\ln{x}} - \frac{1}{2}\sqrt{\pi}\,\textrm{erfi}(\sqrt{\ln{x}} )$
Erfi -- from Wolfram MathWorld
There is an identity which states:
$\displaystyle \int ln^{n}(x)dx=(-1)^{n}\cdot x\sum_{k=0}^{n}\frac{n!}{k!}(-ln(x))^{k}$
This would be interesting to prove.
Also, since $\displaystyle Ex(x,n)=\sum_{k=0}^{n}\frac{x^{k}}{k!}$,
the above can be written as $\displaystyle (-1)^{n}n!xEx(-ln(x),n)$
and $\displaystyle Ex(x,n)=\frac{e^{x}\Gamma(n+1,x)}{\Gamma(n+1)}$
where $\displaystyle {\Gamma}(x,n)$ is the incomplete gamma function.
Just some food for thought.