Intuitively looks like $\log$ doesn't really influence convergence so much and I'd guess that answer is something like $n\geq 4999$, but I'm having trouble formalizing that.
Well, for $n=4999$ we have $\int_0^{0.5} \frac{1}{x (\log(x))^{50}} dx$ and substitution $y=\log(x)$ would prove convergence. For $n \geq 5000$ convergence is obvious.
Thank you.
Ok, I have it now. Put $x=e^{-z}$, $dx=-e^{-z}\,dz$. Then $$ \int _0^{1/2} x^{n-5000}(\log(x))^{-50}\,dx = \int _{\log(2)}^{\infty} e^{(4999-n) z} z^{-50}\,dz $$Clearly if $n\geq 4999$ this converges (power rule if $n=4999$, direct comparison to $z^{-50}$ if $n>4999$). But if $n<4999$ then the integral can't converge because the exponential term dominates the power. So we get convergence exactly when $n\geq 4999$.