$$ F(k)=\bigg|\int_0^1 \frac{e^{\frac{1}{\log(x)}}}{x\log^k(x)}~dx \bigg|=(k-2)! $$ for $\Re(k)\ge2. $
Can you analytically continue $F(k)?$ If so, how?
I was reading about the Gamma function and how it is defined by a convergent integral, which is subsequently analytically continued using complex analysis. I tried fiddling around with the integral and looked for a substitution to change it into the Gamma function but didn't succeed.
I'll do it in several steps for clarity, but you can do it in one fell swoop if you like.
Let $z=\log(x)$. Then $dz=dx/x$ and the new limits of integration are $-\infty,0$: $$ \bigg|\int_0^1 \frac{e^{\frac{1}{\log(x)}}}{x\log^k(x)}~dx \bigg| $$ $$ \Rightarrow \bigg|\int_{-\infty}^0 \frac{e^{\frac{1}{z}}}{z^k}~dz \bigg| $$Now let $1/z=t$ or $z=1/t$: $$ \Rightarrow \bigg|\int_{0}^{-\infty} {e^{t}}t^{k-2}\,dt \bigg| $$Lastly, put $y=-t$ to get rid of the absolute values (for $k>1$): $$ \Rightarrow \int_{0}^{\infty} {e^{-y}}y^{k-2}\,dy=\Gamma(k-1) $$