$\lim_{n\rightarrow +\infty} \int_{1}^{e} (\log x)^n dx $ and $\lim_{n\rightarrow +\infty} \int_{1}^{3} (\log x)^n dx $

Question

I want to calculate this limit.Can I prove the uniformly convergence of the succession so i can exchange limit and integral?

Answer 1

For fixed small $\varepsilon>0$, $$ 0\le \log x\le\log(e-\varepsilon)<1, x\in[1,e-\varepsilon]$$ and hence $$ \int_{1}^{e} (\log x)^n dx=\int_{1}^{e-\varepsilon} (\log x)^n dx +\int_{e-\varepsilon}^{e} (\log x)^n dx \le(\log(e-\varepsilon))^n(e-1)+\varepsilon. $$ Letting $n\to\infty$ gives $$ 0\le\overline{\lim_{n\to\infty}}\int_{1}^{e} (\log x)^n dx\le\varepsilon$$ and thus one has $$ \lim_{n\rightarrow +\infty} \int_{1}^{e} (\log x)^n dx=0.$$ One can use the same to show $$ \lim_{n\rightarrow +\infty} \int_{1}^{3} (\log x)^n dx=\infty.$$

Answer 2

Both integrals can be tackled through the substitution $x=e^{z}$, without switching $\lim$ and $\int$.

$$0\leq \int_{1}^{e}\left(\log x\right)^n\,dx = \int_{0}^{1} z^n e^{z}\,dz \leq e\int_{0}^{1}z^n\,dz = \frac{e}{n+1}\to 0, $$ $$\int_{1}^{3}\left(\log x\right)^n\,dx \geq \int_{1}^{\log 3}z^n e^{z}\,dz \geq e\frac{\left(\log 3\right)^{n+1}-1}{n+1}\to +\infty.$$