Let $$f_n(x)=(\ln x)^n\quad\forall n\in\mathbb{N}$$Prove that $$\lim_{n\to+\infty}{\int_1^e{f_n(x)}dx=0}$$ I studied the convergence of the sequence of functions and I found that, point wise, $$f_n(x) \to f(x)$$ with $$f(x)=\left\{\begin{array}{ccl}0&\text{for}& e^{-1}<x<e\\ 1&\text{for}& x=e\end{array}\right.$$ So there is not uniform convergence on $[1,e]$ because $f(x)$ is discontinuous while $f_n(x)$ is continuous $\forall n\in\mathbb{N}$ and this prevent me from exchanging the limit with the integral.
My idea is to find a function that is greater than $f_n(x)$ over $[1,e]$ and whose integral tends to $0$ when $n\to+{\infty}$ but I have not been able to find it yet. Any Idea?
By the way, we are doing only Riemann-integrals, not Lebesgue-integrals.
You don't need uniform convergence. Since $0 \le \mbox{ln}(x)^n \le 1$ on $[1, e]$, you have that your $f_n$ are all bounded between $0$ and $1$.
Then, from the dominated convergence theorem we know that $\displaystyle{\int_1^e} f_n(x)dx \underset{n \to \infty}{\longrightarrow} \displaystyle{\int_1^e} f(x)dx = 0$.
Other solution: since $\mbox{ln}$ is concave, $\mbox{ln}$ is under its tangent at $x=e$, which equation is $y = \frac{x}{e}$. Thus $$\displaystyle{\int_1^e} f_n(x)dx \le \displaystyle{\int_1^e} \big(\frac{x}{e}\big)^n dx \le \frac{1}{e^n} \Big[ \frac{x^{n+1}}{n+1}\Big]_1^e \le \frac{e}{n+1}$$
so we get the same conclusion.