Evaluate $\lim_{n\to \infty}n\int_2^e{(\ln x)^n}dx$

Question

$$\lim_{n\to \infty}n\int_2^e{(\ln x)^n}dx$$

Any idea on how to approach this problem with elementary methods? The answer should be $e$.

I proved with the mean value theorem for definite integrals that $$\lim_{n\to \infty}\int_2^e{(\ln x)^n}dx=0$$

And I tried $\ln x=t$ but to no use.

Answer 1

By integrating by parts ($u'(x) = \frac{(\ln x)^{n}}{x}$, $v(x) = x$) we have $$ \int (\ln x)^n dx = \frac{(\ln x)^{n+1} x}{n+1} - \frac{1}{n+1} \int (\ln x)^{n+1} dx$$ $$ n\int_2^e (\ln x)^n dx = \frac{n\big(e- 2(\ln 2)^{n+1})\big)}{n+1} - \frac{n}{n+1} \int_2^e (\ln x)^{n+1} dx$$ $$ \lim_{n\rightarrow\infty} n\int_2^e (\ln x)^n dx = e - \lim_{n\rightarrow\infty} \int_2^e (\ln x)^{n+1} dx$$ and you have already proven that the second term is $0$.

Answer 2

Your idea was good. Using $x=e^t$, we end with $$I_n=\int \log^n(x) \,dx=\int e^t\, t^n\,dt=(-1)^{-n} \,\Gamma (n+1,-t)$$ where appears the incomplete gamma function.

So, the definite integral is $$J_n=\int_2^e \log^n(x) \,dx==(-1)^{-n} \, \big(\Gamma (n+1,-1)-\Gamma (n+1,-\log (2))\big)$$ Using the asymptotics, since $\log(2) < 1$, you should have $$J_n \sim \frac e n -\frac {2e}{n^2}$$ and then your result.