This is probably an unremarkable result, but I was curious about exploring the following family of definite integrals to see what asymptotic behaviour it as as $n \rightarrow \infty$
$$I_n = \int_{1}^{e}\left[\ln(x)\right]^n\:dx \mbox{ for } n \in \mathbb{N}$$
The method I took:
Employ integration by parts:
\begin{align} v'(x) &= 1 & u(x) &= \left[\ln(x)\right]^n \\ v(x) &= x & u'(x) &= n\left[\ln(x)\right]^{n - 1}\frac{1}{x} \end{align}
Thus,
\begin{align} I_n &= \int_{1}^{e}\left[\ln(x)\right]^n\:dx = x \cdot \left[\ln(x)\right]^n \Big|_{1}^{e} - \int_{1}^{e} x \cdot n\left[\ln(x)\right]^{n - 1}\frac{1}{x} \\ &= e - n\int_{1}^{e}\left[\ln(x)\right]^{n - 1}\:dx \end{align}
Thus,
\begin{equation} I_n = e - nI_{n - 1} \end{equation}
Where
\begin{equation} I_1 = \int_{1}^{e} \ln(x)\:dx = x\left(\ln(x) - 1\right)\Big|_{1}^{e} = 1 \end{equation}
This yields the solution:
\begin{align} I_n &= n!\left[(-1)^{n - 1} + e\sum_{i = 0}^{n - 2} \frac{(-1)^i}{(n - i)!} \right] = n!\left(-1\right)^n\left[-1 + e\sum_{i = 2}^{n } \frac{(-1)^i}{i!} \right] \\ &= n!\left(-1\right)^n\left[-1 + e\sum_{i = 0}^{n } \frac{(-1)^i}{i!} \right] \end{align}
We see that:
\begin{equation} \lim_{n\rightarrow \infty} I_n = n! (-1)^n \left( -1 + e\cdot e^{-1} \right) = 0 \end{equation}
Or
\begin{equation} \lim_{n\rightarrow \infty}\int_{1}^{e}\left[\ln(x)\right]^n\:dx = 0 \end{equation}
I was wondering if I could have a more technical eye have a look over this to see if the result I've obtained is correct.
Edit - Final result was incorrect and now corrected as per Song's pickup.