Limit Calculation of $I_{n}=\int_{0}^{n}\sqrt[n]{x}\cdot e^{-x}dx$ with dominated converge theorem I considered the series $I_{n}=\int_{0}^{n}\sqrt[n]{x}\cdot e^{-x}dx$ und while calculating $\lim_{n\rightarrow \infty}I_n$ I wasn't sure how to use the dominated converge theorem which allows me to exchange limit and integration. I verified the conditions to use it and I know how to calculate, but the problem is the correct notation because $n$ is also a part of the bounds of integration.

Hint. One may write
$$
\int_{0}^{n}\sqrt[n]{x}\cdot e^{-x}dx=\int_{0}^{\infty}1_{[0,n]}(x)\cdot\sqrt[n]{x}\cdot e^{-x}dx
$$ with $1_{[0,n]}(x)=1$ if $x \in [0,n]$ and $1_{[0,n]}(x)=0$ if $x \notin [0,n]$.
