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.

note that $\sqrt[n]{x}=e^{\frac{1}{n}ln\ x}\leq e^{ln\ x}=x$, if $x\geq 1$. So, $\mathbb{1_{[0,n]}}(x)\sqrt[n]{x}e^{-x}\leq \mathbb{1}_{[0,1]}(x)(1+xe^{-x})$. Since $\mathbb{1}_{[0,1]}(x)(1+xe^{-x})$ is integrable, by the dominated converge theorem we have:
$$
\mathbb{lim}_{n\rightarrow \infty}\int_{0}^{n}\sqrt[n]{x}e^{-x}=\mathbb{lim}_{n\rightarrow \infty}\int_{\mathbb{R}}\mathbb{1_{[0,n]}}(x)\sqrt[n]{x}e^{-x}=\int_{\mathbb{R}}\mathbb{lim}_{n\rightarrow \infty}\mathbb{1_{[0,n]}}(x)\sqrt[n]{x}e^{-x}=\int_{\mathbb{R}}\mathbb{1_{[0,\infty]}}(x)e^{-x}
$$
