Let $f\colon \mathbb{R}\to\mathbb{R}$ be defined as $ \begin{equation} f(x)=\left\{ \begin{array}{@{}ll@{}} xe^{-x^{2}}, & \text{if}\ x\geq0 \\ 0, & \text{if}\ x <0. \\ \end{array}\right. \end{equation} $
$\int_{\mathbb{R}} f \;\mathrm{dμ} = \frac{1}{2}\int_{0}^\infty e^{-y} dy = \frac{1}{2}$, but does this integral converge at all? How can I justify the change of variable in this improper integral?
I was given a hint that monotone converging sequence I can use is $f_n = xe^{-x^2} \cdot I_{[0,n]}$. Why is the following correct
$$\ \int_{\mathbb{R}} f \;\mathrm{dμ} = \lim_{n \to \infty } \int_0^\infty f_n dx =\lim_{n \to \infty } \int_0^n xe^{-x^{2}} dx $$
Let $$f_n(x)=xe^{-x^2} \cdot I_{[0,n]}(x)$$ (where $I_A(x)=1$ if $x\in A$ and $0$ otherwise).
Note that $0\leq f_n(x)\leq f_{n+1}(x)$ and that $f_n \to f$ pointwise. Now, use the monotone convergence theorem.