Let $f:[a,b]\longrightarrow \mathbb R$ a function $\mathcal C^1$. Since $|f(t)e^{ist}|=|f(t)|\in L^1$, using dominated convergence theorem, we should have $$\lim_{s\to \infty }\int_a^b f(t)e^{ist}dt= \int_a^b\lim_{s\to \infty }f(t)e^{its}dt.$$ But why here it doesn't hold ?
Why $\lim_{s\to \infty }\int_a^b f(t)e^{ist}dt\neq \int_a^b\lim_{s\to \infty }f(t)e^{its}dt ?$
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real-analysis
1 Answers
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Because $$\lim_{s\to \infty }f(t)e^{its}$$ doesn't exists. Notice that DCT says :
If $(f_n)\in L^1$ is a sequence s.t. $\color{red}{\displaystyle\lim_{n\to \infty }f_n(x)=f(x)}$ a.e. and $|f_n|\leq g\in L^1$, then $$\lim_{n\to \infty }\int f_n=\int f.$$
Notice that in the previous definition, $f(x)\in \overline{\mathbb R}=\mathbb R\cup\{\pm\infty \}.$ I.e. if $\lim_{n\to \infty }f_n(x)= +\infty $ a.e., then the theorem still work.