$f:\Bbb R \rightarrow \Bbb R $ is a function. given that for all a<b $\in \Bbb R$ , $f \in \mathbf{R}([a,b])$ (Riemann integrable). Also the integral $\int_{-\infty}^ \infty |f| $ $< \infty$ exists and its finite.
first question : show that for all $\epsilon > 0$ , the integral $ \int_{-\infty}^ \infty f(x) e^{-\epsilon |x|}$ exists and its finite.
second question : show that $\lim_{\epsilon \to 0^+} \int_{-\infty}^ \infty f(x) e^{-\epsilon |x|}$ = $\int_{-\infty}^ \infty f(x)$ .
I solves the first question but failed to solve the second one ! , can someone help .
First take $-\varepsilon =-\frac{1}{n}$ and notice that since $e^{-\frac{1}{n}|x|}\leq1$ then $|f(x)e^{-\frac{1}{n}|x|}|\leq|f(x)|$, and by hypothesis is integrable, so we can use Lebesgue's dominated convergence theorem, and if we call $f_n(x)= f(x)e^{-\frac{1}{n}|x|}$ then
$$\lim_{n\to\infty}\int_{-\infty}^{\infty} f_n(x) = \lim_{n\to\infty}\int_{-\infty}^{\infty} f(x)e^{-\frac{1}{n}|x|} = \int_{-\infty}^{\infty} f(x) $$
Because $f_n(x)\to f(x)$ trivially.