%%visits: 2 ## intuition
absolute convergence implies convergence, but the other way round is not true. why? Because oscillatory diminishment is a big factor in convergence.
Case to infinity: Let $f:[a,\infty) \to \mathbb{R}, f \in \fancy{R}[a,b], \forall b > a,$ one has $\int_{a}^{\infty}f(x)dx `:=` \lim_{b \to \infty} \int_{a}^{b}f(x)dx$
Case to -infinity: Let $f:(-\infty,b] \to \mathbb{R}, f \in \fancy{R}[a,b], \forall a> b,$ one has $\int_{-\infty}^{b}f(x)dx `:=` \lim_{a \to -\infty} \int_{a}^{b}f(x)dx$
Now to consider the case of from −∞ to ∞ we have to combine the two definitions, but doing so carefully.
Case from -infinity to infinity: Let $f:(-\infty,\infty] \to \mathbb{R}, f \in \fancy{R}[a,b], \forall a< b,$ one has $\int_{-\infty}^{\infty}f(x)dx `:=` lim_{b \to \infty} \int_{0}^{b} f(x)dx + \lim_{a \to -\infty} \int_{a}^{0}f(x)dx$
∫a∞f(x)dx converges absolutely if ∫a∞|f(x)|dx converges
If two improper_integrals converge, then their sums also converge.
Let f and g be continuous on [a, ∞) st (i) the integral ∫axf(t)dt is bounded and (ii) is continuously differentiable as f(x) → ∞ as x → ∞ and is monotone then the integral. ∫a∞ converges
absolute converge imples
$\sum _{k=1}^{\infty} f(k)$ converges ⇔ ∫a∞f(x)dx converges and the value can be estimated by $\int_{1}^{\infty}f(x) dx \le \sum _{k=1}^{\infty} f(k) \le \int_{0}^{\infty}f(x) dx$
tags :math: