# Thread: Divergence of improper integrals

1. ## Divergence of improper integrals

Let $\displaystyle f\in\mathrm{C}([\sigma,\infty),[0,\infty))$, where $\displaystyle \sigma\in(-\infty,\infty)$, and $\displaystyle \tau\in(0,\infty)$.
Prove that
$\displaystyle \displaystyle\int_{\sigma}^{\infty}\eta f(\eta)\,\mathrm{d}\eta=\infty$ if and only if $\displaystyle \displaystyle\sum_{\ell=0}^{\infty}\displaystyle\i nt_{\sigma+\ell\tau}^{\infty}f(\eta)\,\mathrm{d}\e ta=\infty$.

2. I want to give the proof for those who were interested.
First of all, a very indirect proof of this result can be found in [1, Lemma 2].

References
[1] B.G. Zhang and J.S. Yu, Existence of positive solutions for neutral differential equations,
Science in China Series A 35 (1992), no. 11, 1306--1313. View PDF

I will here give a shorter proof for a more generalized form.

Lemma 1. Let $\displaystyle \{\xi_{k}\}\subset\mathbb{R}$ be an increasing divergent sequence and $\displaystyle f\in\cnt([\xi_{0},\infty),\mathbb{R}_{0}^{+})$.
Then
$\displaystyle \displaystyle\int_{\xi_{0}}^{\infty}\eta f(\eta)\,\mathrm{d}\eta=\infty$ if and only if $\displaystyle \displaystyle\sum_{\ell=0}^{\infty}\bigg(\int_{\xi _{\ell}}^{\infty}f(\eta)\,\mathrm{d}\eta\bigg)\big[\xi_{\ell+1}-\xi_{\ell}\big]=\infty$
provided that
$\displaystyle \displaystyle\limsup_{k\to\infty}\dfrac{\xi_{k+1}-\xi_{k}}{\xi_{k}-\xi_{k-1}}<\infty$.____________________________(*)

Proof. We shall consider two different cases.
1. Let
$\displaystyle \displaystyle\int_{\xi_{0}}^{\infty}f(\eta)\,\math rm{d}\eta=\infty$.
In this case, the claim is true since both the integral and the sum diverge.
2. Let
$\displaystyle \displaystyle\int_{\xi_{0}}^{\infty}f(\eta)\,\math rm{d}\eta<\infty$.
If we define
$\displaystyle g(t):=\displaystyle\int_{t}^{\infty}f(\eta)\,\math rm{d}\eta$ for $\displaystyle t\in[\xi_{0},\infty)$,
then $\displaystyle g$ is decreasing on $\displaystyle [\xi_{0},\infty)$.
We compute that
$\displaystyle \displaystyle\int_{\xi_{0}}^{\infty}(\eta-\xi_{0})f(\eta)\,\mathrm{d}\eta=\displaystyle\int_ {\xi_{0}}^{\infty}\int_{\xi_{0}}^{\eta}f(\eta)\,\m athrm{d}\zeta\,\mathrm{d}\eta=\displaystyle\int_{\ xi_{0}}^{\infty}g(\zeta)\,\mathrm{d}\zeta,$
where we have changed the order of integration in the last step.
Then an equivalent claim reads as follows.
$\displaystyle \displaystyle\int_{\xi_{0}}^{\infty}g(\eta)\,\math rm{d}\eta=\infty$ if and only if $\displaystyle \displaystyle\sum_{\ell=0}^{\infty}g(\xi_{\ell})\D elta\xi_{\ell}=\infty$.
Define $\displaystyle \nu:[\xi_{0},\infty)\to\mathbb{N}$ by $\displaystyle \nu(t):=\max\{\ell\in\mathbb{N}:\ t<\xi_{\ell}\}$ for $\displaystyle t\in[\xi_{0},\infty)$.
Adopting the convention that the empty sum is $\displaystyle 0$, we get
$\displaystyle \displaystyle\sum_{\ell=1}^{\nu(t)-1}g(\xi_{\ell})\big[\xi_{\ell}-\xi_{\ell-1}\big]<\displaystyle\int_{\xi_{0}}^{t}g(\eta)\,\mathrm{d }\eta\leq\displaystyle\sum_{\ell=0}^{\nu(t)-1}g(\xi_{\ell})\big[\xi_{\ell+1}-\xi_{\ell}\big]$ for all $\displaystyle t\in[\xi_{0},\infty)$,
which proves that
$\displaystyle \displaystyle\int_{\xi_{0}}^{\infty}g(\eta)\,\math rm{d}\eta$ and $\displaystyle \displaystyle\sum_{\ell=0}^{\infty}g(\xi_{\ell})\b ig[\xi_{\ell+1}-\xi_{\ell}\big]$
diverge or converge together since by the assumption (*), we learn that
$\displaystyle \displaystyle\sum_{\ell=1}^{\infty}g(\xi_{\ell})\b ig[\xi_{\ell}-\xi_{\ell-1}\big]$ and $\displaystyle \displaystyle\sum_{\ell=0}^{\infty}g(\xi_{\ell})\b ig[\xi_{\ell+1}-\xi_{\ell}\big]$
diverge or converge together.

This completes the proof.________________________________$\displaystyle \rule{0.2cm}{0.2cm}$

Corollary 1. If in addition to the assumptions of Lemma 1, we have
$\displaystyle \displaystyle\limsup_{k\to\infty}\big[\xi_{k+1}-\xi_{k}\big]<\infty$.___________________________(**)
Then
$\displaystyle \displaystyle\int_{\xi_{0}}^{\infty}\eta f(\eta)\,\mathrm{d}\eta=\infty$ if and only if $\displaystyle \displaystyle\sum_{\ell=0}^{\infty}\int_{\xi_{\ell }}^{\infty}f(\eta)\,\mathrm{d}\eta=\infty$.

Remark. In Corollary 1, if we let $\displaystyle \xi_{0}=\sigma$ and $\displaystyle \xi_{k}=\sigma+k\tau$ for $\displaystyle k\in\mathbb{N}$, we get
the result in the first post. Trivially, in this case (*) and (**) hold.