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Math Help - Divergence of improper integrals

  1. #1
    Senior Member bkarpuz's Avatar
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    Divergence of improper integrals

    Let f\in\mathrm{C}([\sigma,\infty),[0,\infty)), where \sigma\in(-\infty,\infty), and \tau\in(0,\infty).
    Prove that
    \displaystyle\int_{\sigma}^{\infty}\eta f(\eta)\,\mathrm{d}\eta=\infty if and only if \displaystyle\sum_{\ell=0}^{\infty}\displaystyle\i  nt_{\sigma+\ell\tau}^{\infty}f(\eta)\,\mathrm{d}\e  ta=\infty.
    Last edited by bkarpuz; March 26th 2011 at 10:48 PM.
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  2. #2
    Senior Member bkarpuz's Avatar
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    Smile

    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 \{\xi_{k}\}\subset\mathbb{R} be an increasing divergent sequence and f\in\cnt([\xi_{0},\infty),\mathbb{R}_{0}^{+}).
    Then
    \displaystyle\int_{\xi_{0}}^{\infty}\eta f(\eta)\,\mathrm{d}\eta=\infty if and only if \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\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\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\int_{\xi_{0}}^{\infty}f(\eta)\,\math  rm{d}\eta<\infty.
      If we define
      g(t):=\displaystyle\int_{t}^{\infty}f(\eta)\,\math  rm{d}\eta for t\in[\xi_{0},\infty),
      then g is decreasing on [\xi_{0},\infty).
      We compute that
      \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\int_{\xi_{0}}^{\infty}g(\eta)\,\math  rm{d}\eta=\infty if and only if \displaystyle\sum_{\ell=0}^{\infty}g(\xi_{\ell})\D  elta\xi_{\ell}=\infty.
      Define \nu:[\xi_{0},\infty)\to\mathbb{N} by \nu(t):=\max\{\ell\in\mathbb{N}:\ t<\xi_{\ell}\} for t\in[\xi_{0},\infty).
      Adopting the convention that the empty sum is 0, we get
      \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 t\in[\xi_{0},\infty),
      which proves that
      \displaystyle\int_{\xi_{0}}^{\infty}g(\eta)\,\math  rm{d}\eta and \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\sum_{\ell=1}^{\infty}g(\xi_{\ell})\b  ig[\xi_{\ell}-\xi_{\ell-1}\big] and \displaystyle\sum_{\ell=0}^{\infty}g(\xi_{\ell})\b  ig[\xi_{\ell+1}-\xi_{\ell}\big]
      diverge or converge together.

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

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

    Remark. In Corollary 1, if we let \xi_{0}=\sigma and \xi_{k}=\sigma+k\tau for k\in\mathbb{N}, we get
    the result in the first post. Trivially, in this case (*) and (**) hold.
    Last edited by bkarpuz; April 2nd 2011 at 09:56 AM.
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