Convergence of an improper integral,depending on parameters












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$begingroup$


I have to show how the convergence of $$int_2^infty frac {1}{x^alpha (ln x)^beta} mathrm{ d}x $$ depends on parameters $$alpha,betagt0$$
And considering the case $alphagt1$,my textbook says $xgeq2$ implies $ln xgeq ln 2$ and hence $$frac {1}{x^alpha (ln x)^beta}leqfrac {1}{x^alpha (ln2)^beta}$$ $$forall xgeq2$$
And it is easy to see that $frac {1}{x^alpha (ln2)^beta}$ converges,so does $frac {1}{x^alpha (ln x)^beta}$,by Comparison Theorem.
But $$frac {1}{x^alpha (ln x)^beta}leqfrac {1}{x^alpha (ln2)^beta}$$
Implies $$x^alpha(ln x)^betageq x^alpha(ln2)^beta$$ $forall xgeq2$ and $alphagt1$.
But I struggle to see why $$frac {1}{x^alpha (ln x)^beta}leqfrac {1}{x^alpha (ln2)^beta}$$ and therefore $$x^alpha(ln x)^betageq x^alpha(ln2)^beta$$ doesn't hold for $0ltalphaleq1$






improper-integrals







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    0












    $begingroup$


    I have to show how the convergence of $$int_2^infty frac {1}{x^alpha (ln x)^beta} mathrm{ d}x $$ depends on parameters $$alpha,betagt0$$
    And considering the case $alphagt1$,my textbook says $xgeq2$ implies $ln xgeq ln 2$ and hence $$frac {1}{x^alpha (ln x)^beta}leqfrac {1}{x^alpha (ln2)^beta}$$ $$forall xgeq2$$
    And it is easy to see that $frac {1}{x^alpha (ln2)^beta}$ converges,so does $frac {1}{x^alpha (ln x)^beta}$,by Comparison Theorem.
    But $$frac {1}{x^alpha (ln x)^beta}leqfrac {1}{x^alpha (ln2)^beta}$$
    Implies $$x^alpha(ln x)^betageq x^alpha(ln2)^beta$$ $forall xgeq2$ and $alphagt1$.
    But I struggle to see why $$frac {1}{x^alpha (ln x)^beta}leqfrac {1}{x^alpha (ln2)^beta}$$ and therefore $$x^alpha(ln x)^betageq x^alpha(ln2)^beta$$ doesn't hold for $0ltalphaleq1$






    improper-integrals







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      I have to show how the convergence of $$int_2^infty frac {1}{x^alpha (ln x)^beta} mathrm{ d}x $$ depends on parameters $$alpha,betagt0$$
      And considering the case $alphagt1$,my textbook says $xgeq2$ implies $ln xgeq ln 2$ and hence $$frac {1}{x^alpha (ln x)^beta}leqfrac {1}{x^alpha (ln2)^beta}$$ $$forall xgeq2$$
      And it is easy to see that $frac {1}{x^alpha (ln2)^beta}$ converges,so does $frac {1}{x^alpha (ln x)^beta}$,by Comparison Theorem.
      But $$frac {1}{x^alpha (ln x)^beta}leqfrac {1}{x^alpha (ln2)^beta}$$
      Implies $$x^alpha(ln x)^betageq x^alpha(ln2)^beta$$ $forall xgeq2$ and $alphagt1$.
      But I struggle to see why $$frac {1}{x^alpha (ln x)^beta}leqfrac {1}{x^alpha (ln2)^beta}$$ and therefore $$x^alpha(ln x)^betageq x^alpha(ln2)^beta$$ doesn't hold for $0ltalphaleq1$






      improper-integrals







      share|cite|improve this question









      $endgroup$




      I have to show how the convergence of $$int_2^infty frac {1}{x^alpha (ln x)^beta} mathrm{ d}x $$ depends on parameters $$alpha,betagt0$$
      And considering the case $alphagt1$,my textbook says $xgeq2$ implies $ln xgeq ln 2$ and hence $$frac {1}{x^alpha (ln x)^beta}leqfrac {1}{x^alpha (ln2)^beta}$$ $$forall xgeq2$$
      And it is easy to see that $frac {1}{x^alpha (ln2)^beta}$ converges,so does $frac {1}{x^alpha (ln x)^beta}$,by Comparison Theorem.
      But $$frac {1}{x^alpha (ln x)^beta}leqfrac {1}{x^alpha (ln2)^beta}$$
      Implies $$x^alpha(ln x)^betageq x^alpha(ln2)^beta$$ $forall xgeq2$ and $alphagt1$.
      But I struggle to see why $$frac {1}{x^alpha (ln x)^beta}leqfrac {1}{x^alpha (ln2)^beta}$$ and therefore $$x^alpha(ln x)^betageq x^alpha(ln2)^beta$$ doesn't hold for $0ltalphaleq1$






      improper-integrals




      improper-integrals






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      share|cite|improve this question











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      asked Jan 8 at 7:39









      Turan NasibliTuran Nasibli

      846




      846






















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          0












          $begingroup$

          Your inequality does hold, but you cannot use the comparison theorem anymore, since $int_2^{infty}{frac{1}{x^{alpha}ln(2)^{beta}}} = infty$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you.But,then if we want to show if integral diverges or converges for $alphalt1$ we need to use Comparison theorem again,but I am having trouble with finding a greater function that converges,or smaller one that diverges.Or is there another way to decide?
            $endgroup$
            – Turan Nasibli
            Jan 8 at 8:36












          • $begingroup$
            Hint: when $x$ is large enough, and $0 < alpha < 1$, $x^{-alpha}(ln{x})^{beta} geq x^{-(1+alpha)/2}$. For $alpha=1$, you can compute exactly an antiderivative.
            $endgroup$
            – Mindlack
            Jan 8 at 9:38










          • $begingroup$
            So, if I got it right, we must have $$frac{1}{x^frac{alpha +1}{2}}leqfrac{1}{x^alpha (ln x)^beta}$$ and the former is divergent,so is our integral.I would be greatful if u shared with me how exactly u figured out this function.Thanks in advance
            $endgroup$
            – Turan Nasibli
            Jan 8 at 10:47












          • $begingroup$
            First, note that your inequality only holds for every sufficiently large $x$. Second, what we wanted is a function $f > 0$ such that $int_2^{infty}{f}=infty$ and $f(x) leq x^{-alpha}(ln{x})^{-beta}$ for all large enough $x$. It is natural to try for $f(x)=x^{-gamma}$. The conditions then translate to $gamma leq 1$, $gamma geq alpha$ (and $gamma > alpha$ if $beta > 0$). So with $gamma=frac{1+alpha}{2}$ we are good in all cases.
            $endgroup$
            – Mindlack
            Jan 8 at 11:00










          • $begingroup$
            I still don't understand how the fact that our inequality holds only for sufficiently large $x$ is acceptable.I mean,in the previous case, where $alphagt1$ we picked up a function that held true for $forall xgeq2$.
            $endgroup$
            – Turan Nasibli
            Jan 8 at 11:06














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          0












          $begingroup$

          Your inequality does hold, but you cannot use the comparison theorem anymore, since $int_2^{infty}{frac{1}{x^{alpha}ln(2)^{beta}}} = infty$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you.But,then if we want to show if integral diverges or converges for $alphalt1$ we need to use Comparison theorem again,but I am having trouble with finding a greater function that converges,or smaller one that diverges.Or is there another way to decide?
            $endgroup$
            – Turan Nasibli
            Jan 8 at 8:36












          • $begingroup$
            Hint: when $x$ is large enough, and $0 < alpha < 1$, $x^{-alpha}(ln{x})^{beta} geq x^{-(1+alpha)/2}$. For $alpha=1$, you can compute exactly an antiderivative.
            $endgroup$
            – Mindlack
            Jan 8 at 9:38










          • $begingroup$
            So, if I got it right, we must have $$frac{1}{x^frac{alpha +1}{2}}leqfrac{1}{x^alpha (ln x)^beta}$$ and the former is divergent,so is our integral.I would be greatful if u shared with me how exactly u figured out this function.Thanks in advance
            $endgroup$
            – Turan Nasibli
            Jan 8 at 10:47












          • $begingroup$
            First, note that your inequality only holds for every sufficiently large $x$. Second, what we wanted is a function $f > 0$ such that $int_2^{infty}{f}=infty$ and $f(x) leq x^{-alpha}(ln{x})^{-beta}$ for all large enough $x$. It is natural to try for $f(x)=x^{-gamma}$. The conditions then translate to $gamma leq 1$, $gamma geq alpha$ (and $gamma > alpha$ if $beta > 0$). So with $gamma=frac{1+alpha}{2}$ we are good in all cases.
            $endgroup$
            – Mindlack
            Jan 8 at 11:00










          • $begingroup$
            I still don't understand how the fact that our inequality holds only for sufficiently large $x$ is acceptable.I mean,in the previous case, where $alphagt1$ we picked up a function that held true for $forall xgeq2$.
            $endgroup$
            – Turan Nasibli
            Jan 8 at 11:06


















          0












          $begingroup$

          Your inequality does hold, but you cannot use the comparison theorem anymore, since $int_2^{infty}{frac{1}{x^{alpha}ln(2)^{beta}}} = infty$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you.But,then if we want to show if integral diverges or converges for $alphalt1$ we need to use Comparison theorem again,but I am having trouble with finding a greater function that converges,or smaller one that diverges.Or is there another way to decide?
            $endgroup$
            – Turan Nasibli
            Jan 8 at 8:36












          • $begingroup$
            Hint: when $x$ is large enough, and $0 < alpha < 1$, $x^{-alpha}(ln{x})^{beta} geq x^{-(1+alpha)/2}$. For $alpha=1$, you can compute exactly an antiderivative.
            $endgroup$
            – Mindlack
            Jan 8 at 9:38










          • $begingroup$
            So, if I got it right, we must have $$frac{1}{x^frac{alpha +1}{2}}leqfrac{1}{x^alpha (ln x)^beta}$$ and the former is divergent,so is our integral.I would be greatful if u shared with me how exactly u figured out this function.Thanks in advance
            $endgroup$
            – Turan Nasibli
            Jan 8 at 10:47












          • $begingroup$
            First, note that your inequality only holds for every sufficiently large $x$. Second, what we wanted is a function $f > 0$ such that $int_2^{infty}{f}=infty$ and $f(x) leq x^{-alpha}(ln{x})^{-beta}$ for all large enough $x$. It is natural to try for $f(x)=x^{-gamma}$. The conditions then translate to $gamma leq 1$, $gamma geq alpha$ (and $gamma > alpha$ if $beta > 0$). So with $gamma=frac{1+alpha}{2}$ we are good in all cases.
            $endgroup$
            – Mindlack
            Jan 8 at 11:00










          • $begingroup$
            I still don't understand how the fact that our inequality holds only for sufficiently large $x$ is acceptable.I mean,in the previous case, where $alphagt1$ we picked up a function that held true for $forall xgeq2$.
            $endgroup$
            – Turan Nasibli
            Jan 8 at 11:06
















          0












          0








          0





          $begingroup$

          Your inequality does hold, but you cannot use the comparison theorem anymore, since $int_2^{infty}{frac{1}{x^{alpha}ln(2)^{beta}}} = infty$.






          share|cite|improve this answer











          $endgroup$



          Your inequality does hold, but you cannot use the comparison theorem anymore, since $int_2^{infty}{frac{1}{x^{alpha}ln(2)^{beta}}} = infty$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 8 at 9:36

























          answered Jan 8 at 7:41









          MindlackMindlack

          4,910211




          4,910211












          • $begingroup$
            Thank you.But,then if we want to show if integral diverges or converges for $alphalt1$ we need to use Comparison theorem again,but I am having trouble with finding a greater function that converges,or smaller one that diverges.Or is there another way to decide?
            $endgroup$
            – Turan Nasibli
            Jan 8 at 8:36












          • $begingroup$
            Hint: when $x$ is large enough, and $0 < alpha < 1$, $x^{-alpha}(ln{x})^{beta} geq x^{-(1+alpha)/2}$. For $alpha=1$, you can compute exactly an antiderivative.
            $endgroup$
            – Mindlack
            Jan 8 at 9:38










          • $begingroup$
            So, if I got it right, we must have $$frac{1}{x^frac{alpha +1}{2}}leqfrac{1}{x^alpha (ln x)^beta}$$ and the former is divergent,so is our integral.I would be greatful if u shared with me how exactly u figured out this function.Thanks in advance
            $endgroup$
            – Turan Nasibli
            Jan 8 at 10:47












          • $begingroup$
            First, note that your inequality only holds for every sufficiently large $x$. Second, what we wanted is a function $f > 0$ such that $int_2^{infty}{f}=infty$ and $f(x) leq x^{-alpha}(ln{x})^{-beta}$ for all large enough $x$. It is natural to try for $f(x)=x^{-gamma}$. The conditions then translate to $gamma leq 1$, $gamma geq alpha$ (and $gamma > alpha$ if $beta > 0$). So with $gamma=frac{1+alpha}{2}$ we are good in all cases.
            $endgroup$
            – Mindlack
            Jan 8 at 11:00










          • $begingroup$
            I still don't understand how the fact that our inequality holds only for sufficiently large $x$ is acceptable.I mean,in the previous case, where $alphagt1$ we picked up a function that held true for $forall xgeq2$.
            $endgroup$
            – Turan Nasibli
            Jan 8 at 11:06




















          • $begingroup$
            Thank you.But,then if we want to show if integral diverges or converges for $alphalt1$ we need to use Comparison theorem again,but I am having trouble with finding a greater function that converges,or smaller one that diverges.Or is there another way to decide?
            $endgroup$
            – Turan Nasibli
            Jan 8 at 8:36












          • $begingroup$
            Hint: when $x$ is large enough, and $0 < alpha < 1$, $x^{-alpha}(ln{x})^{beta} geq x^{-(1+alpha)/2}$. For $alpha=1$, you can compute exactly an antiderivative.
            $endgroup$
            – Mindlack
            Jan 8 at 9:38










          • $begingroup$
            So, if I got it right, we must have $$frac{1}{x^frac{alpha +1}{2}}leqfrac{1}{x^alpha (ln x)^beta}$$ and the former is divergent,so is our integral.I would be greatful if u shared with me how exactly u figured out this function.Thanks in advance
            $endgroup$
            – Turan Nasibli
            Jan 8 at 10:47












          • $begingroup$
            First, note that your inequality only holds for every sufficiently large $x$. Second, what we wanted is a function $f > 0$ such that $int_2^{infty}{f}=infty$ and $f(x) leq x^{-alpha}(ln{x})^{-beta}$ for all large enough $x$. It is natural to try for $f(x)=x^{-gamma}$. The conditions then translate to $gamma leq 1$, $gamma geq alpha$ (and $gamma > alpha$ if $beta > 0$). So with $gamma=frac{1+alpha}{2}$ we are good in all cases.
            $endgroup$
            – Mindlack
            Jan 8 at 11:00










          • $begingroup$
            I still don't understand how the fact that our inequality holds only for sufficiently large $x$ is acceptable.I mean,in the previous case, where $alphagt1$ we picked up a function that held true for $forall xgeq2$.
            $endgroup$
            – Turan Nasibli
            Jan 8 at 11:06


















          $begingroup$
          Thank you.But,then if we want to show if integral diverges or converges for $alphalt1$ we need to use Comparison theorem again,but I am having trouble with finding a greater function that converges,or smaller one that diverges.Or is there another way to decide?
          $endgroup$
          – Turan Nasibli
          Jan 8 at 8:36






          $begingroup$
          Thank you.But,then if we want to show if integral diverges or converges for $alphalt1$ we need to use Comparison theorem again,but I am having trouble with finding a greater function that converges,or smaller one that diverges.Or is there another way to decide?
          $endgroup$
          – Turan Nasibli
          Jan 8 at 8:36














          $begingroup$
          Hint: when $x$ is large enough, and $0 < alpha < 1$, $x^{-alpha}(ln{x})^{beta} geq x^{-(1+alpha)/2}$. For $alpha=1$, you can compute exactly an antiderivative.
          $endgroup$
          – Mindlack
          Jan 8 at 9:38




          $begingroup$
          Hint: when $x$ is large enough, and $0 < alpha < 1$, $x^{-alpha}(ln{x})^{beta} geq x^{-(1+alpha)/2}$. For $alpha=1$, you can compute exactly an antiderivative.
          $endgroup$
          – Mindlack
          Jan 8 at 9:38












          $begingroup$
          So, if I got it right, we must have $$frac{1}{x^frac{alpha +1}{2}}leqfrac{1}{x^alpha (ln x)^beta}$$ and the former is divergent,so is our integral.I would be greatful if u shared with me how exactly u figured out this function.Thanks in advance
          $endgroup$
          – Turan Nasibli
          Jan 8 at 10:47






          $begingroup$
          So, if I got it right, we must have $$frac{1}{x^frac{alpha +1}{2}}leqfrac{1}{x^alpha (ln x)^beta}$$ and the former is divergent,so is our integral.I would be greatful if u shared with me how exactly u figured out this function.Thanks in advance
          $endgroup$
          – Turan Nasibli
          Jan 8 at 10:47














          $begingroup$
          First, note that your inequality only holds for every sufficiently large $x$. Second, what we wanted is a function $f > 0$ such that $int_2^{infty}{f}=infty$ and $f(x) leq x^{-alpha}(ln{x})^{-beta}$ for all large enough $x$. It is natural to try for $f(x)=x^{-gamma}$. The conditions then translate to $gamma leq 1$, $gamma geq alpha$ (and $gamma > alpha$ if $beta > 0$). So with $gamma=frac{1+alpha}{2}$ we are good in all cases.
          $endgroup$
          – Mindlack
          Jan 8 at 11:00




          $begingroup$
          First, note that your inequality only holds for every sufficiently large $x$. Second, what we wanted is a function $f > 0$ such that $int_2^{infty}{f}=infty$ and $f(x) leq x^{-alpha}(ln{x})^{-beta}$ for all large enough $x$. It is natural to try for $f(x)=x^{-gamma}$. The conditions then translate to $gamma leq 1$, $gamma geq alpha$ (and $gamma > alpha$ if $beta > 0$). So with $gamma=frac{1+alpha}{2}$ we are good in all cases.
          $endgroup$
          – Mindlack
          Jan 8 at 11:00












          $begingroup$
          I still don't understand how the fact that our inequality holds only for sufficiently large $x$ is acceptable.I mean,in the previous case, where $alphagt1$ we picked up a function that held true for $forall xgeq2$.
          $endgroup$
          – Turan Nasibli
          Jan 8 at 11:06






          $begingroup$
          I still don't understand how the fact that our inequality holds only for sufficiently large $x$ is acceptable.I mean,in the previous case, where $alphagt1$ we picked up a function that held true for $forall xgeq2$.
          $endgroup$
          – Turan Nasibli
          Jan 8 at 11:06




















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