New bounds for $int_{2}^{x-2}frac{t}{log(t)}frac{1}{log(x-t)}mathrm{dt}$












0














I have revised this interesting stackexchange question with solution, $int_{2}^{x-2}frac{t}{log(t)}frac{1}{log(x-t)}mathrm{dt} tag{1}$
in the form of this approximation :



$frac{x}{log(x-2)}int_{2}^{x/2}frac{dt}{log(t)} tag{2}$



now the rational it seems behind using $frac{x}{log(x-2)}$ outside the LogIntegral seems to ensure a bound in one direction but I'd like to get tighter bounds in both directions.



The question is how do I 'tighten' up the bounds on this particular approximation (2)?



What are the strategies for going about this sort of thing?






integration definite-integrals asymptotics approximation







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    0














    I have revised this interesting stackexchange question with solution, $int_{2}^{x-2}frac{t}{log(t)}frac{1}{log(x-t)}mathrm{dt} tag{1}$
    in the form of this approximation :



    $frac{x}{log(x-2)}int_{2}^{x/2}frac{dt}{log(t)} tag{2}$



    now the rational it seems behind using $frac{x}{log(x-2)}$ outside the LogIntegral seems to ensure a bound in one direction but I'd like to get tighter bounds in both directions.



    The question is how do I 'tighten' up the bounds on this particular approximation (2)?



    What are the strategies for going about this sort of thing?






    integration definite-integrals asymptotics approximation







    share|cite|improve this question



























      0












      0








      0







      I have revised this interesting stackexchange question with solution, $int_{2}^{x-2}frac{t}{log(t)}frac{1}{log(x-t)}mathrm{dt} tag{1}$
      in the form of this approximation :



      $frac{x}{log(x-2)}int_{2}^{x/2}frac{dt}{log(t)} tag{2}$



      now the rational it seems behind using $frac{x}{log(x-2)}$ outside the LogIntegral seems to ensure a bound in one direction but I'd like to get tighter bounds in both directions.



      The question is how do I 'tighten' up the bounds on this particular approximation (2)?



      What are the strategies for going about this sort of thing?






      integration definite-integrals asymptotics approximation







      share|cite|improve this question















      I have revised this interesting stackexchange question with solution, $int_{2}^{x-2}frac{t}{log(t)}frac{1}{log(x-t)}mathrm{dt} tag{1}$
      in the form of this approximation :



      $frac{x}{log(x-2)}int_{2}^{x/2}frac{dt}{log(t)} tag{2}$



      now the rational it seems behind using $frac{x}{log(x-2)}$ outside the LogIntegral seems to ensure a bound in one direction but I'd like to get tighter bounds in both directions.



      The question is how do I 'tighten' up the bounds on this particular approximation (2)?



      What are the strategies for going about this sort of thing?






      integration definite-integrals asymptotics approximation




      integration definite-integrals asymptotics approximation






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 4 at 14:27







      onepound

















      asked Jan 4 at 14:08









      onepoundonepound

      331216




      331216






















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          To help get things along since posting the question a couple of days ago, I've considered a loose bound as a first attempt. Maybe somebody can improve or tighten the bounds by some other method.



          Using $frac{x}{log(x-2)}$ (as suggested by @Jack D'Aurizio ) outside the Log Integral one bound can be established and looking at the plot of $frac{x}{log(x)}$ and $frac{x}{log(x-t)}$, $frac{x}{log(x/2)}$ ensures the other bound such that:



          $frac{x}{log(x-2)}int_{2}^{x/2}frac{dt}{log(t)}<int_{2}^{x-2}frac{t}{log(t)}frac{1}{log(x-t)}mathrm{dt}<frac{x}{log(x/2)}int_{2}^{x-2}frac{dt}{log(t)}$



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            1 Answer
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            1 Answer
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            active

            oldest

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            0














            To help get things along since posting the question a couple of days ago, I've considered a loose bound as a first attempt. Maybe somebody can improve or tighten the bounds by some other method.



            Using $frac{x}{log(x-2)}$ (as suggested by @Jack D'Aurizio ) outside the Log Integral one bound can be established and looking at the plot of $frac{x}{log(x)}$ and $frac{x}{log(x-t)}$, $frac{x}{log(x/2)}$ ensures the other bound such that:



            $frac{x}{log(x-2)}int_{2}^{x/2}frac{dt}{log(t)}<int_{2}^{x-2}frac{t}{log(t)}frac{1}{log(x-t)}mathrm{dt}<frac{x}{log(x/2)}int_{2}^{x-2}frac{dt}{log(t)}$



            w






            share|cite|improve this answer


























              0














              To help get things along since posting the question a couple of days ago, I've considered a loose bound as a first attempt. Maybe somebody can improve or tighten the bounds by some other method.



              Using $frac{x}{log(x-2)}$ (as suggested by @Jack D'Aurizio ) outside the Log Integral one bound can be established and looking at the plot of $frac{x}{log(x)}$ and $frac{x}{log(x-t)}$, $frac{x}{log(x/2)}$ ensures the other bound such that:



              $frac{x}{log(x-2)}int_{2}^{x/2}frac{dt}{log(t)}<int_{2}^{x-2}frac{t}{log(t)}frac{1}{log(x-t)}mathrm{dt}<frac{x}{log(x/2)}int_{2}^{x-2}frac{dt}{log(t)}$



              w






              share|cite|improve this answer
























                0












                0








                0






                To help get things along since posting the question a couple of days ago, I've considered a loose bound as a first attempt. Maybe somebody can improve or tighten the bounds by some other method.



                Using $frac{x}{log(x-2)}$ (as suggested by @Jack D'Aurizio ) outside the Log Integral one bound can be established and looking at the plot of $frac{x}{log(x)}$ and $frac{x}{log(x-t)}$, $frac{x}{log(x/2)}$ ensures the other bound such that:



                $frac{x}{log(x-2)}int_{2}^{x/2}frac{dt}{log(t)}<int_{2}^{x-2}frac{t}{log(t)}frac{1}{log(x-t)}mathrm{dt}<frac{x}{log(x/2)}int_{2}^{x-2}frac{dt}{log(t)}$



                w






                share|cite|improve this answer












                To help get things along since posting the question a couple of days ago, I've considered a loose bound as a first attempt. Maybe somebody can improve or tighten the bounds by some other method.



                Using $frac{x}{log(x-2)}$ (as suggested by @Jack D'Aurizio ) outside the Log Integral one bound can be established and looking at the plot of $frac{x}{log(x)}$ and $frac{x}{log(x-t)}$, $frac{x}{log(x/2)}$ ensures the other bound such that:



                $frac{x}{log(x-2)}int_{2}^{x/2}frac{dt}{log(t)}<int_{2}^{x-2}frac{t}{log(t)}frac{1}{log(x-t)}mathrm{dt}<frac{x}{log(x/2)}int_{2}^{x-2}frac{dt}{log(t)}$



                w







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 4 at 14:09









                onepoundonepound

                331216




                331216






























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