Integral of 1/x - base of logarithm












0












$begingroup$


I see a proof in https://arxiv.org/abs/1805.11965 (equation 3.36) that uses the following.



$log x = int_0^{infty} ds left(frac{1}{1+s} - frac{1}{s+x}right)$.



This seems to hinge on $int frac{1}{x} = log_2 x$ (the context is information theory), as opposed to $log_e(x)$. Why is this true?






integration logarithms







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$endgroup$












  • $begingroup$
    It isn't? It is true up to a multiplied constant, though, since $log_2(x) = ln(x)/ln(2)$
    $endgroup$
    – Juan Sebastian Lozano
    Dec 11 '18 at 15:41










  • $begingroup$
    What is the "this" in "this seems to hinge"? The equation $log x = int_0^{infty} ds left(frac{1}{1+s} - frac{1}{s+x}right)$? If so, why do you say it seems to hinge on $int frac 1x$[sic] $= log_2x$? If something else in the article, what?
    $endgroup$
    – fleablood
    Dec 11 '18 at 16:01






  • 1




    $begingroup$
    The notation in the pape is a little confusing because in the classical part entropy is measured in bits and $log$ represents the base 2 logarithm, however part 3 deals with quantum entropy, and the definition by Von Newmann uses natural log - in fact the unit of entropy when using natural logarithm has a name, nat, nit, or nepit - see e.g. en.wikipedia.org/wiki/Nat_(unit)
    $endgroup$
    – mlerma54
    Dec 11 '18 at 16:08












  • $begingroup$
    Ah @mlerma54, I think that's the error in my assumption then. I assumed quantum entropy also stayed in $log_2$. If you put your comment as an answer, I can accept it. Thank you
    $endgroup$
    – user1936752
    Dec 11 '18 at 17:23
















0












$begingroup$


I see a proof in https://arxiv.org/abs/1805.11965 (equation 3.36) that uses the following.



$log x = int_0^{infty} ds left(frac{1}{1+s} - frac{1}{s+x}right)$.



This seems to hinge on $int frac{1}{x} = log_2 x$ (the context is information theory), as opposed to $log_e(x)$. Why is this true?






integration logarithms







share|cite|improve this question











$endgroup$












  • $begingroup$
    It isn't? It is true up to a multiplied constant, though, since $log_2(x) = ln(x)/ln(2)$
    $endgroup$
    – Juan Sebastian Lozano
    Dec 11 '18 at 15:41










  • $begingroup$
    What is the "this" in "this seems to hinge"? The equation $log x = int_0^{infty} ds left(frac{1}{1+s} - frac{1}{s+x}right)$? If so, why do you say it seems to hinge on $int frac 1x$[sic] $= log_2x$? If something else in the article, what?
    $endgroup$
    – fleablood
    Dec 11 '18 at 16:01






  • 1




    $begingroup$
    The notation in the pape is a little confusing because in the classical part entropy is measured in bits and $log$ represents the base 2 logarithm, however part 3 deals with quantum entropy, and the definition by Von Newmann uses natural log - in fact the unit of entropy when using natural logarithm has a name, nat, nit, or nepit - see e.g. en.wikipedia.org/wiki/Nat_(unit)
    $endgroup$
    – mlerma54
    Dec 11 '18 at 16:08












  • $begingroup$
    Ah @mlerma54, I think that's the error in my assumption then. I assumed quantum entropy also stayed in $log_2$. If you put your comment as an answer, I can accept it. Thank you
    $endgroup$
    – user1936752
    Dec 11 '18 at 17:23














0












0








0





$begingroup$


I see a proof in https://arxiv.org/abs/1805.11965 (equation 3.36) that uses the following.



$log x = int_0^{infty} ds left(frac{1}{1+s} - frac{1}{s+x}right)$.



This seems to hinge on $int frac{1}{x} = log_2 x$ (the context is information theory), as opposed to $log_e(x)$. Why is this true?






integration logarithms







share|cite|improve this question











$endgroup$




I see a proof in https://arxiv.org/abs/1805.11965 (equation 3.36) that uses the following.



$log x = int_0^{infty} ds left(frac{1}{1+s} - frac{1}{s+x}right)$.



This seems to hinge on $int frac{1}{x} = log_2 x$ (the context is information theory), as opposed to $log_e(x)$. Why is this true?






integration logarithms




integration logarithms






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 11 '18 at 15:45







user1936752

















asked Dec 11 '18 at 15:35









user1936752user1936752

5531513




5531513












  • $begingroup$
    It isn't? It is true up to a multiplied constant, though, since $log_2(x) = ln(x)/ln(2)$
    $endgroup$
    – Juan Sebastian Lozano
    Dec 11 '18 at 15:41










  • $begingroup$
    What is the "this" in "this seems to hinge"? The equation $log x = int_0^{infty} ds left(frac{1}{1+s} - frac{1}{s+x}right)$? If so, why do you say it seems to hinge on $int frac 1x$[sic] $= log_2x$? If something else in the article, what?
    $endgroup$
    – fleablood
    Dec 11 '18 at 16:01






  • 1




    $begingroup$
    The notation in the pape is a little confusing because in the classical part entropy is measured in bits and $log$ represents the base 2 logarithm, however part 3 deals with quantum entropy, and the definition by Von Newmann uses natural log - in fact the unit of entropy when using natural logarithm has a name, nat, nit, or nepit - see e.g. en.wikipedia.org/wiki/Nat_(unit)
    $endgroup$
    – mlerma54
    Dec 11 '18 at 16:08












  • $begingroup$
    Ah @mlerma54, I think that's the error in my assumption then. I assumed quantum entropy also stayed in $log_2$. If you put your comment as an answer, I can accept it. Thank you
    $endgroup$
    – user1936752
    Dec 11 '18 at 17:23


















  • $begingroup$
    It isn't? It is true up to a multiplied constant, though, since $log_2(x) = ln(x)/ln(2)$
    $endgroup$
    – Juan Sebastian Lozano
    Dec 11 '18 at 15:41










  • $begingroup$
    What is the "this" in "this seems to hinge"? The equation $log x = int_0^{infty} ds left(frac{1}{1+s} - frac{1}{s+x}right)$? If so, why do you say it seems to hinge on $int frac 1x$[sic] $= log_2x$? If something else in the article, what?
    $endgroup$
    – fleablood
    Dec 11 '18 at 16:01






  • 1




    $begingroup$
    The notation in the pape is a little confusing because in the classical part entropy is measured in bits and $log$ represents the base 2 logarithm, however part 3 deals with quantum entropy, and the definition by Von Newmann uses natural log - in fact the unit of entropy when using natural logarithm has a name, nat, nit, or nepit - see e.g. en.wikipedia.org/wiki/Nat_(unit)
    $endgroup$
    – mlerma54
    Dec 11 '18 at 16:08












  • $begingroup$
    Ah @mlerma54, I think that's the error in my assumption then. I assumed quantum entropy also stayed in $log_2$. If you put your comment as an answer, I can accept it. Thank you
    $endgroup$
    – user1936752
    Dec 11 '18 at 17:23
















$begingroup$
It isn't? It is true up to a multiplied constant, though, since $log_2(x) = ln(x)/ln(2)$
$endgroup$
– Juan Sebastian Lozano
Dec 11 '18 at 15:41




$begingroup$
It isn't? It is true up to a multiplied constant, though, since $log_2(x) = ln(x)/ln(2)$
$endgroup$
– Juan Sebastian Lozano
Dec 11 '18 at 15:41












$begingroup$
What is the "this" in "this seems to hinge"? The equation $log x = int_0^{infty} ds left(frac{1}{1+s} - frac{1}{s+x}right)$? If so, why do you say it seems to hinge on $int frac 1x$[sic] $= log_2x$? If something else in the article, what?
$endgroup$
– fleablood
Dec 11 '18 at 16:01




$begingroup$
What is the "this" in "this seems to hinge"? The equation $log x = int_0^{infty} ds left(frac{1}{1+s} - frac{1}{s+x}right)$? If so, why do you say it seems to hinge on $int frac 1x$[sic] $= log_2x$? If something else in the article, what?
$endgroup$
– fleablood
Dec 11 '18 at 16:01




1




1




$begingroup$
The notation in the pape is a little confusing because in the classical part entropy is measured in bits and $log$ represents the base 2 logarithm, however part 3 deals with quantum entropy, and the definition by Von Newmann uses natural log - in fact the unit of entropy when using natural logarithm has a name, nat, nit, or nepit - see e.g. en.wikipedia.org/wiki/Nat_(unit)
$endgroup$
– mlerma54
Dec 11 '18 at 16:08






$begingroup$
The notation in the pape is a little confusing because in the classical part entropy is measured in bits and $log$ represents the base 2 logarithm, however part 3 deals with quantum entropy, and the definition by Von Newmann uses natural log - in fact the unit of entropy when using natural logarithm has a name, nat, nit, or nepit - see e.g. en.wikipedia.org/wiki/Nat_(unit)
$endgroup$
– mlerma54
Dec 11 '18 at 16:08














$begingroup$
Ah @mlerma54, I think that's the error in my assumption then. I assumed quantum entropy also stayed in $log_2$. If you put your comment as an answer, I can accept it. Thank you
$endgroup$
– user1936752
Dec 11 '18 at 17:23




$begingroup$
Ah @mlerma54, I think that's the error in my assumption then. I assumed quantum entropy also stayed in $log_2$. If you put your comment as an answer, I can accept it. Thank you
$endgroup$
– user1936752
Dec 11 '18 at 17:23










2 Answers
2






active

oldest

votes


















1












$begingroup$

The notation in the paper is a little confusing because in the classical part entropy is measured in bits and $log$ represents the base 2 logarithm, however part 3 deals with quantum entropy, and the definition by Von Newmann uses natural log - in fact the unit of entropy when using natural logarithm has a name, nat, nit, or nepit - see e.g. en.wikipedia.org/wiki/Nat_(unit)






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Note that for $a > 0$,
    $$
    int_0^M frac{1}{a+s}ds = int_{a}^{M+a} frac{1}{s} ds = ln (M+a) - ln a.
    $$
    Therefore,
    $$
    int_0^M left(frac{1}{1+s} - frac{1}{x+s}right) ds =
    ln (M+1) - ln 1 - (ln (M+x) - ln x).
    $$
    Therefore, by letting $M to infty$, we have
    $$
    int_0^infty left(frac{1}{1+s} - frac{1}{x+s}right) ds = ln x.
    $$






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

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      2 Answers
      2






      active

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      active

      oldest

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      active

      oldest

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      1












      $begingroup$

      The notation in the paper is a little confusing because in the classical part entropy is measured in bits and $log$ represents the base 2 logarithm, however part 3 deals with quantum entropy, and the definition by Von Newmann uses natural log - in fact the unit of entropy when using natural logarithm has a name, nat, nit, or nepit - see e.g. en.wikipedia.org/wiki/Nat_(unit)






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        The notation in the paper is a little confusing because in the classical part entropy is measured in bits and $log$ represents the base 2 logarithm, however part 3 deals with quantum entropy, and the definition by Von Newmann uses natural log - in fact the unit of entropy when using natural logarithm has a name, nat, nit, or nepit - see e.g. en.wikipedia.org/wiki/Nat_(unit)






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          The notation in the paper is a little confusing because in the classical part entropy is measured in bits and $log$ represents the base 2 logarithm, however part 3 deals with quantum entropy, and the definition by Von Newmann uses natural log - in fact the unit of entropy when using natural logarithm has a name, nat, nit, or nepit - see e.g. en.wikipedia.org/wiki/Nat_(unit)






          share|cite|improve this answer









          $endgroup$



          The notation in the paper is a little confusing because in the classical part entropy is measured in bits and $log$ represents the base 2 logarithm, however part 3 deals with quantum entropy, and the definition by Von Newmann uses natural log - in fact the unit of entropy when using natural logarithm has a name, nat, nit, or nepit - see e.g. en.wikipedia.org/wiki/Nat_(unit)







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 11 '18 at 19:45









          mlerma54mlerma54

          1,177148




          1,177148























              0












              $begingroup$

              Note that for $a > 0$,
              $$
              int_0^M frac{1}{a+s}ds = int_{a}^{M+a} frac{1}{s} ds = ln (M+a) - ln a.
              $$
              Therefore,
              $$
              int_0^M left(frac{1}{1+s} - frac{1}{x+s}right) ds =
              ln (M+1) - ln 1 - (ln (M+x) - ln x).
              $$
              Therefore, by letting $M to infty$, we have
              $$
              int_0^infty left(frac{1}{1+s} - frac{1}{x+s}right) ds = ln x.
              $$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Note that for $a > 0$,
                $$
                int_0^M frac{1}{a+s}ds = int_{a}^{M+a} frac{1}{s} ds = ln (M+a) - ln a.
                $$
                Therefore,
                $$
                int_0^M left(frac{1}{1+s} - frac{1}{x+s}right) ds =
                ln (M+1) - ln 1 - (ln (M+x) - ln x).
                $$
                Therefore, by letting $M to infty$, we have
                $$
                int_0^infty left(frac{1}{1+s} - frac{1}{x+s}right) ds = ln x.
                $$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Note that for $a > 0$,
                  $$
                  int_0^M frac{1}{a+s}ds = int_{a}^{M+a} frac{1}{s} ds = ln (M+a) - ln a.
                  $$
                  Therefore,
                  $$
                  int_0^M left(frac{1}{1+s} - frac{1}{x+s}right) ds =
                  ln (M+1) - ln 1 - (ln (M+x) - ln x).
                  $$
                  Therefore, by letting $M to infty$, we have
                  $$
                  int_0^infty left(frac{1}{1+s} - frac{1}{x+s}right) ds = ln x.
                  $$






                  share|cite|improve this answer









                  $endgroup$



                  Note that for $a > 0$,
                  $$
                  int_0^M frac{1}{a+s}ds = int_{a}^{M+a} frac{1}{s} ds = ln (M+a) - ln a.
                  $$
                  Therefore,
                  $$
                  int_0^M left(frac{1}{1+s} - frac{1}{x+s}right) ds =
                  ln (M+1) - ln 1 - (ln (M+x) - ln x).
                  $$
                  Therefore, by letting $M to infty$, we have
                  $$
                  int_0^infty left(frac{1}{1+s} - frac{1}{x+s}right) ds = ln x.
                  $$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 11 '18 at 15:51









                  induction601induction601

                  1,276314




                  1,276314






























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