Integrating $int_2^{infty}({{log(x+n)}/{log(x)}-1})dx$












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I was thinking about functions $f(x)$ such that $lim_{x to infty}{frac{f(x+n)}{f(x)}=1}$, where $n geq 0$. I was looking for some special properties of them, and decided to first focus on $log(x)$. My first natural question was on the rate of divergence of $f(x+n)$ vs $f(x)$, and I'm now trying to solve the following integral:



$$int_2^{infty}left({frac{log(x+n)}{log(x)}-1}right)dx$$



So far, I've tried the following simplifications: $int_2^{infty}left({frac{log(x+n)-log(x)}{log(x)}}right)dx = int_2^{infty}left({frac{logleft(1+frac{n}{x}right)}{log(x)}}right)dx = int_2^{infty}log_xleft(1+frac{n}{x}right)dx$. I examined each form and did various substitutions, none of which led to any obvious findings. Does anyone know a better way to approach this integral?



Would also appreciate any ideas about the class of functions I'm thinking about more generally. Thanks!






integration limits functions definite-integrals







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    $begingroup$


    I was thinking about functions $f(x)$ such that $lim_{x to infty}{frac{f(x+n)}{f(x)}=1}$, where $n geq 0$. I was looking for some special properties of them, and decided to first focus on $log(x)$. My first natural question was on the rate of divergence of $f(x+n)$ vs $f(x)$, and I'm now trying to solve the following integral:



    $$int_2^{infty}left({frac{log(x+n)}{log(x)}-1}right)dx$$



    So far, I've tried the following simplifications: $int_2^{infty}left({frac{log(x+n)-log(x)}{log(x)}}right)dx = int_2^{infty}left({frac{logleft(1+frac{n}{x}right)}{log(x)}}right)dx = int_2^{infty}log_xleft(1+frac{n}{x}right)dx$. I examined each form and did various substitutions, none of which led to any obvious findings. Does anyone know a better way to approach this integral?



    Would also appreciate any ideas about the class of functions I'm thinking about more generally. Thanks!






    integration limits functions definite-integrals







    share|cite|improve this question











    $endgroup$















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      1








      1


      0



      $begingroup$


      I was thinking about functions $f(x)$ such that $lim_{x to infty}{frac{f(x+n)}{f(x)}=1}$, where $n geq 0$. I was looking for some special properties of them, and decided to first focus on $log(x)$. My first natural question was on the rate of divergence of $f(x+n)$ vs $f(x)$, and I'm now trying to solve the following integral:



      $$int_2^{infty}left({frac{log(x+n)}{log(x)}-1}right)dx$$



      So far, I've tried the following simplifications: $int_2^{infty}left({frac{log(x+n)-log(x)}{log(x)}}right)dx = int_2^{infty}left({frac{logleft(1+frac{n}{x}right)}{log(x)}}right)dx = int_2^{infty}log_xleft(1+frac{n}{x}right)dx$. I examined each form and did various substitutions, none of which led to any obvious findings. Does anyone know a better way to approach this integral?



      Would also appreciate any ideas about the class of functions I'm thinking about more generally. Thanks!






      integration limits functions definite-integrals







      share|cite|improve this question











      $endgroup$




      I was thinking about functions $f(x)$ such that $lim_{x to infty}{frac{f(x+n)}{f(x)}=1}$, where $n geq 0$. I was looking for some special properties of them, and decided to first focus on $log(x)$. My first natural question was on the rate of divergence of $f(x+n)$ vs $f(x)$, and I'm now trying to solve the following integral:



      $$int_2^{infty}left({frac{log(x+n)}{log(x)}-1}right)dx$$



      So far, I've tried the following simplifications: $int_2^{infty}left({frac{log(x+n)-log(x)}{log(x)}}right)dx = int_2^{infty}left({frac{logleft(1+frac{n}{x}right)}{log(x)}}right)dx = int_2^{infty}log_xleft(1+frac{n}{x}right)dx$. I examined each form and did various substitutions, none of which led to any obvious findings. Does anyone know a better way to approach this integral?



      Would also appreciate any ideas about the class of functions I'm thinking about more generally. Thanks!






      integration limits functions definite-integrals




      integration limits functions definite-integrals






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      edited Dec 1 '18 at 16:42









      Henning Makholm

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      asked Jun 8 '17 at 14:08









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          $begingroup$

          The integral does not converge for $n>0$. Note that:



          $$frac{ln(x+n)}{ln(x)}-1>frac n{(x+n)ln(x)}>frac n{(x+n)ln(x+n)}$$



          Which can be proven by multiplying both sides by $ln(x)$ to see that



          $$ln(x+n)-ln(x)=int_x^{x+n}frac1t~mathrm dt>int_x^{x+n}frac1{x+n}~mathrm dt=frac n{x+n}$$



          From here it is very easy to show that it diverges,



          $$begin{align}int_2^inftyfrac{ln(x+n)}{ln(x)}-1~mathrm dx&>int_2^inftyfrac n{(x+n)ln(x+n)}~mathrm dx\&=int_{2+n}^inftyfrac n{xln(x)}~mathrm dx\&=int_{ln(2+n)}^inftyfrac nx~mathrm dx\&=+inftyend{align}$$






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            $begingroup$

            We have
            $$
            frac{ln(x+n)}{ln x}-1=frac{ln x+ln(1+n/x)}{ln x}-1=frac{ln(1+n/x)}{ln x}>0
            $$
            Since
            $$
            lim_{tto 0}frac{ln(1+t)}{t}=1
            $$
            we get
            $$
            lim_{xto+infty}biggl(frac{ln(1+n/x)}{ln x}biggr)biggm/frac{n}{xln x}=1
            $$
            Now, by the comparison theorem (in the form of limit) we find that
            $$
            int_2^{+infty}Bigl(frac{ln(x+n)}{ln x}-1Bigr),dxquadtext{diverges}
            $$
            since
            $$
            int_2^{R}frac{n}{xln x},dx=bigl[nln(ln x)bigr]_2^{R}to+inftyquadtext{as $Rto+infty$},
            $$
            i.e. $int_2^{+infty} n/(xln x),dx$ diverges.






            share|cite|improve this answer









            $endgroup$













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              2 Answers
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              10












              $begingroup$

              The integral does not converge for $n>0$. Note that:



              $$frac{ln(x+n)}{ln(x)}-1>frac n{(x+n)ln(x)}>frac n{(x+n)ln(x+n)}$$



              Which can be proven by multiplying both sides by $ln(x)$ to see that



              $$ln(x+n)-ln(x)=int_x^{x+n}frac1t~mathrm dt>int_x^{x+n}frac1{x+n}~mathrm dt=frac n{x+n}$$



              From here it is very easy to show that it diverges,



              $$begin{align}int_2^inftyfrac{ln(x+n)}{ln(x)}-1~mathrm dx&>int_2^inftyfrac n{(x+n)ln(x+n)}~mathrm dx\&=int_{2+n}^inftyfrac n{xln(x)}~mathrm dx\&=int_{ln(2+n)}^inftyfrac nx~mathrm dx\&=+inftyend{align}$$






              share|cite|improve this answer









              $endgroup$


















                10












                $begingroup$

                The integral does not converge for $n>0$. Note that:



                $$frac{ln(x+n)}{ln(x)}-1>frac n{(x+n)ln(x)}>frac n{(x+n)ln(x+n)}$$



                Which can be proven by multiplying both sides by $ln(x)$ to see that



                $$ln(x+n)-ln(x)=int_x^{x+n}frac1t~mathrm dt>int_x^{x+n}frac1{x+n}~mathrm dt=frac n{x+n}$$



                From here it is very easy to show that it diverges,



                $$begin{align}int_2^inftyfrac{ln(x+n)}{ln(x)}-1~mathrm dx&>int_2^inftyfrac n{(x+n)ln(x+n)}~mathrm dx\&=int_{2+n}^inftyfrac n{xln(x)}~mathrm dx\&=int_{ln(2+n)}^inftyfrac nx~mathrm dx\&=+inftyend{align}$$






                share|cite|improve this answer









                $endgroup$
















                  10












                  10








                  10





                  $begingroup$

                  The integral does not converge for $n>0$. Note that:



                  $$frac{ln(x+n)}{ln(x)}-1>frac n{(x+n)ln(x)}>frac n{(x+n)ln(x+n)}$$



                  Which can be proven by multiplying both sides by $ln(x)$ to see that



                  $$ln(x+n)-ln(x)=int_x^{x+n}frac1t~mathrm dt>int_x^{x+n}frac1{x+n}~mathrm dt=frac n{x+n}$$



                  From here it is very easy to show that it diverges,



                  $$begin{align}int_2^inftyfrac{ln(x+n)}{ln(x)}-1~mathrm dx&>int_2^inftyfrac n{(x+n)ln(x+n)}~mathrm dx\&=int_{2+n}^inftyfrac n{xln(x)}~mathrm dx\&=int_{ln(2+n)}^inftyfrac nx~mathrm dx\&=+inftyend{align}$$






                  share|cite|improve this answer









                  $endgroup$



                  The integral does not converge for $n>0$. Note that:



                  $$frac{ln(x+n)}{ln(x)}-1>frac n{(x+n)ln(x)}>frac n{(x+n)ln(x+n)}$$



                  Which can be proven by multiplying both sides by $ln(x)$ to see that



                  $$ln(x+n)-ln(x)=int_x^{x+n}frac1t~mathrm dt>int_x^{x+n}frac1{x+n}~mathrm dt=frac n{x+n}$$



                  From here it is very easy to show that it diverges,



                  $$begin{align}int_2^inftyfrac{ln(x+n)}{ln(x)}-1~mathrm dx&>int_2^inftyfrac n{(x+n)ln(x+n)}~mathrm dx\&=int_{2+n}^inftyfrac n{xln(x)}~mathrm dx\&=int_{ln(2+n)}^inftyfrac nx~mathrm dx\&=+inftyend{align}$$







                  share|cite|improve this answer












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                  share|cite|improve this answer










                  answered Jun 8 '17 at 14:25









                  Simply Beautiful ArtSimply Beautiful Art

                  50.5k578181




                  50.5k578181























                      0












                      $begingroup$

                      We have
                      $$
                      frac{ln(x+n)}{ln x}-1=frac{ln x+ln(1+n/x)}{ln x}-1=frac{ln(1+n/x)}{ln x}>0
                      $$
                      Since
                      $$
                      lim_{tto 0}frac{ln(1+t)}{t}=1
                      $$
                      we get
                      $$
                      lim_{xto+infty}biggl(frac{ln(1+n/x)}{ln x}biggr)biggm/frac{n}{xln x}=1
                      $$
                      Now, by the comparison theorem (in the form of limit) we find that
                      $$
                      int_2^{+infty}Bigl(frac{ln(x+n)}{ln x}-1Bigr),dxquadtext{diverges}
                      $$
                      since
                      $$
                      int_2^{R}frac{n}{xln x},dx=bigl[nln(ln x)bigr]_2^{R}to+inftyquadtext{as $Rto+infty$},
                      $$
                      i.e. $int_2^{+infty} n/(xln x),dx$ diverges.






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        We have
                        $$
                        frac{ln(x+n)}{ln x}-1=frac{ln x+ln(1+n/x)}{ln x}-1=frac{ln(1+n/x)}{ln x}>0
                        $$
                        Since
                        $$
                        lim_{tto 0}frac{ln(1+t)}{t}=1
                        $$
                        we get
                        $$
                        lim_{xto+infty}biggl(frac{ln(1+n/x)}{ln x}biggr)biggm/frac{n}{xln x}=1
                        $$
                        Now, by the comparison theorem (in the form of limit) we find that
                        $$
                        int_2^{+infty}Bigl(frac{ln(x+n)}{ln x}-1Bigr),dxquadtext{diverges}
                        $$
                        since
                        $$
                        int_2^{R}frac{n}{xln x},dx=bigl[nln(ln x)bigr]_2^{R}to+inftyquadtext{as $Rto+infty$},
                        $$
                        i.e. $int_2^{+infty} n/(xln x),dx$ diverges.






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          We have
                          $$
                          frac{ln(x+n)}{ln x}-1=frac{ln x+ln(1+n/x)}{ln x}-1=frac{ln(1+n/x)}{ln x}>0
                          $$
                          Since
                          $$
                          lim_{tto 0}frac{ln(1+t)}{t}=1
                          $$
                          we get
                          $$
                          lim_{xto+infty}biggl(frac{ln(1+n/x)}{ln x}biggr)biggm/frac{n}{xln x}=1
                          $$
                          Now, by the comparison theorem (in the form of limit) we find that
                          $$
                          int_2^{+infty}Bigl(frac{ln(x+n)}{ln x}-1Bigr),dxquadtext{diverges}
                          $$
                          since
                          $$
                          int_2^{R}frac{n}{xln x},dx=bigl[nln(ln x)bigr]_2^{R}to+inftyquadtext{as $Rto+infty$},
                          $$
                          i.e. $int_2^{+infty} n/(xln x),dx$ diverges.






                          share|cite|improve this answer









                          $endgroup$



                          We have
                          $$
                          frac{ln(x+n)}{ln x}-1=frac{ln x+ln(1+n/x)}{ln x}-1=frac{ln(1+n/x)}{ln x}>0
                          $$
                          Since
                          $$
                          lim_{tto 0}frac{ln(1+t)}{t}=1
                          $$
                          we get
                          $$
                          lim_{xto+infty}biggl(frac{ln(1+n/x)}{ln x}biggr)biggm/frac{n}{xln x}=1
                          $$
                          Now, by the comparison theorem (in the form of limit) we find that
                          $$
                          int_2^{+infty}Bigl(frac{ln(x+n)}{ln x}-1Bigr),dxquadtext{diverges}
                          $$
                          since
                          $$
                          int_2^{R}frac{n}{xln x},dx=bigl[nln(ln x)bigr]_2^{R}to+inftyquadtext{as $Rto+infty$},
                          $$
                          i.e. $int_2^{+infty} n/(xln x),dx$ diverges.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jun 8 '17 at 16:40









                          mickepmickep

                          18.5k12250




                          18.5k12250






























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