Definition of $int_a^b f(x) ,mathrm{d}x$ for $f$ continuous on $[ a,b )$ and unbounded at the right-hand...












1














How would I formulate a definition of the integral $int_a^b f(x) ,mathrm{d}x$ for a function continuous on $[a,b)$ and unbounded at the right-hand endpoint?



Could anyone provide an example?



This is the definition I have for a closed and bounded interval:




Let $f$ be a continuous function on an interval $[a,b]$ Then
$$
intlimits_a^b f(x) ,mathrm{d}x = limlimits_{ntoinfty} dfrac{b-a}{n}sumlimits_{k=0}^{n-1} f Bigg( a+dfrac{k}{n}(b-a) Bigg)
$$




I am thinking I need to use the left-hand endpoint to do this because the right-hand endpoint is unbounded. Thus the definition will have
$$
sumlimits_{k=0}^{n-1}f(ldots)
$$



I am not really sure where to start though.






real-analysis definite-integrals







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    1














    How would I formulate a definition of the integral $int_a^b f(x) ,mathrm{d}x$ for a function continuous on $[a,b)$ and unbounded at the right-hand endpoint?



    Could anyone provide an example?



    This is the definition I have for a closed and bounded interval:




    Let $f$ be a continuous function on an interval $[a,b]$ Then
    $$
    intlimits_a^b f(x) ,mathrm{d}x = limlimits_{ntoinfty} dfrac{b-a}{n}sumlimits_{k=0}^{n-1} f Bigg( a+dfrac{k}{n}(b-a) Bigg)
    $$




    I am thinking I need to use the left-hand endpoint to do this because the right-hand endpoint is unbounded. Thus the definition will have
    $$
    sumlimits_{k=0}^{n-1}f(ldots)
    $$



    I am not really sure where to start though.






    real-analysis definite-integrals







    share|cite|improve this question



























      1












      1








      1







      How would I formulate a definition of the integral $int_a^b f(x) ,mathrm{d}x$ for a function continuous on $[a,b)$ and unbounded at the right-hand endpoint?



      Could anyone provide an example?



      This is the definition I have for a closed and bounded interval:




      Let $f$ be a continuous function on an interval $[a,b]$ Then
      $$
      intlimits_a^b f(x) ,mathrm{d}x = limlimits_{ntoinfty} dfrac{b-a}{n}sumlimits_{k=0}^{n-1} f Bigg( a+dfrac{k}{n}(b-a) Bigg)
      $$




      I am thinking I need to use the left-hand endpoint to do this because the right-hand endpoint is unbounded. Thus the definition will have
      $$
      sumlimits_{k=0}^{n-1}f(ldots)
      $$



      I am not really sure where to start though.






      real-analysis definite-integrals







      share|cite|improve this question















      How would I formulate a definition of the integral $int_a^b f(x) ,mathrm{d}x$ for a function continuous on $[a,b)$ and unbounded at the right-hand endpoint?



      Could anyone provide an example?



      This is the definition I have for a closed and bounded interval:




      Let $f$ be a continuous function on an interval $[a,b]$ Then
      $$
      intlimits_a^b f(x) ,mathrm{d}x = limlimits_{ntoinfty} dfrac{b-a}{n}sumlimits_{k=0}^{n-1} f Bigg( a+dfrac{k}{n}(b-a) Bigg)
      $$




      I am thinking I need to use the left-hand endpoint to do this because the right-hand endpoint is unbounded. Thus the definition will have
      $$
      sumlimits_{k=0}^{n-1}f(ldots)
      $$



      I am not really sure where to start though.






      real-analysis definite-integrals




      real-analysis definite-integrals






      share|cite|improve this question















      share|cite|improve this question













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      edited Dec 5 '18 at 12:41









      Björn Friedrich

      2,58961831




      2,58961831










      asked Dec 5 '18 at 12:12









      kaisakaisa

      1019




      1019






















          2 Answers
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          The integral of Riemann doesn't exist for unbounded functions. This is the reason why mathematicians created the improper integral of Riemann.



          The improper integral of Riemann is defined as a limit of proper integrals, in your case



          $$int_a^b f(x), dx:=lim_{yto b^-}int_a^y f(x), dx$$




          To show why the proper integral of Riemann doesn't exists for unbounded functions WLOG suppose that we have a function $f:[a,b]toBbb R$ such that it is continuous to the left of some point $cin[a,b]$ but $lim_{xto c^-}f(x)=infty$.



          Then we can choose a sequence of partitions $(P_k)$ of $[a,b]$ with decreasing mesh $Delta_{P_k}=delta_k$ for some sequence $(delta_k)downarrow 0$, such that in each $P_k$ there is an interval of the kind $(c-delta_k,c]$. Then, because $f$ is continuous and unbounded to the left of $c$, for each $k$ there is some $x_kin(c-delta_k,c]$ such that $f(x_k)ge k/delta_k$.



          Then if, for the interval $(c-delta_k,c]$, we choose alternatively the tag $c$, when $k$ is odd, and the tag $x_k$ when $k$ is even, we find that the sequence of Riemann sums is not convergent, it will have two cluster points, hence the integral of Riemann is not defined for this function in $[a,b]$.




          Note that the (proper) integral of Riemann is defined only in compact intervals, so an integral over a set of the kind $[a,b)$ is outside of the original definition of Riemann. To make sense we need to integrate on $[a,b]$ or $[a,c]$ for some $cin[a,b)$.



          If we try to apply the original definition of Riemann in an interval of the kind $[a,b)$ where $lim_{xto b^-}f(x)=infty$ and $f$ is continuous in $[a,b)$ then we will get an integral that diverges to infinity.






          share|cite|improve this answer































            3














            Answer.



            Definition Improper Riemann integral
            $$
            int_a^b f(x),dx=lim_{varepsilonsearrow 0}int_a^{b-varepsilon} f(x),dx
            $$






            share|cite|improve this answer





















            • This is the thing to use. Riemann sums may or may not converge to the right value.
              – GEdgar
              Dec 5 '18 at 15:19











            Your Answer





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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            4














            The integral of Riemann doesn't exist for unbounded functions. This is the reason why mathematicians created the improper integral of Riemann.



            The improper integral of Riemann is defined as a limit of proper integrals, in your case



            $$int_a^b f(x), dx:=lim_{yto b^-}int_a^y f(x), dx$$




            To show why the proper integral of Riemann doesn't exists for unbounded functions WLOG suppose that we have a function $f:[a,b]toBbb R$ such that it is continuous to the left of some point $cin[a,b]$ but $lim_{xto c^-}f(x)=infty$.



            Then we can choose a sequence of partitions $(P_k)$ of $[a,b]$ with decreasing mesh $Delta_{P_k}=delta_k$ for some sequence $(delta_k)downarrow 0$, such that in each $P_k$ there is an interval of the kind $(c-delta_k,c]$. Then, because $f$ is continuous and unbounded to the left of $c$, for each $k$ there is some $x_kin(c-delta_k,c]$ such that $f(x_k)ge k/delta_k$.



            Then if, for the interval $(c-delta_k,c]$, we choose alternatively the tag $c$, when $k$ is odd, and the tag $x_k$ when $k$ is even, we find that the sequence of Riemann sums is not convergent, it will have two cluster points, hence the integral of Riemann is not defined for this function in $[a,b]$.




            Note that the (proper) integral of Riemann is defined only in compact intervals, so an integral over a set of the kind $[a,b)$ is outside of the original definition of Riemann. To make sense we need to integrate on $[a,b]$ or $[a,c]$ for some $cin[a,b)$.



            If we try to apply the original definition of Riemann in an interval of the kind $[a,b)$ where $lim_{xto b^-}f(x)=infty$ and $f$ is continuous in $[a,b)$ then we will get an integral that diverges to infinity.






            share|cite|improve this answer




























              4














              The integral of Riemann doesn't exist for unbounded functions. This is the reason why mathematicians created the improper integral of Riemann.



              The improper integral of Riemann is defined as a limit of proper integrals, in your case



              $$int_a^b f(x), dx:=lim_{yto b^-}int_a^y f(x), dx$$




              To show why the proper integral of Riemann doesn't exists for unbounded functions WLOG suppose that we have a function $f:[a,b]toBbb R$ such that it is continuous to the left of some point $cin[a,b]$ but $lim_{xto c^-}f(x)=infty$.



              Then we can choose a sequence of partitions $(P_k)$ of $[a,b]$ with decreasing mesh $Delta_{P_k}=delta_k$ for some sequence $(delta_k)downarrow 0$, such that in each $P_k$ there is an interval of the kind $(c-delta_k,c]$. Then, because $f$ is continuous and unbounded to the left of $c$, for each $k$ there is some $x_kin(c-delta_k,c]$ such that $f(x_k)ge k/delta_k$.



              Then if, for the interval $(c-delta_k,c]$, we choose alternatively the tag $c$, when $k$ is odd, and the tag $x_k$ when $k$ is even, we find that the sequence of Riemann sums is not convergent, it will have two cluster points, hence the integral of Riemann is not defined for this function in $[a,b]$.




              Note that the (proper) integral of Riemann is defined only in compact intervals, so an integral over a set of the kind $[a,b)$ is outside of the original definition of Riemann. To make sense we need to integrate on $[a,b]$ or $[a,c]$ for some $cin[a,b)$.



              If we try to apply the original definition of Riemann in an interval of the kind $[a,b)$ where $lim_{xto b^-}f(x)=infty$ and $f$ is continuous in $[a,b)$ then we will get an integral that diverges to infinity.






              share|cite|improve this answer


























                4












                4








                4






                The integral of Riemann doesn't exist for unbounded functions. This is the reason why mathematicians created the improper integral of Riemann.



                The improper integral of Riemann is defined as a limit of proper integrals, in your case



                $$int_a^b f(x), dx:=lim_{yto b^-}int_a^y f(x), dx$$




                To show why the proper integral of Riemann doesn't exists for unbounded functions WLOG suppose that we have a function $f:[a,b]toBbb R$ such that it is continuous to the left of some point $cin[a,b]$ but $lim_{xto c^-}f(x)=infty$.



                Then we can choose a sequence of partitions $(P_k)$ of $[a,b]$ with decreasing mesh $Delta_{P_k}=delta_k$ for some sequence $(delta_k)downarrow 0$, such that in each $P_k$ there is an interval of the kind $(c-delta_k,c]$. Then, because $f$ is continuous and unbounded to the left of $c$, for each $k$ there is some $x_kin(c-delta_k,c]$ such that $f(x_k)ge k/delta_k$.



                Then if, for the interval $(c-delta_k,c]$, we choose alternatively the tag $c$, when $k$ is odd, and the tag $x_k$ when $k$ is even, we find that the sequence of Riemann sums is not convergent, it will have two cluster points, hence the integral of Riemann is not defined for this function in $[a,b]$.




                Note that the (proper) integral of Riemann is defined only in compact intervals, so an integral over a set of the kind $[a,b)$ is outside of the original definition of Riemann. To make sense we need to integrate on $[a,b]$ or $[a,c]$ for some $cin[a,b)$.



                If we try to apply the original definition of Riemann in an interval of the kind $[a,b)$ where $lim_{xto b^-}f(x)=infty$ and $f$ is continuous in $[a,b)$ then we will get an integral that diverges to infinity.






                share|cite|improve this answer














                The integral of Riemann doesn't exist for unbounded functions. This is the reason why mathematicians created the improper integral of Riemann.



                The improper integral of Riemann is defined as a limit of proper integrals, in your case



                $$int_a^b f(x), dx:=lim_{yto b^-}int_a^y f(x), dx$$




                To show why the proper integral of Riemann doesn't exists for unbounded functions WLOG suppose that we have a function $f:[a,b]toBbb R$ such that it is continuous to the left of some point $cin[a,b]$ but $lim_{xto c^-}f(x)=infty$.



                Then we can choose a sequence of partitions $(P_k)$ of $[a,b]$ with decreasing mesh $Delta_{P_k}=delta_k$ for some sequence $(delta_k)downarrow 0$, such that in each $P_k$ there is an interval of the kind $(c-delta_k,c]$. Then, because $f$ is continuous and unbounded to the left of $c$, for each $k$ there is some $x_kin(c-delta_k,c]$ such that $f(x_k)ge k/delta_k$.



                Then if, for the interval $(c-delta_k,c]$, we choose alternatively the tag $c$, when $k$ is odd, and the tag $x_k$ when $k$ is even, we find that the sequence of Riemann sums is not convergent, it will have two cluster points, hence the integral of Riemann is not defined for this function in $[a,b]$.




                Note that the (proper) integral of Riemann is defined only in compact intervals, so an integral over a set of the kind $[a,b)$ is outside of the original definition of Riemann. To make sense we need to integrate on $[a,b]$ or $[a,c]$ for some $cin[a,b)$.



                If we try to apply the original definition of Riemann in an interval of the kind $[a,b)$ where $lim_{xto b^-}f(x)=infty$ and $f$ is continuous in $[a,b)$ then we will get an integral that diverges to infinity.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 5 '18 at 17:49

























                answered Dec 5 '18 at 14:00









                MasacrosoMasacroso

                13k41746




                13k41746























                    3














                    Answer.



                    Definition Improper Riemann integral
                    $$
                    int_a^b f(x),dx=lim_{varepsilonsearrow 0}int_a^{b-varepsilon} f(x),dx
                    $$






                    share|cite|improve this answer





















                    • This is the thing to use. Riemann sums may or may not converge to the right value.
                      – GEdgar
                      Dec 5 '18 at 15:19
















                    3














                    Answer.



                    Definition Improper Riemann integral
                    $$
                    int_a^b f(x),dx=lim_{varepsilonsearrow 0}int_a^{b-varepsilon} f(x),dx
                    $$






                    share|cite|improve this answer





















                    • This is the thing to use. Riemann sums may or may not converge to the right value.
                      – GEdgar
                      Dec 5 '18 at 15:19














                    3












                    3








                    3






                    Answer.



                    Definition Improper Riemann integral
                    $$
                    int_a^b f(x),dx=lim_{varepsilonsearrow 0}int_a^{b-varepsilon} f(x),dx
                    $$






                    share|cite|improve this answer












                    Answer.



                    Definition Improper Riemann integral
                    $$
                    int_a^b f(x),dx=lim_{varepsilonsearrow 0}int_a^{b-varepsilon} f(x),dx
                    $$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 5 '18 at 12:36









                    Yiorgos S. SmyrlisYiorgos S. Smyrlis

                    62.8k1383163




                    62.8k1383163












                    • This is the thing to use. Riemann sums may or may not converge to the right value.
                      – GEdgar
                      Dec 5 '18 at 15:19


















                    • This is the thing to use. Riemann sums may or may not converge to the right value.
                      – GEdgar
                      Dec 5 '18 at 15:19
















                    This is the thing to use. Riemann sums may or may not converge to the right value.
                    – GEdgar
                    Dec 5 '18 at 15:19




                    This is the thing to use. Riemann sums may or may not converge to the right value.
                    – GEdgar
                    Dec 5 '18 at 15:19


















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