Interesting Integration with respect to $[x]$ where $[cdot]$ is greatest integer function












3












$begingroup$


$$int_{0}^{3} {(x^2+1)}d[x]$$ is equal to ___________ .



Attempt



$[x]$ is constant for every interval. So for that intervals $d[x]=0$ so intergral given is zero. Am I right? Though the answer given is $dfrac{17}{2}$.






integration definite-integrals self-learning







share|cite|improve this question











$endgroup$

















    3












    $begingroup$


    $$int_{0}^{3} {(x^2+1)}d[x]$$ is equal to ___________ .



    Attempt



    $[x]$ is constant for every interval. So for that intervals $d[x]=0$ so intergral given is zero. Am I right? Though the answer given is $dfrac{17}{2}$.






    integration definite-integrals self-learning







    share|cite|improve this question











    $endgroup$















      3












      3








      3


      1



      $begingroup$


      $$int_{0}^{3} {(x^2+1)}d[x]$$ is equal to ___________ .



      Attempt



      $[x]$ is constant for every interval. So for that intervals $d[x]=0$ so intergral given is zero. Am I right? Though the answer given is $dfrac{17}{2}$.






      integration definite-integrals self-learning







      share|cite|improve this question











      $endgroup$




      $$int_{0}^{3} {(x^2+1)}d[x]$$ is equal to ___________ .



      Attempt



      $[x]$ is constant for every interval. So for that intervals $d[x]=0$ so intergral given is zero. Am I right? Though the answer given is $dfrac{17}{2}$.






      integration definite-integrals self-learning




      integration definite-integrals self-learning






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 2 '18 at 7:41









      Tianlalu

      3,08621038




      3,08621038










      asked Nov 28 '18 at 17:36









      jayant98jayant98

      513116




      513116






















          2 Answers
          2






          active

          oldest

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          3












          $begingroup$

          You must be careful about how you define your integral. I am guessing that you are defining
          $$ int_a^b f(x) dg(x) = lim sum_i f(c_i) (g(x_{i+1}) - g(x_i)),$$
          where $c_i in [x_i, x_{i+1}]$, and the limit is really a limit over partitions of the interval $[a,b]$. This is typically called the Riemann-Stieltjes integral.



          Then in your case, $d lfloor x rfloor$ is $0$ when its defined, but at integer values one must be a bit delicate. For instance, looking just at
          $$ int_{1/2}^{3/2} (x^2 + 1) dlfloor x rfloor = lim sum (x_i^2 + 1)Big( lfloor x_{i+1} rfloor - lfloor x_i rfloorBig),$$
          and all summands are zero except for the one summand where $x_i < 1$ and $x_{i+1} geq 1$. For that one term, $lfloor x_{i+1} rfloor - lfloor x_i rfloor = 1$. And as the partitions of $[1/2, 3/2]$ become finer, the endpoints $x_i$ and $x_{i+1}$ (by which I mean the two partition points surrounding $1$ in the corresponding partition --- there is a minor abuse of notation here) both approach $1$. Thus
          $$ int_{1/2}^{3/2} (x^2 + 1) dlfloor x rfloor = lim (x_i^2 + 1) (1) = 1^2 + 1 = 2.$$



          In fact, what this really is is exactly the value of $x^2 + 1$ at $1$, and more generally
          $$ int_a^b f(x) d lfloor x rfloor = sum_{a < n leq b} f(n)$$
          for a continuous function $f$.




          Having described this, I hope it is now not so hard to compute the entire integral.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            So for whole we can say that for x=2 the value will be there and so for the x=3. And following this I got the answer as 17. Is it now correct?
            $endgroup$
            – jayant98
            Nov 28 '18 at 18:17












          • $begingroup$
            @jayant98 Yes, it should be $17$.
            $endgroup$
            – davidlowryduda♦
            Nov 28 '18 at 18:23



















          3












          $begingroup$

          In your reasoning you forget that at integers the value of $d[x] $ diverges. To write the integral in a tractable way you should use that
          $$
          int_a^b f(x) dy = int_a^b f(x) frac{dy}{dx} dx
          $$
          and apply it to your integral, yielding
          $$
          int_0^3 (x^2+1) frac{d[x] }{dx} dx.
          $$
          The value of $frac{d[x] }{dx}$ is a sum of Dirac deltas. However, there is a delta at each of the endpoints of the integration domain, and the value of an integral who's integrand contains a delta at an endpoint is not defined, so you cannot assign a value to this integral.






          share|cite|improve this answer









          $endgroup$













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            2 Answers
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            active

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            2 Answers
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            3












            $begingroup$

            You must be careful about how you define your integral. I am guessing that you are defining
            $$ int_a^b f(x) dg(x) = lim sum_i f(c_i) (g(x_{i+1}) - g(x_i)),$$
            where $c_i in [x_i, x_{i+1}]$, and the limit is really a limit over partitions of the interval $[a,b]$. This is typically called the Riemann-Stieltjes integral.



            Then in your case, $d lfloor x rfloor$ is $0$ when its defined, but at integer values one must be a bit delicate. For instance, looking just at
            $$ int_{1/2}^{3/2} (x^2 + 1) dlfloor x rfloor = lim sum (x_i^2 + 1)Big( lfloor x_{i+1} rfloor - lfloor x_i rfloorBig),$$
            and all summands are zero except for the one summand where $x_i < 1$ and $x_{i+1} geq 1$. For that one term, $lfloor x_{i+1} rfloor - lfloor x_i rfloor = 1$. And as the partitions of $[1/2, 3/2]$ become finer, the endpoints $x_i$ and $x_{i+1}$ (by which I mean the two partition points surrounding $1$ in the corresponding partition --- there is a minor abuse of notation here) both approach $1$. Thus
            $$ int_{1/2}^{3/2} (x^2 + 1) dlfloor x rfloor = lim (x_i^2 + 1) (1) = 1^2 + 1 = 2.$$



            In fact, what this really is is exactly the value of $x^2 + 1$ at $1$, and more generally
            $$ int_a^b f(x) d lfloor x rfloor = sum_{a < n leq b} f(n)$$
            for a continuous function $f$.




            Having described this, I hope it is now not so hard to compute the entire integral.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              So for whole we can say that for x=2 the value will be there and so for the x=3. And following this I got the answer as 17. Is it now correct?
              $endgroup$
              – jayant98
              Nov 28 '18 at 18:17












            • $begingroup$
              @jayant98 Yes, it should be $17$.
              $endgroup$
              – davidlowryduda♦
              Nov 28 '18 at 18:23
















            3












            $begingroup$

            You must be careful about how you define your integral. I am guessing that you are defining
            $$ int_a^b f(x) dg(x) = lim sum_i f(c_i) (g(x_{i+1}) - g(x_i)),$$
            where $c_i in [x_i, x_{i+1}]$, and the limit is really a limit over partitions of the interval $[a,b]$. This is typically called the Riemann-Stieltjes integral.



            Then in your case, $d lfloor x rfloor$ is $0$ when its defined, but at integer values one must be a bit delicate. For instance, looking just at
            $$ int_{1/2}^{3/2} (x^2 + 1) dlfloor x rfloor = lim sum (x_i^2 + 1)Big( lfloor x_{i+1} rfloor - lfloor x_i rfloorBig),$$
            and all summands are zero except for the one summand where $x_i < 1$ and $x_{i+1} geq 1$. For that one term, $lfloor x_{i+1} rfloor - lfloor x_i rfloor = 1$. And as the partitions of $[1/2, 3/2]$ become finer, the endpoints $x_i$ and $x_{i+1}$ (by which I mean the two partition points surrounding $1$ in the corresponding partition --- there is a minor abuse of notation here) both approach $1$. Thus
            $$ int_{1/2}^{3/2} (x^2 + 1) dlfloor x rfloor = lim (x_i^2 + 1) (1) = 1^2 + 1 = 2.$$



            In fact, what this really is is exactly the value of $x^2 + 1$ at $1$, and more generally
            $$ int_a^b f(x) d lfloor x rfloor = sum_{a < n leq b} f(n)$$
            for a continuous function $f$.




            Having described this, I hope it is now not so hard to compute the entire integral.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              So for whole we can say that for x=2 the value will be there and so for the x=3. And following this I got the answer as 17. Is it now correct?
              $endgroup$
              – jayant98
              Nov 28 '18 at 18:17












            • $begingroup$
              @jayant98 Yes, it should be $17$.
              $endgroup$
              – davidlowryduda♦
              Nov 28 '18 at 18:23














            3












            3








            3





            $begingroup$

            You must be careful about how you define your integral. I am guessing that you are defining
            $$ int_a^b f(x) dg(x) = lim sum_i f(c_i) (g(x_{i+1}) - g(x_i)),$$
            where $c_i in [x_i, x_{i+1}]$, and the limit is really a limit over partitions of the interval $[a,b]$. This is typically called the Riemann-Stieltjes integral.



            Then in your case, $d lfloor x rfloor$ is $0$ when its defined, but at integer values one must be a bit delicate. For instance, looking just at
            $$ int_{1/2}^{3/2} (x^2 + 1) dlfloor x rfloor = lim sum (x_i^2 + 1)Big( lfloor x_{i+1} rfloor - lfloor x_i rfloorBig),$$
            and all summands are zero except for the one summand where $x_i < 1$ and $x_{i+1} geq 1$. For that one term, $lfloor x_{i+1} rfloor - lfloor x_i rfloor = 1$. And as the partitions of $[1/2, 3/2]$ become finer, the endpoints $x_i$ and $x_{i+1}$ (by which I mean the two partition points surrounding $1$ in the corresponding partition --- there is a minor abuse of notation here) both approach $1$. Thus
            $$ int_{1/2}^{3/2} (x^2 + 1) dlfloor x rfloor = lim (x_i^2 + 1) (1) = 1^2 + 1 = 2.$$



            In fact, what this really is is exactly the value of $x^2 + 1$ at $1$, and more generally
            $$ int_a^b f(x) d lfloor x rfloor = sum_{a < n leq b} f(n)$$
            for a continuous function $f$.




            Having described this, I hope it is now not so hard to compute the entire integral.






            share|cite|improve this answer











            $endgroup$



            You must be careful about how you define your integral. I am guessing that you are defining
            $$ int_a^b f(x) dg(x) = lim sum_i f(c_i) (g(x_{i+1}) - g(x_i)),$$
            where $c_i in [x_i, x_{i+1}]$, and the limit is really a limit over partitions of the interval $[a,b]$. This is typically called the Riemann-Stieltjes integral.



            Then in your case, $d lfloor x rfloor$ is $0$ when its defined, but at integer values one must be a bit delicate. For instance, looking just at
            $$ int_{1/2}^{3/2} (x^2 + 1) dlfloor x rfloor = lim sum (x_i^2 + 1)Big( lfloor x_{i+1} rfloor - lfloor x_i rfloorBig),$$
            and all summands are zero except for the one summand where $x_i < 1$ and $x_{i+1} geq 1$. For that one term, $lfloor x_{i+1} rfloor - lfloor x_i rfloor = 1$. And as the partitions of $[1/2, 3/2]$ become finer, the endpoints $x_i$ and $x_{i+1}$ (by which I mean the two partition points surrounding $1$ in the corresponding partition --- there is a minor abuse of notation here) both approach $1$. Thus
            $$ int_{1/2}^{3/2} (x^2 + 1) dlfloor x rfloor = lim (x_i^2 + 1) (1) = 1^2 + 1 = 2.$$



            In fact, what this really is is exactly the value of $x^2 + 1$ at $1$, and more generally
            $$ int_a^b f(x) d lfloor x rfloor = sum_{a < n leq b} f(n)$$
            for a continuous function $f$.




            Having described this, I hope it is now not so hard to compute the entire integral.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 28 '18 at 18:21

























            answered Nov 28 '18 at 17:50









            davidlowryduda♦davidlowryduda

            74.5k7118253




            74.5k7118253












            • $begingroup$
              So for whole we can say that for x=2 the value will be there and so for the x=3. And following this I got the answer as 17. Is it now correct?
              $endgroup$
              – jayant98
              Nov 28 '18 at 18:17












            • $begingroup$
              @jayant98 Yes, it should be $17$.
              $endgroup$
              – davidlowryduda♦
              Nov 28 '18 at 18:23


















            • $begingroup$
              So for whole we can say that for x=2 the value will be there and so for the x=3. And following this I got the answer as 17. Is it now correct?
              $endgroup$
              – jayant98
              Nov 28 '18 at 18:17












            • $begingroup$
              @jayant98 Yes, it should be $17$.
              $endgroup$
              – davidlowryduda♦
              Nov 28 '18 at 18:23
















            $begingroup$
            So for whole we can say that for x=2 the value will be there and so for the x=3. And following this I got the answer as 17. Is it now correct?
            $endgroup$
            – jayant98
            Nov 28 '18 at 18:17






            $begingroup$
            So for whole we can say that for x=2 the value will be there and so for the x=3. And following this I got the answer as 17. Is it now correct?
            $endgroup$
            – jayant98
            Nov 28 '18 at 18:17














            $begingroup$
            @jayant98 Yes, it should be $17$.
            $endgroup$
            – davidlowryduda♦
            Nov 28 '18 at 18:23




            $begingroup$
            @jayant98 Yes, it should be $17$.
            $endgroup$
            – davidlowryduda♦
            Nov 28 '18 at 18:23











            3












            $begingroup$

            In your reasoning you forget that at integers the value of $d[x] $ diverges. To write the integral in a tractable way you should use that
            $$
            int_a^b f(x) dy = int_a^b f(x) frac{dy}{dx} dx
            $$
            and apply it to your integral, yielding
            $$
            int_0^3 (x^2+1) frac{d[x] }{dx} dx.
            $$
            The value of $frac{d[x] }{dx}$ is a sum of Dirac deltas. However, there is a delta at each of the endpoints of the integration domain, and the value of an integral who's integrand contains a delta at an endpoint is not defined, so you cannot assign a value to this integral.






            share|cite|improve this answer









            $endgroup$


















              3












              $begingroup$

              In your reasoning you forget that at integers the value of $d[x] $ diverges. To write the integral in a tractable way you should use that
              $$
              int_a^b f(x) dy = int_a^b f(x) frac{dy}{dx} dx
              $$
              and apply it to your integral, yielding
              $$
              int_0^3 (x^2+1) frac{d[x] }{dx} dx.
              $$
              The value of $frac{d[x] }{dx}$ is a sum of Dirac deltas. However, there is a delta at each of the endpoints of the integration domain, and the value of an integral who's integrand contains a delta at an endpoint is not defined, so you cannot assign a value to this integral.






              share|cite|improve this answer









              $endgroup$
















                3












                3








                3





                $begingroup$

                In your reasoning you forget that at integers the value of $d[x] $ diverges. To write the integral in a tractable way you should use that
                $$
                int_a^b f(x) dy = int_a^b f(x) frac{dy}{dx} dx
                $$
                and apply it to your integral, yielding
                $$
                int_0^3 (x^2+1) frac{d[x] }{dx} dx.
                $$
                The value of $frac{d[x] }{dx}$ is a sum of Dirac deltas. However, there is a delta at each of the endpoints of the integration domain, and the value of an integral who's integrand contains a delta at an endpoint is not defined, so you cannot assign a value to this integral.






                share|cite|improve this answer









                $endgroup$



                In your reasoning you forget that at integers the value of $d[x] $ diverges. To write the integral in a tractable way you should use that
                $$
                int_a^b f(x) dy = int_a^b f(x) frac{dy}{dx} dx
                $$
                and apply it to your integral, yielding
                $$
                int_0^3 (x^2+1) frac{d[x] }{dx} dx.
                $$
                The value of $frac{d[x] }{dx}$ is a sum of Dirac deltas. However, there is a delta at each of the endpoints of the integration domain, and the value of an integral who's integrand contains a delta at an endpoint is not defined, so you cannot assign a value to this integral.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 28 '18 at 17:53









                EddyEddy

                884612




                884612






























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