Proof That $sum_{k=1}^{n-1}int_{k}^{k+1}left{xright}f´(x)dx=int_{1}^{n}left{xright}f´(x)dx$












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I am Reading the following notes: Ramanujan summation of divergent series by B Candelpergher (https://hal.univ-cotedazur.fr/hal-01150208v2/document). There, the author derives the Euler-MacLaurin summation formula in the beginning of the first chapter.
In one of the steps (on page 18 of the file) he states that
$$sum_{k=1}^{n-1}int_{k}^{k+1}[x]f´(x)dx=int_{1}^{n}xf´(x)dx+int_{1}^{n}left{xright}f´(x)dx$$
$left{xright}$ is the fractional part of $x$.



The first integral on the right side I understand, but the second one involving the fractional part of $x$, I can´t really see how it´s true. In other words, he is saying that



$$sum_{k=1}^{n-1}int_{k}^{k+1}left{xright}f´(x)dx=int_{1}^{n}left{xright}f´(x)dx$$
Can someone kindly show the proof or the intuition behind it?
Thank you in advance.






integration summation euler-maclaurin ramanujan-summation







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    I am Reading the following notes: Ramanujan summation of divergent series by B Candelpergher (https://hal.univ-cotedazur.fr/hal-01150208v2/document). There, the author derives the Euler-MacLaurin summation formula in the beginning of the first chapter.
    In one of the steps (on page 18 of the file) he states that
    $$sum_{k=1}^{n-1}int_{k}^{k+1}[x]f´(x)dx=int_{1}^{n}xf´(x)dx+int_{1}^{n}left{xright}f´(x)dx$$
    $left{xright}$ is the fractional part of $x$.



    The first integral on the right side I understand, but the second one involving the fractional part of $x$, I can´t really see how it´s true. In other words, he is saying that



    $$sum_{k=1}^{n-1}int_{k}^{k+1}left{xright}f´(x)dx=int_{1}^{n}left{xright}f´(x)dx$$
    Can someone kindly show the proof or the intuition behind it?
    Thank you in advance.






    integration summation euler-maclaurin ramanujan-summation







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      I am Reading the following notes: Ramanujan summation of divergent series by B Candelpergher (https://hal.univ-cotedazur.fr/hal-01150208v2/document). There, the author derives the Euler-MacLaurin summation formula in the beginning of the first chapter.
      In one of the steps (on page 18 of the file) he states that
      $$sum_{k=1}^{n-1}int_{k}^{k+1}[x]f´(x)dx=int_{1}^{n}xf´(x)dx+int_{1}^{n}left{xright}f´(x)dx$$
      $left{xright}$ is the fractional part of $x$.



      The first integral on the right side I understand, but the second one involving the fractional part of $x$, I can´t really see how it´s true. In other words, he is saying that



      $$sum_{k=1}^{n-1}int_{k}^{k+1}left{xright}f´(x)dx=int_{1}^{n}left{xright}f´(x)dx$$
      Can someone kindly show the proof or the intuition behind it?
      Thank you in advance.






      integration summation euler-maclaurin ramanujan-summation







      share|cite|improve this question















      I am Reading the following notes: Ramanujan summation of divergent series by B Candelpergher (https://hal.univ-cotedazur.fr/hal-01150208v2/document). There, the author derives the Euler-MacLaurin summation formula in the beginning of the first chapter.
      In one of the steps (on page 18 of the file) he states that
      $$sum_{k=1}^{n-1}int_{k}^{k+1}[x]f´(x)dx=int_{1}^{n}xf´(x)dx+int_{1}^{n}left{xright}f´(x)dx$$
      $left{xright}$ is the fractional part of $x$.



      The first integral on the right side I understand, but the second one involving the fractional part of $x$, I can´t really see how it´s true. In other words, he is saying that



      $$sum_{k=1}^{n-1}int_{k}^{k+1}left{xright}f´(x)dx=int_{1}^{n}left{xright}f´(x)dx$$
      Can someone kindly show the proof or the intuition behind it?
      Thank you in advance.






      integration summation euler-maclaurin ramanujan-summation




      integration summation euler-maclaurin ramanujan-summation






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      edited Jan 4 at 14:31







      Ricardo770

















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      Ricardo770Ricardo770

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          Just write out the sum: begin{align*}
          sum_{k=1}^{n-1} int_k^{k+1} text{whatever}, dx &= int_1^2 text{whatever}, dx + int_2^3 text{whatever}, dx + cdots + int_{n-1}^n text{whatever}, dx \
          &= int_1^n text{whatever}, dx.
          end{align*}






          share|cite|improve this answer

















          • 1




            ...where $text{whatever}={x}f´(x)$ :)
            – Wojowu
            Jan 4 at 14:38










          • @Connor Harris I tried what you say, but I get the following: $int_{n-1}^{n} left{xright}f´(x)dx=int_{n-1}^{n}(x-left[xright])f´(x)dx$ integrating the first integral by parts I get $int_{n-1}^{n}(x-left[xright])f´(x)dx=f(n)-int_{n-1}^{n}f(x)dx$. summing up all the integrals I got $sum_{k=1}^{n-1}int_{k}^{k+1}left{xright}f´(x)dx= sum_{k=1}^{n-1}f(k)-int_{1}^{n} f(x)dx$
            – Ricardo770
            Jan 4 at 15:31












          • Why do you need to do any integrals?
            – Connor Harris
            Jan 4 at 16:22



















          0














          Ok, I got it!



          I took the reverse way to prove it.
          $$int_{1}^{n}left{xright}f´(x)dx= int_{1}^{n}(x-left[xright])f´(x)dx$$
          $$ =int_{1}^{n}x f´(x)dx -int_{1}^{n} left[xright]f´(x)dx$$
          $$ =int_{1}^{n}x f´(x)dx -sum_{k=1}^{n-1}int_{k}^{k+1} left[xright]f´(x)dx$$
          $$=sum_{k=1}^{n-1} int_{k}^{k+1}x f´(x)dx -sum_{k=1}^{n-1}int_{k}^{k+1} left[xright]f´(x)dx$$
          $$=sum_{k=1}^{n-1} int_{k}^{k+1}(x - left[xright])f´(x)dx$$
          $$=sum_{k=1}^{n-1} int_{k}^{k+1}left{xright}f´(x)dx$$






          share|cite|improve this answer





















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            2 Answers
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            Just write out the sum: begin{align*}
            sum_{k=1}^{n-1} int_k^{k+1} text{whatever}, dx &= int_1^2 text{whatever}, dx + int_2^3 text{whatever}, dx + cdots + int_{n-1}^n text{whatever}, dx \
            &= int_1^n text{whatever}, dx.
            end{align*}






            share|cite|improve this answer

















            • 1




              ...where $text{whatever}={x}f´(x)$ :)
              – Wojowu
              Jan 4 at 14:38










            • @Connor Harris I tried what you say, but I get the following: $int_{n-1}^{n} left{xright}f´(x)dx=int_{n-1}^{n}(x-left[xright])f´(x)dx$ integrating the first integral by parts I get $int_{n-1}^{n}(x-left[xright])f´(x)dx=f(n)-int_{n-1}^{n}f(x)dx$. summing up all the integrals I got $sum_{k=1}^{n-1}int_{k}^{k+1}left{xright}f´(x)dx= sum_{k=1}^{n-1}f(k)-int_{1}^{n} f(x)dx$
              – Ricardo770
              Jan 4 at 15:31












            • Why do you need to do any integrals?
              – Connor Harris
              Jan 4 at 16:22
















            1














            Just write out the sum: begin{align*}
            sum_{k=1}^{n-1} int_k^{k+1} text{whatever}, dx &= int_1^2 text{whatever}, dx + int_2^3 text{whatever}, dx + cdots + int_{n-1}^n text{whatever}, dx \
            &= int_1^n text{whatever}, dx.
            end{align*}






            share|cite|improve this answer

















            • 1




              ...where $text{whatever}={x}f´(x)$ :)
              – Wojowu
              Jan 4 at 14:38










            • @Connor Harris I tried what you say, but I get the following: $int_{n-1}^{n} left{xright}f´(x)dx=int_{n-1}^{n}(x-left[xright])f´(x)dx$ integrating the first integral by parts I get $int_{n-1}^{n}(x-left[xright])f´(x)dx=f(n)-int_{n-1}^{n}f(x)dx$. summing up all the integrals I got $sum_{k=1}^{n-1}int_{k}^{k+1}left{xright}f´(x)dx= sum_{k=1}^{n-1}f(k)-int_{1}^{n} f(x)dx$
              – Ricardo770
              Jan 4 at 15:31












            • Why do you need to do any integrals?
              – Connor Harris
              Jan 4 at 16:22














            1












            1








            1






            Just write out the sum: begin{align*}
            sum_{k=1}^{n-1} int_k^{k+1} text{whatever}, dx &= int_1^2 text{whatever}, dx + int_2^3 text{whatever}, dx + cdots + int_{n-1}^n text{whatever}, dx \
            &= int_1^n text{whatever}, dx.
            end{align*}






            share|cite|improve this answer












            Just write out the sum: begin{align*}
            sum_{k=1}^{n-1} int_k^{k+1} text{whatever}, dx &= int_1^2 text{whatever}, dx + int_2^3 text{whatever}, dx + cdots + int_{n-1}^n text{whatever}, dx \
            &= int_1^n text{whatever}, dx.
            end{align*}







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 4 at 14:34









            Connor HarrisConnor Harris

            4,350723




            4,350723








            • 1




              ...where $text{whatever}={x}f´(x)$ :)
              – Wojowu
              Jan 4 at 14:38










            • @Connor Harris I tried what you say, but I get the following: $int_{n-1}^{n} left{xright}f´(x)dx=int_{n-1}^{n}(x-left[xright])f´(x)dx$ integrating the first integral by parts I get $int_{n-1}^{n}(x-left[xright])f´(x)dx=f(n)-int_{n-1}^{n}f(x)dx$. summing up all the integrals I got $sum_{k=1}^{n-1}int_{k}^{k+1}left{xright}f´(x)dx= sum_{k=1}^{n-1}f(k)-int_{1}^{n} f(x)dx$
              – Ricardo770
              Jan 4 at 15:31












            • Why do you need to do any integrals?
              – Connor Harris
              Jan 4 at 16:22














            • 1




              ...where $text{whatever}={x}f´(x)$ :)
              – Wojowu
              Jan 4 at 14:38










            • @Connor Harris I tried what you say, but I get the following: $int_{n-1}^{n} left{xright}f´(x)dx=int_{n-1}^{n}(x-left[xright])f´(x)dx$ integrating the first integral by parts I get $int_{n-1}^{n}(x-left[xright])f´(x)dx=f(n)-int_{n-1}^{n}f(x)dx$. summing up all the integrals I got $sum_{k=1}^{n-1}int_{k}^{k+1}left{xright}f´(x)dx= sum_{k=1}^{n-1}f(k)-int_{1}^{n} f(x)dx$
              – Ricardo770
              Jan 4 at 15:31












            • Why do you need to do any integrals?
              – Connor Harris
              Jan 4 at 16:22








            1




            1




            ...where $text{whatever}={x}f´(x)$ :)
            – Wojowu
            Jan 4 at 14:38




            ...where $text{whatever}={x}f´(x)$ :)
            – Wojowu
            Jan 4 at 14:38












            @Connor Harris I tried what you say, but I get the following: $int_{n-1}^{n} left{xright}f´(x)dx=int_{n-1}^{n}(x-left[xright])f´(x)dx$ integrating the first integral by parts I get $int_{n-1}^{n}(x-left[xright])f´(x)dx=f(n)-int_{n-1}^{n}f(x)dx$. summing up all the integrals I got $sum_{k=1}^{n-1}int_{k}^{k+1}left{xright}f´(x)dx= sum_{k=1}^{n-1}f(k)-int_{1}^{n} f(x)dx$
            – Ricardo770
            Jan 4 at 15:31






            @Connor Harris I tried what you say, but I get the following: $int_{n-1}^{n} left{xright}f´(x)dx=int_{n-1}^{n}(x-left[xright])f´(x)dx$ integrating the first integral by parts I get $int_{n-1}^{n}(x-left[xright])f´(x)dx=f(n)-int_{n-1}^{n}f(x)dx$. summing up all the integrals I got $sum_{k=1}^{n-1}int_{k}^{k+1}left{xright}f´(x)dx= sum_{k=1}^{n-1}f(k)-int_{1}^{n} f(x)dx$
            – Ricardo770
            Jan 4 at 15:31














            Why do you need to do any integrals?
            – Connor Harris
            Jan 4 at 16:22




            Why do you need to do any integrals?
            – Connor Harris
            Jan 4 at 16:22











            0














            Ok, I got it!



            I took the reverse way to prove it.
            $$int_{1}^{n}left{xright}f´(x)dx= int_{1}^{n}(x-left[xright])f´(x)dx$$
            $$ =int_{1}^{n}x f´(x)dx -int_{1}^{n} left[xright]f´(x)dx$$
            $$ =int_{1}^{n}x f´(x)dx -sum_{k=1}^{n-1}int_{k}^{k+1} left[xright]f´(x)dx$$
            $$=sum_{k=1}^{n-1} int_{k}^{k+1}x f´(x)dx -sum_{k=1}^{n-1}int_{k}^{k+1} left[xright]f´(x)dx$$
            $$=sum_{k=1}^{n-1} int_{k}^{k+1}(x - left[xright])f´(x)dx$$
            $$=sum_{k=1}^{n-1} int_{k}^{k+1}left{xright}f´(x)dx$$






            share|cite|improve this answer


























              0














              Ok, I got it!



              I took the reverse way to prove it.
              $$int_{1}^{n}left{xright}f´(x)dx= int_{1}^{n}(x-left[xright])f´(x)dx$$
              $$ =int_{1}^{n}x f´(x)dx -int_{1}^{n} left[xright]f´(x)dx$$
              $$ =int_{1}^{n}x f´(x)dx -sum_{k=1}^{n-1}int_{k}^{k+1} left[xright]f´(x)dx$$
              $$=sum_{k=1}^{n-1} int_{k}^{k+1}x f´(x)dx -sum_{k=1}^{n-1}int_{k}^{k+1} left[xright]f´(x)dx$$
              $$=sum_{k=1}^{n-1} int_{k}^{k+1}(x - left[xright])f´(x)dx$$
              $$=sum_{k=1}^{n-1} int_{k}^{k+1}left{xright}f´(x)dx$$






              share|cite|improve this answer
























                0












                0








                0






                Ok, I got it!



                I took the reverse way to prove it.
                $$int_{1}^{n}left{xright}f´(x)dx= int_{1}^{n}(x-left[xright])f´(x)dx$$
                $$ =int_{1}^{n}x f´(x)dx -int_{1}^{n} left[xright]f´(x)dx$$
                $$ =int_{1}^{n}x f´(x)dx -sum_{k=1}^{n-1}int_{k}^{k+1} left[xright]f´(x)dx$$
                $$=sum_{k=1}^{n-1} int_{k}^{k+1}x f´(x)dx -sum_{k=1}^{n-1}int_{k}^{k+1} left[xright]f´(x)dx$$
                $$=sum_{k=1}^{n-1} int_{k}^{k+1}(x - left[xright])f´(x)dx$$
                $$=sum_{k=1}^{n-1} int_{k}^{k+1}left{xright}f´(x)dx$$






                share|cite|improve this answer












                Ok, I got it!



                I took the reverse way to prove it.
                $$int_{1}^{n}left{xright}f´(x)dx= int_{1}^{n}(x-left[xright])f´(x)dx$$
                $$ =int_{1}^{n}x f´(x)dx -int_{1}^{n} left[xright]f´(x)dx$$
                $$ =int_{1}^{n}x f´(x)dx -sum_{k=1}^{n-1}int_{k}^{k+1} left[xright]f´(x)dx$$
                $$=sum_{k=1}^{n-1} int_{k}^{k+1}x f´(x)dx -sum_{k=1}^{n-1}int_{k}^{k+1} left[xright]f´(x)dx$$
                $$=sum_{k=1}^{n-1} int_{k}^{k+1}(x - left[xright])f´(x)dx$$
                $$=sum_{k=1}^{n-1} int_{k}^{k+1}left{xright}f´(x)dx$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 4 at 16:33









                Ricardo770Ricardo770

                8017




                8017






























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