Integral of following fractional part












2














How would I evaluate this integral, where $nin R$ and ${,.}$ is the fractional part?



$$int_{-4}^4{nx},dx$$



I want to use it in one of problems of the ellipse. Thanks. I haven't shown any effort as I got no good step towards the solution.






integration definite-integrals fractional-part







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    The indefinite integral cannot be calculated as such, only the definite integral can be, so you need to supply the upper and lower bounds.
    – SchrodingersCat
    Jul 21 '16 at 14:40






  • 1




    If it was a definite integral, we could write ${nx}=nx-lfloor nx rfloor$ and then integrate the floor function with proper range.
    – StubbornAtom
    Jul 21 '16 at 14:44


















2














How would I evaluate this integral, where $nin R$ and ${,.}$ is the fractional part?



$$int_{-4}^4{nx},dx$$



I want to use it in one of problems of the ellipse. Thanks. I haven't shown any effort as I got no good step towards the solution.






integration definite-integrals fractional-part







share|cite|improve this question




















  • 1




    The indefinite integral cannot be calculated as such, only the definite integral can be, so you need to supply the upper and lower bounds.
    – SchrodingersCat
    Jul 21 '16 at 14:40






  • 1




    If it was a definite integral, we could write ${nx}=nx-lfloor nx rfloor$ and then integrate the floor function with proper range.
    – StubbornAtom
    Jul 21 '16 at 14:44
















2












2








2







How would I evaluate this integral, where $nin R$ and ${,.}$ is the fractional part?



$$int_{-4}^4{nx},dx$$



I want to use it in one of problems of the ellipse. Thanks. I haven't shown any effort as I got no good step towards the solution.






integration definite-integrals fractional-part







share|cite|improve this question















How would I evaluate this integral, where $nin R$ and ${,.}$ is the fractional part?



$$int_{-4}^4{nx},dx$$



I want to use it in one of problems of the ellipse. Thanks. I haven't shown any effort as I got no good step towards the solution.






integration definite-integrals fractional-part




integration definite-integrals fractional-part






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share|cite|improve this question













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edited Jul 19 at 16:51









Robert Howard

1,9161822




1,9161822










asked Jul 21 '16 at 14:37









Archis Welankar

12k41640




12k41640








  • 1




    The indefinite integral cannot be calculated as such, only the definite integral can be, so you need to supply the upper and lower bounds.
    – SchrodingersCat
    Jul 21 '16 at 14:40






  • 1




    If it was a definite integral, we could write ${nx}=nx-lfloor nx rfloor$ and then integrate the floor function with proper range.
    – StubbornAtom
    Jul 21 '16 at 14:44
















  • 1




    The indefinite integral cannot be calculated as such, only the definite integral can be, so you need to supply the upper and lower bounds.
    – SchrodingersCat
    Jul 21 '16 at 14:40






  • 1




    If it was a definite integral, we could write ${nx}=nx-lfloor nx rfloor$ and then integrate the floor function with proper range.
    – StubbornAtom
    Jul 21 '16 at 14:44










1




1




The indefinite integral cannot be calculated as such, only the definite integral can be, so you need to supply the upper and lower bounds.
– SchrodingersCat
Jul 21 '16 at 14:40




The indefinite integral cannot be calculated as such, only the definite integral can be, so you need to supply the upper and lower bounds.
– SchrodingersCat
Jul 21 '16 at 14:40




1




1




If it was a definite integral, we could write ${nx}=nx-lfloor nx rfloor$ and then integrate the floor function with proper range.
– StubbornAtom
Jul 21 '16 at 14:44






If it was a definite integral, we could write ${nx}=nx-lfloor nx rfloor$ and then integrate the floor function with proper range.
– StubbornAtom
Jul 21 '16 at 14:44












1 Answer
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Let's start. $$begin{aligned}int_{-4}^4{nx},dx&=left[begin{matrix}t=n x\dx=frac{dt}nend{matrix}right]=int_{-4n}^{4n}frac{{t}}n,dt=frac1nint_{-4n}^{4n}{t},dt=frac1nint_{-4n}^{4n}left(t-lfloor trfloorright),dt=\
&=frac1nleft(int_{-4n}^{4n}t~dt-int_{-4n}^{4n}lfloor trfloor~dtright)=-frac1nint_{-4n}^{4n}lfloor trfloor~dt=Iend{aligned}$$
Denote $N=lfloor 4nrfloor$. Then $$lfloor -4nrfloor=begin{cases}-N, & 4ninmathbb{Z} & (1)\-N-1, & 4ninmathbb{R}setminusmathbb{Z} & (2)end{cases}$$
When $4ninmathbb{Z}$ $$begin{aligned}I^{(1)}&=-frac1nint_{-4n}^{4n}lfloor trfloor~dt=-frac1nsum_{k=-N}^{N-1}int_k^{k+1}lfloor trfloor~dt=-frac1nsum_{k=-N}^{N-1}int_k^{k+1}k~dt=-frac1nsum_{k=-N}^{N-1}k~dt=\
&=-frac1n(-N)=frac{N}n=frac{lfloor 4nrfloor}{n}=left[4ninmathbb{Z},lfloor 4nrfloor=4nright]=frac{4n}n=4end{aligned}$$
When $4ninmathbb{R}setminusmathbb{Z}$ $$begin{aligned}I^{(2)}&=-frac1nint_{-4n}^{4n}lfloor trfloor~dt=-frac1nleft(int_{-4n}^{-N}lfloor trfloor~dt+int_{-N}^{N}lfloor trfloor~dt+int_{N}^{4n}lfloor trfloor~dtright)=\
&=-frac1nleft(int_{-4n}^{-N}(-N-1)~dt+sum_{k=-N}^{N-1}int_k^{k+1}lfloor trfloor~dt+int_{N}^{4n}N~dtright)=\
&=-frac1nleft((-N-1)(-N+4n)+sum_{k=-N}^{N-1}int_k^{k+1}k~dt+N(4n-N)right)=\
&=-frac1nleft((N+1)(N-4n)+sum_{k=-N}^{N-1}k+4nN-N^2right)=\
&=-frac1nleft(N^2+N-4nN-4n-N+4nN-N^2right)=\
&=-frac1n(-4n)=4end{aligned}$$
Thus $$I=I^{(1)}=I^{(2)}=4$$






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    Let's start. $$begin{aligned}int_{-4}^4{nx},dx&=left[begin{matrix}t=n x\dx=frac{dt}nend{matrix}right]=int_{-4n}^{4n}frac{{t}}n,dt=frac1nint_{-4n}^{4n}{t},dt=frac1nint_{-4n}^{4n}left(t-lfloor trfloorright),dt=\
    &=frac1nleft(int_{-4n}^{4n}t~dt-int_{-4n}^{4n}lfloor trfloor~dtright)=-frac1nint_{-4n}^{4n}lfloor trfloor~dt=Iend{aligned}$$
    Denote $N=lfloor 4nrfloor$. Then $$lfloor -4nrfloor=begin{cases}-N, & 4ninmathbb{Z} & (1)\-N-1, & 4ninmathbb{R}setminusmathbb{Z} & (2)end{cases}$$
    When $4ninmathbb{Z}$ $$begin{aligned}I^{(1)}&=-frac1nint_{-4n}^{4n}lfloor trfloor~dt=-frac1nsum_{k=-N}^{N-1}int_k^{k+1}lfloor trfloor~dt=-frac1nsum_{k=-N}^{N-1}int_k^{k+1}k~dt=-frac1nsum_{k=-N}^{N-1}k~dt=\
    &=-frac1n(-N)=frac{N}n=frac{lfloor 4nrfloor}{n}=left[4ninmathbb{Z},lfloor 4nrfloor=4nright]=frac{4n}n=4end{aligned}$$
    When $4ninmathbb{R}setminusmathbb{Z}$ $$begin{aligned}I^{(2)}&=-frac1nint_{-4n}^{4n}lfloor trfloor~dt=-frac1nleft(int_{-4n}^{-N}lfloor trfloor~dt+int_{-N}^{N}lfloor trfloor~dt+int_{N}^{4n}lfloor trfloor~dtright)=\
    &=-frac1nleft(int_{-4n}^{-N}(-N-1)~dt+sum_{k=-N}^{N-1}int_k^{k+1}lfloor trfloor~dt+int_{N}^{4n}N~dtright)=\
    &=-frac1nleft((-N-1)(-N+4n)+sum_{k=-N}^{N-1}int_k^{k+1}k~dt+N(4n-N)right)=\
    &=-frac1nleft((N+1)(N-4n)+sum_{k=-N}^{N-1}k+4nN-N^2right)=\
    &=-frac1nleft(N^2+N-4nN-4n-N+4nN-N^2right)=\
    &=-frac1n(-4n)=4end{aligned}$$
    Thus $$I=I^{(1)}=I^{(2)}=4$$






    share|cite|improve this answer


























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      Let's start. $$begin{aligned}int_{-4}^4{nx},dx&=left[begin{matrix}t=n x\dx=frac{dt}nend{matrix}right]=int_{-4n}^{4n}frac{{t}}n,dt=frac1nint_{-4n}^{4n}{t},dt=frac1nint_{-4n}^{4n}left(t-lfloor trfloorright),dt=\
      &=frac1nleft(int_{-4n}^{4n}t~dt-int_{-4n}^{4n}lfloor trfloor~dtright)=-frac1nint_{-4n}^{4n}lfloor trfloor~dt=Iend{aligned}$$
      Denote $N=lfloor 4nrfloor$. Then $$lfloor -4nrfloor=begin{cases}-N, & 4ninmathbb{Z} & (1)\-N-1, & 4ninmathbb{R}setminusmathbb{Z} & (2)end{cases}$$
      When $4ninmathbb{Z}$ $$begin{aligned}I^{(1)}&=-frac1nint_{-4n}^{4n}lfloor trfloor~dt=-frac1nsum_{k=-N}^{N-1}int_k^{k+1}lfloor trfloor~dt=-frac1nsum_{k=-N}^{N-1}int_k^{k+1}k~dt=-frac1nsum_{k=-N}^{N-1}k~dt=\
      &=-frac1n(-N)=frac{N}n=frac{lfloor 4nrfloor}{n}=left[4ninmathbb{Z},lfloor 4nrfloor=4nright]=frac{4n}n=4end{aligned}$$
      When $4ninmathbb{R}setminusmathbb{Z}$ $$begin{aligned}I^{(2)}&=-frac1nint_{-4n}^{4n}lfloor trfloor~dt=-frac1nleft(int_{-4n}^{-N}lfloor trfloor~dt+int_{-N}^{N}lfloor trfloor~dt+int_{N}^{4n}lfloor trfloor~dtright)=\
      &=-frac1nleft(int_{-4n}^{-N}(-N-1)~dt+sum_{k=-N}^{N-1}int_k^{k+1}lfloor trfloor~dt+int_{N}^{4n}N~dtright)=\
      &=-frac1nleft((-N-1)(-N+4n)+sum_{k=-N}^{N-1}int_k^{k+1}k~dt+N(4n-N)right)=\
      &=-frac1nleft((N+1)(N-4n)+sum_{k=-N}^{N-1}k+4nN-N^2right)=\
      &=-frac1nleft(N^2+N-4nN-4n-N+4nN-N^2right)=\
      &=-frac1n(-4n)=4end{aligned}$$
      Thus $$I=I^{(1)}=I^{(2)}=4$$






      share|cite|improve this answer
























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        Let's start. $$begin{aligned}int_{-4}^4{nx},dx&=left[begin{matrix}t=n x\dx=frac{dt}nend{matrix}right]=int_{-4n}^{4n}frac{{t}}n,dt=frac1nint_{-4n}^{4n}{t},dt=frac1nint_{-4n}^{4n}left(t-lfloor trfloorright),dt=\
        &=frac1nleft(int_{-4n}^{4n}t~dt-int_{-4n}^{4n}lfloor trfloor~dtright)=-frac1nint_{-4n}^{4n}lfloor trfloor~dt=Iend{aligned}$$
        Denote $N=lfloor 4nrfloor$. Then $$lfloor -4nrfloor=begin{cases}-N, & 4ninmathbb{Z} & (1)\-N-1, & 4ninmathbb{R}setminusmathbb{Z} & (2)end{cases}$$
        When $4ninmathbb{Z}$ $$begin{aligned}I^{(1)}&=-frac1nint_{-4n}^{4n}lfloor trfloor~dt=-frac1nsum_{k=-N}^{N-1}int_k^{k+1}lfloor trfloor~dt=-frac1nsum_{k=-N}^{N-1}int_k^{k+1}k~dt=-frac1nsum_{k=-N}^{N-1}k~dt=\
        &=-frac1n(-N)=frac{N}n=frac{lfloor 4nrfloor}{n}=left[4ninmathbb{Z},lfloor 4nrfloor=4nright]=frac{4n}n=4end{aligned}$$
        When $4ninmathbb{R}setminusmathbb{Z}$ $$begin{aligned}I^{(2)}&=-frac1nint_{-4n}^{4n}lfloor trfloor~dt=-frac1nleft(int_{-4n}^{-N}lfloor trfloor~dt+int_{-N}^{N}lfloor trfloor~dt+int_{N}^{4n}lfloor trfloor~dtright)=\
        &=-frac1nleft(int_{-4n}^{-N}(-N-1)~dt+sum_{k=-N}^{N-1}int_k^{k+1}lfloor trfloor~dt+int_{N}^{4n}N~dtright)=\
        &=-frac1nleft((-N-1)(-N+4n)+sum_{k=-N}^{N-1}int_k^{k+1}k~dt+N(4n-N)right)=\
        &=-frac1nleft((N+1)(N-4n)+sum_{k=-N}^{N-1}k+4nN-N^2right)=\
        &=-frac1nleft(N^2+N-4nN-4n-N+4nN-N^2right)=\
        &=-frac1n(-4n)=4end{aligned}$$
        Thus $$I=I^{(1)}=I^{(2)}=4$$






        share|cite|improve this answer












        Let's start. $$begin{aligned}int_{-4}^4{nx},dx&=left[begin{matrix}t=n x\dx=frac{dt}nend{matrix}right]=int_{-4n}^{4n}frac{{t}}n,dt=frac1nint_{-4n}^{4n}{t},dt=frac1nint_{-4n}^{4n}left(t-lfloor trfloorright),dt=\
        &=frac1nleft(int_{-4n}^{4n}t~dt-int_{-4n}^{4n}lfloor trfloor~dtright)=-frac1nint_{-4n}^{4n}lfloor trfloor~dt=Iend{aligned}$$
        Denote $N=lfloor 4nrfloor$. Then $$lfloor -4nrfloor=begin{cases}-N, & 4ninmathbb{Z} & (1)\-N-1, & 4ninmathbb{R}setminusmathbb{Z} & (2)end{cases}$$
        When $4ninmathbb{Z}$ $$begin{aligned}I^{(1)}&=-frac1nint_{-4n}^{4n}lfloor trfloor~dt=-frac1nsum_{k=-N}^{N-1}int_k^{k+1}lfloor trfloor~dt=-frac1nsum_{k=-N}^{N-1}int_k^{k+1}k~dt=-frac1nsum_{k=-N}^{N-1}k~dt=\
        &=-frac1n(-N)=frac{N}n=frac{lfloor 4nrfloor}{n}=left[4ninmathbb{Z},lfloor 4nrfloor=4nright]=frac{4n}n=4end{aligned}$$
        When $4ninmathbb{R}setminusmathbb{Z}$ $$begin{aligned}I^{(2)}&=-frac1nint_{-4n}^{4n}lfloor trfloor~dt=-frac1nleft(int_{-4n}^{-N}lfloor trfloor~dt+int_{-N}^{N}lfloor trfloor~dt+int_{N}^{4n}lfloor trfloor~dtright)=\
        &=-frac1nleft(int_{-4n}^{-N}(-N-1)~dt+sum_{k=-N}^{N-1}int_k^{k+1}lfloor trfloor~dt+int_{N}^{4n}N~dtright)=\
        &=-frac1nleft((-N-1)(-N+4n)+sum_{k=-N}^{N-1}int_k^{k+1}k~dt+N(4n-N)right)=\
        &=-frac1nleft((N+1)(N-4n)+sum_{k=-N}^{N-1}k+4nN-N^2right)=\
        &=-frac1nleft(N^2+N-4nN-4n-N+4nN-N^2right)=\
        &=-frac1n(-4n)=4end{aligned}$$
        Thus $$I=I^{(1)}=I^{(2)}=4$$







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        answered Nov 23 at 7:00









        Mikalai Parshutsich

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