Reduction formula of $int left(frac{x^{n-2}}{x^n-1}right)dx$












2












$begingroup$


I need help in finding reduction formula of the following:



$$ I_n=int left(frac{x^{n-2}}{x^n-1}right)dx $$



Any hint or a complete solution would be very helpful.






calculus integration proof-writing reduction-formula







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  • $begingroup$
    @AlexFrancisco no, reduction formulas are a thing, often used in contexts such as this. You solve for $I_n$ in general usually by solving $I_n$ in terms of $I_{n-1}$ and computing $I_0$ directly (for example).
    $endgroup$
    – David Bowman
    Jan 1 '18 at 4:09






  • 3




    $begingroup$
    What are the bounds of the integral?
    $endgroup$
    – Abishanka Saha
    Jan 1 '18 at 4:35










  • $begingroup$
    Nothing major, but this doesn't work for $n = 0$. I'm guessing this is for $n geq 1$ ?
    $endgroup$
    – DavidG
    Jan 6 at 6:15










  • $begingroup$
    Understand you are seeking a recurrence relationship (I had a quick go and couldn't find anything). You can certainly evaluate your integral by factorising the denominator and applying a partial fraction decomposition. If you would like some further detail on that, let me know and I will post up.
    $endgroup$
    – DavidG
    Jan 6 at 8:32
















2












$begingroup$


I need help in finding reduction formula of the following:



$$ I_n=int left(frac{x^{n-2}}{x^n-1}right)dx $$



Any hint or a complete solution would be very helpful.






calculus integration proof-writing reduction-formula







share|cite|improve this question











$endgroup$












  • $begingroup$
    @AlexFrancisco no, reduction formulas are a thing, often used in contexts such as this. You solve for $I_n$ in general usually by solving $I_n$ in terms of $I_{n-1}$ and computing $I_0$ directly (for example).
    $endgroup$
    – David Bowman
    Jan 1 '18 at 4:09






  • 3




    $begingroup$
    What are the bounds of the integral?
    $endgroup$
    – Abishanka Saha
    Jan 1 '18 at 4:35










  • $begingroup$
    Nothing major, but this doesn't work for $n = 0$. I'm guessing this is for $n geq 1$ ?
    $endgroup$
    – DavidG
    Jan 6 at 6:15










  • $begingroup$
    Understand you are seeking a recurrence relationship (I had a quick go and couldn't find anything). You can certainly evaluate your integral by factorising the denominator and applying a partial fraction decomposition. If you would like some further detail on that, let me know and I will post up.
    $endgroup$
    – DavidG
    Jan 6 at 8:32














2












2








2





$begingroup$


I need help in finding reduction formula of the following:



$$ I_n=int left(frac{x^{n-2}}{x^n-1}right)dx $$



Any hint or a complete solution would be very helpful.






calculus integration proof-writing reduction-formula







share|cite|improve this question











$endgroup$




I need help in finding reduction formula of the following:



$$ I_n=int left(frac{x^{n-2}}{x^n-1}right)dx $$



Any hint or a complete solution would be very helpful.






calculus integration proof-writing reduction-formula




calculus integration proof-writing reduction-formula






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 6 at 5:43









clathratus

5,1701338




5,1701338










asked Jan 1 '18 at 3:48









Mrigank Shekhar PathakMrigank Shekhar Pathak

52229




52229












  • $begingroup$
    @AlexFrancisco no, reduction formulas are a thing, often used in contexts such as this. You solve for $I_n$ in general usually by solving $I_n$ in terms of $I_{n-1}$ and computing $I_0$ directly (for example).
    $endgroup$
    – David Bowman
    Jan 1 '18 at 4:09






  • 3




    $begingroup$
    What are the bounds of the integral?
    $endgroup$
    – Abishanka Saha
    Jan 1 '18 at 4:35










  • $begingroup$
    Nothing major, but this doesn't work for $n = 0$. I'm guessing this is for $n geq 1$ ?
    $endgroup$
    – DavidG
    Jan 6 at 6:15










  • $begingroup$
    Understand you are seeking a recurrence relationship (I had a quick go and couldn't find anything). You can certainly evaluate your integral by factorising the denominator and applying a partial fraction decomposition. If you would like some further detail on that, let me know and I will post up.
    $endgroup$
    – DavidG
    Jan 6 at 8:32


















  • $begingroup$
    @AlexFrancisco no, reduction formulas are a thing, often used in contexts such as this. You solve for $I_n$ in general usually by solving $I_n$ in terms of $I_{n-1}$ and computing $I_0$ directly (for example).
    $endgroup$
    – David Bowman
    Jan 1 '18 at 4:09






  • 3




    $begingroup$
    What are the bounds of the integral?
    $endgroup$
    – Abishanka Saha
    Jan 1 '18 at 4:35










  • $begingroup$
    Nothing major, but this doesn't work for $n = 0$. I'm guessing this is for $n geq 1$ ?
    $endgroup$
    – DavidG
    Jan 6 at 6:15










  • $begingroup$
    Understand you are seeking a recurrence relationship (I had a quick go and couldn't find anything). You can certainly evaluate your integral by factorising the denominator and applying a partial fraction decomposition. If you would like some further detail on that, let me know and I will post up.
    $endgroup$
    – DavidG
    Jan 6 at 8:32
















$begingroup$
@AlexFrancisco no, reduction formulas are a thing, often used in contexts such as this. You solve for $I_n$ in general usually by solving $I_n$ in terms of $I_{n-1}$ and computing $I_0$ directly (for example).
$endgroup$
– David Bowman
Jan 1 '18 at 4:09




$begingroup$
@AlexFrancisco no, reduction formulas are a thing, often used in contexts such as this. You solve for $I_n$ in general usually by solving $I_n$ in terms of $I_{n-1}$ and computing $I_0$ directly (for example).
$endgroup$
– David Bowman
Jan 1 '18 at 4:09




3




3




$begingroup$
What are the bounds of the integral?
$endgroup$
– Abishanka Saha
Jan 1 '18 at 4:35




$begingroup$
What are the bounds of the integral?
$endgroup$
– Abishanka Saha
Jan 1 '18 at 4:35












$begingroup$
Nothing major, but this doesn't work for $n = 0$. I'm guessing this is for $n geq 1$ ?
$endgroup$
– DavidG
Jan 6 at 6:15




$begingroup$
Nothing major, but this doesn't work for $n = 0$. I'm guessing this is for $n geq 1$ ?
$endgroup$
– DavidG
Jan 6 at 6:15












$begingroup$
Understand you are seeking a recurrence relationship (I had a quick go and couldn't find anything). You can certainly evaluate your integral by factorising the denominator and applying a partial fraction decomposition. If you would like some further detail on that, let me know and I will post up.
$endgroup$
– DavidG
Jan 6 at 8:32




$begingroup$
Understand you are seeking a recurrence relationship (I had a quick go and couldn't find anything). You can certainly evaluate your integral by factorising the denominator and applying a partial fraction decomposition. If you would like some further detail on that, let me know and I will post up.
$endgroup$
– DavidG
Jan 6 at 8:32










1 Answer
1






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$begingroup$

NOT A FULL SOLUTION:



I'm not sure is this will be of help:



begin{align}
I_n &= intfrac{x^{n - 2}}{x^n - 1}:dx = int frac{x^n}{x^2left(x^n - 1right)}:dx \
&= int frac{1}{x^2}frac{x^n - 1 + 1}{left(x^n - 1right)}:dx = int frac{1}{x^2}left(1 - frac{1}{x^n - 1}right):dx = int frac{1}{x^2}:dx - int frac{1}{x^2left(x^n - 1right)}:dx \
&= -frac{1}{x} - underbrace{int frac{1}{x^2left(x^n - 1right)}}_{J_n}:dx
end{align}



So, instead of investigating $I_n$ it may be better to investigate $J_n$ instead.



Alternatively you can easily factor $x^n - 1$ for both odd and even $n$. As above, if you would like detail on that method, let me know and I will post up.






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    0












    $begingroup$

    NOT A FULL SOLUTION:



    I'm not sure is this will be of help:



    begin{align}
    I_n &= intfrac{x^{n - 2}}{x^n - 1}:dx = int frac{x^n}{x^2left(x^n - 1right)}:dx \
    &= int frac{1}{x^2}frac{x^n - 1 + 1}{left(x^n - 1right)}:dx = int frac{1}{x^2}left(1 - frac{1}{x^n - 1}right):dx = int frac{1}{x^2}:dx - int frac{1}{x^2left(x^n - 1right)}:dx \
    &= -frac{1}{x} - underbrace{int frac{1}{x^2left(x^n - 1right)}}_{J_n}:dx
    end{align}



    So, instead of investigating $I_n$ it may be better to investigate $J_n$ instead.



    Alternatively you can easily factor $x^n - 1$ for both odd and even $n$. As above, if you would like detail on that method, let me know and I will post up.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      NOT A FULL SOLUTION:



      I'm not sure is this will be of help:



      begin{align}
      I_n &= intfrac{x^{n - 2}}{x^n - 1}:dx = int frac{x^n}{x^2left(x^n - 1right)}:dx \
      &= int frac{1}{x^2}frac{x^n - 1 + 1}{left(x^n - 1right)}:dx = int frac{1}{x^2}left(1 - frac{1}{x^n - 1}right):dx = int frac{1}{x^2}:dx - int frac{1}{x^2left(x^n - 1right)}:dx \
      &= -frac{1}{x} - underbrace{int frac{1}{x^2left(x^n - 1right)}}_{J_n}:dx
      end{align}



      So, instead of investigating $I_n$ it may be better to investigate $J_n$ instead.



      Alternatively you can easily factor $x^n - 1$ for both odd and even $n$. As above, if you would like detail on that method, let me know and I will post up.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        NOT A FULL SOLUTION:



        I'm not sure is this will be of help:



        begin{align}
        I_n &= intfrac{x^{n - 2}}{x^n - 1}:dx = int frac{x^n}{x^2left(x^n - 1right)}:dx \
        &= int frac{1}{x^2}frac{x^n - 1 + 1}{left(x^n - 1right)}:dx = int frac{1}{x^2}left(1 - frac{1}{x^n - 1}right):dx = int frac{1}{x^2}:dx - int frac{1}{x^2left(x^n - 1right)}:dx \
        &= -frac{1}{x} - underbrace{int frac{1}{x^2left(x^n - 1right)}}_{J_n}:dx
        end{align}



        So, instead of investigating $I_n$ it may be better to investigate $J_n$ instead.



        Alternatively you can easily factor $x^n - 1$ for both odd and even $n$. As above, if you would like detail on that method, let me know and I will post up.






        share|cite|improve this answer









        $endgroup$



        NOT A FULL SOLUTION:



        I'm not sure is this will be of help:



        begin{align}
        I_n &= intfrac{x^{n - 2}}{x^n - 1}:dx = int frac{x^n}{x^2left(x^n - 1right)}:dx \
        &= int frac{1}{x^2}frac{x^n - 1 + 1}{left(x^n - 1right)}:dx = int frac{1}{x^2}left(1 - frac{1}{x^n - 1}right):dx = int frac{1}{x^2}:dx - int frac{1}{x^2left(x^n - 1right)}:dx \
        &= -frac{1}{x} - underbrace{int frac{1}{x^2left(x^n - 1right)}}_{J_n}:dx
        end{align}



        So, instead of investigating $I_n$ it may be better to investigate $J_n$ instead.



        Alternatively you can easily factor $x^n - 1$ for both odd and even $n$. As above, if you would like detail on that method, let me know and I will post up.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 6 at 11:37









        DavidGDavidG

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