$intlimits_0^infty {x^4 over (x^4-x^2+1)^4} dx$












2












$begingroup$


I want to calculate
$$intlimits_0^infty frac{x^4}{(x^4-x^2+1)^4}dx$$



I have searched with keywords "frac{x^4}{(x^4-x^2+1)^4}" and "x^4/(x^4-x^2+1)^4". But there are no results






definite-integrals







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$endgroup$












  • $begingroup$
    The integral is same as$$int_{0}^{infty} frac{x^{4}(x^{2}+1)^{4}}{(x^{6}+1)^{4}}dx$$
    $endgroup$
    – Seewoo Lee
    Jan 22 at 5:54










  • $begingroup$
    Partial fractions?
    $endgroup$
    – Lord Shark the Unknown
    Jan 22 at 5:54






  • 4




    $begingroup$
    I remember the days when people just did integrations rather than Google them.
    $endgroup$
    – Lord Shark the Unknown
    Jan 22 at 5:55










  • $begingroup$
    These days are over!
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 22 at 6:00






  • 2




    $begingroup$
    Note that in $0$ and $infty$ the rational and logarithmic contribution is zero. Remains only the $arctan$ value.
    $endgroup$
    – zwim
    Jan 22 at 6:41


















2












$begingroup$


I want to calculate
$$intlimits_0^infty frac{x^4}{(x^4-x^2+1)^4}dx$$



I have searched with keywords "frac{x^4}{(x^4-x^2+1)^4}" and "x^4/(x^4-x^2+1)^4". But there are no results






definite-integrals







share|cite|improve this question











$endgroup$












  • $begingroup$
    The integral is same as$$int_{0}^{infty} frac{x^{4}(x^{2}+1)^{4}}{(x^{6}+1)^{4}}dx$$
    $endgroup$
    – Seewoo Lee
    Jan 22 at 5:54










  • $begingroup$
    Partial fractions?
    $endgroup$
    – Lord Shark the Unknown
    Jan 22 at 5:54






  • 4




    $begingroup$
    I remember the days when people just did integrations rather than Google them.
    $endgroup$
    – Lord Shark the Unknown
    Jan 22 at 5:55










  • $begingroup$
    These days are over!
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 22 at 6:00






  • 2




    $begingroup$
    Note that in $0$ and $infty$ the rational and logarithmic contribution is zero. Remains only the $arctan$ value.
    $endgroup$
    – zwim
    Jan 22 at 6:41
















2












2








2


3



$begingroup$


I want to calculate
$$intlimits_0^infty frac{x^4}{(x^4-x^2+1)^4}dx$$



I have searched with keywords "frac{x^4}{(x^4-x^2+1)^4}" and "x^4/(x^4-x^2+1)^4". But there are no results






definite-integrals







share|cite|improve this question











$endgroup$




I want to calculate
$$intlimits_0^infty frac{x^4}{(x^4-x^2+1)^4}dx$$



I have searched with keywords "frac{x^4}{(x^4-x^2+1)^4}" and "x^4/(x^4-x^2+1)^4". But there are no results






definite-integrals




definite-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 22 at 6:32









David G. Stork

11.1k41432




11.1k41432










asked Jan 22 at 5:52









Reynan HenryReynan Henry

751




751












  • $begingroup$
    The integral is same as$$int_{0}^{infty} frac{x^{4}(x^{2}+1)^{4}}{(x^{6}+1)^{4}}dx$$
    $endgroup$
    – Seewoo Lee
    Jan 22 at 5:54










  • $begingroup$
    Partial fractions?
    $endgroup$
    – Lord Shark the Unknown
    Jan 22 at 5:54






  • 4




    $begingroup$
    I remember the days when people just did integrations rather than Google them.
    $endgroup$
    – Lord Shark the Unknown
    Jan 22 at 5:55










  • $begingroup$
    These days are over!
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 22 at 6:00






  • 2




    $begingroup$
    Note that in $0$ and $infty$ the rational and logarithmic contribution is zero. Remains only the $arctan$ value.
    $endgroup$
    – zwim
    Jan 22 at 6:41




















  • $begingroup$
    The integral is same as$$int_{0}^{infty} frac{x^{4}(x^{2}+1)^{4}}{(x^{6}+1)^{4}}dx$$
    $endgroup$
    – Seewoo Lee
    Jan 22 at 5:54










  • $begingroup$
    Partial fractions?
    $endgroup$
    – Lord Shark the Unknown
    Jan 22 at 5:54






  • 4




    $begingroup$
    I remember the days when people just did integrations rather than Google them.
    $endgroup$
    – Lord Shark the Unknown
    Jan 22 at 5:55










  • $begingroup$
    These days are over!
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 22 at 6:00






  • 2




    $begingroup$
    Note that in $0$ and $infty$ the rational and logarithmic contribution is zero. Remains only the $arctan$ value.
    $endgroup$
    – zwim
    Jan 22 at 6:41


















$begingroup$
The integral is same as$$int_{0}^{infty} frac{x^{4}(x^{2}+1)^{4}}{(x^{6}+1)^{4}}dx$$
$endgroup$
– Seewoo Lee
Jan 22 at 5:54




$begingroup$
The integral is same as$$int_{0}^{infty} frac{x^{4}(x^{2}+1)^{4}}{(x^{6}+1)^{4}}dx$$
$endgroup$
– Seewoo Lee
Jan 22 at 5:54












$begingroup$
Partial fractions?
$endgroup$
– Lord Shark the Unknown
Jan 22 at 5:54




$begingroup$
Partial fractions?
$endgroup$
– Lord Shark the Unknown
Jan 22 at 5:54




4




4




$begingroup$
I remember the days when people just did integrations rather than Google them.
$endgroup$
– Lord Shark the Unknown
Jan 22 at 5:55




$begingroup$
I remember the days when people just did integrations rather than Google them.
$endgroup$
– Lord Shark the Unknown
Jan 22 at 5:55












$begingroup$
These days are over!
$endgroup$
– Dr. Sonnhard Graubner
Jan 22 at 6:00




$begingroup$
These days are over!
$endgroup$
– Dr. Sonnhard Graubner
Jan 22 at 6:00




2




2




$begingroup$
Note that in $0$ and $infty$ the rational and logarithmic contribution is zero. Remains only the $arctan$ value.
$endgroup$
– zwim
Jan 22 at 6:41






$begingroup$
Note that in $0$ and $infty$ the rational and logarithmic contribution is zero. Remains only the $arctan$ value.
$endgroup$
– zwim
Jan 22 at 6:41












2 Answers
2






active

oldest

votes


















5












$begingroup$

For $b > 0$, define $I_b(a)$ by



$$ I_b(a)
= int_{0}^{infty} frac{x^{a}}{((x - x^{-1})^2 + 1)^b} , mathrm{d}x
= int_{0}^{infty} frac{x^{a+2b}}{(x^4 - x^2 + 1)^b} , mathrm{d}x. $$



This integral converges if $|a+1| < 2b$. We can also prove that $I_b(a) = I_b(-a-2)$ holds, by using the substitution $x mapsto 1/x$. Then



begin{align*}
int_{0}^{infty} frac{x^4}{(x^4 - x^2 + 1)^4} , mathrm{d}x
&= I_4(-4)
= I_4(2) \
&= I_4(2) - I_4(0) + I_4(-2) \
&= int_{0}^{infty} frac{1}{((x - x^{-1})^2 + 1)^3} , mathrm{d}x
end{align*}



So, by the Glasser's master theorem,



begin{align*}
int_{0}^{infty} frac{x^4}{(x^4 - x^2 + 1)^4} , mathrm{d}x
&= int_{0}^{infty} frac{1}{(u^2 + 1)^3} , mathrm{d}u \
&= int_{0}^{frac{pi}{2}} cos^4theta , mathrm{d}theta tag{$(u=tantheta)$} \
&= frac{3pi}{16}
approx 0.58904862254808623221 cdots.
end{align*}






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    +1. I knew there'd be a Glasser solution somewhere.
    $endgroup$
    – J.G.
    Jan 22 at 7:31



















4












$begingroup$

Here is another way to get to the same point as what @Sangchul Lee gives.



Let
$$I = int_0^infty frac{x^4}{(x^4 - x^2 + 1)^4} , dx.$$
Then
$$I = int_0^1 frac{x^4}{(x^4 - x^2 + 1)^4} , dx + int_1^infty frac{x^4}{(x^4 - x^2 + 1)^4} , dx.$$
Enforcing a substitution of $x mapsto 1/x$ in the right most integral leads to
begin{align}
I &= int_0^1 frac{x^4 (1 + x^6)}{(x^4 - x^2 + 1)^4} , dx\
&= int_0^1 frac{x^4 (1 + x^2)(x^4 - x^2 + 1)}{(x^4 - x^2 + 1)^4} , dx\
&= int_0^1 frac{x^4 (1 + x^2)}{(x^4 - x^2 + 1)^3} , dx\
&= int_0^1 frac{1 + 1/x^2}{left [left (x - frac{1}{x} right )^2 + 1 right ]^3} , dx.
end{align}
On setting $-u = x - 1/x$, $-du = (1 + 1/x^2) , dx$ one has
begin{align}
I &= int_0^infty frac{du}{(u^2 + 1)^3}\
&= int_0^{frac{pi}{2}} cos^4 theta , dtheta\
&= int_0^{frac{pi}{2}} left (frac{1}{2} cos 2theta + frac{1}{8} cos 4theta + frac{3}{8} right ) , dtheta\
&= frac{3pi}{16},
end{align}
as expected.






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    5












    $begingroup$

    For $b > 0$, define $I_b(a)$ by



    $$ I_b(a)
    = int_{0}^{infty} frac{x^{a}}{((x - x^{-1})^2 + 1)^b} , mathrm{d}x
    = int_{0}^{infty} frac{x^{a+2b}}{(x^4 - x^2 + 1)^b} , mathrm{d}x. $$



    This integral converges if $|a+1| < 2b$. We can also prove that $I_b(a) = I_b(-a-2)$ holds, by using the substitution $x mapsto 1/x$. Then



    begin{align*}
    int_{0}^{infty} frac{x^4}{(x^4 - x^2 + 1)^4} , mathrm{d}x
    &= I_4(-4)
    = I_4(2) \
    &= I_4(2) - I_4(0) + I_4(-2) \
    &= int_{0}^{infty} frac{1}{((x - x^{-1})^2 + 1)^3} , mathrm{d}x
    end{align*}



    So, by the Glasser's master theorem,



    begin{align*}
    int_{0}^{infty} frac{x^4}{(x^4 - x^2 + 1)^4} , mathrm{d}x
    &= int_{0}^{infty} frac{1}{(u^2 + 1)^3} , mathrm{d}u \
    &= int_{0}^{frac{pi}{2}} cos^4theta , mathrm{d}theta tag{$(u=tantheta)$} \
    &= frac{3pi}{16}
    approx 0.58904862254808623221 cdots.
    end{align*}






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      +1. I knew there'd be a Glasser solution somewhere.
      $endgroup$
      – J.G.
      Jan 22 at 7:31
















    5












    $begingroup$

    For $b > 0$, define $I_b(a)$ by



    $$ I_b(a)
    = int_{0}^{infty} frac{x^{a}}{((x - x^{-1})^2 + 1)^b} , mathrm{d}x
    = int_{0}^{infty} frac{x^{a+2b}}{(x^4 - x^2 + 1)^b} , mathrm{d}x. $$



    This integral converges if $|a+1| < 2b$. We can also prove that $I_b(a) = I_b(-a-2)$ holds, by using the substitution $x mapsto 1/x$. Then



    begin{align*}
    int_{0}^{infty} frac{x^4}{(x^4 - x^2 + 1)^4} , mathrm{d}x
    &= I_4(-4)
    = I_4(2) \
    &= I_4(2) - I_4(0) + I_4(-2) \
    &= int_{0}^{infty} frac{1}{((x - x^{-1})^2 + 1)^3} , mathrm{d}x
    end{align*}



    So, by the Glasser's master theorem,



    begin{align*}
    int_{0}^{infty} frac{x^4}{(x^4 - x^2 + 1)^4} , mathrm{d}x
    &= int_{0}^{infty} frac{1}{(u^2 + 1)^3} , mathrm{d}u \
    &= int_{0}^{frac{pi}{2}} cos^4theta , mathrm{d}theta tag{$(u=tantheta)$} \
    &= frac{3pi}{16}
    approx 0.58904862254808623221 cdots.
    end{align*}






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      +1. I knew there'd be a Glasser solution somewhere.
      $endgroup$
      – J.G.
      Jan 22 at 7:31














    5












    5








    5





    $begingroup$

    For $b > 0$, define $I_b(a)$ by



    $$ I_b(a)
    = int_{0}^{infty} frac{x^{a}}{((x - x^{-1})^2 + 1)^b} , mathrm{d}x
    = int_{0}^{infty} frac{x^{a+2b}}{(x^4 - x^2 + 1)^b} , mathrm{d}x. $$



    This integral converges if $|a+1| < 2b$. We can also prove that $I_b(a) = I_b(-a-2)$ holds, by using the substitution $x mapsto 1/x$. Then



    begin{align*}
    int_{0}^{infty} frac{x^4}{(x^4 - x^2 + 1)^4} , mathrm{d}x
    &= I_4(-4)
    = I_4(2) \
    &= I_4(2) - I_4(0) + I_4(-2) \
    &= int_{0}^{infty} frac{1}{((x - x^{-1})^2 + 1)^3} , mathrm{d}x
    end{align*}



    So, by the Glasser's master theorem,



    begin{align*}
    int_{0}^{infty} frac{x^4}{(x^4 - x^2 + 1)^4} , mathrm{d}x
    &= int_{0}^{infty} frac{1}{(u^2 + 1)^3} , mathrm{d}u \
    &= int_{0}^{frac{pi}{2}} cos^4theta , mathrm{d}theta tag{$(u=tantheta)$} \
    &= frac{3pi}{16}
    approx 0.58904862254808623221 cdots.
    end{align*}






    share|cite|improve this answer









    $endgroup$



    For $b > 0$, define $I_b(a)$ by



    $$ I_b(a)
    = int_{0}^{infty} frac{x^{a}}{((x - x^{-1})^2 + 1)^b} , mathrm{d}x
    = int_{0}^{infty} frac{x^{a+2b}}{(x^4 - x^2 + 1)^b} , mathrm{d}x. $$



    This integral converges if $|a+1| < 2b$. We can also prove that $I_b(a) = I_b(-a-2)$ holds, by using the substitution $x mapsto 1/x$. Then



    begin{align*}
    int_{0}^{infty} frac{x^4}{(x^4 - x^2 + 1)^4} , mathrm{d}x
    &= I_4(-4)
    = I_4(2) \
    &= I_4(2) - I_4(0) + I_4(-2) \
    &= int_{0}^{infty} frac{1}{((x - x^{-1})^2 + 1)^3} , mathrm{d}x
    end{align*}



    So, by the Glasser's master theorem,



    begin{align*}
    int_{0}^{infty} frac{x^4}{(x^4 - x^2 + 1)^4} , mathrm{d}x
    &= int_{0}^{infty} frac{1}{(u^2 + 1)^3} , mathrm{d}u \
    &= int_{0}^{frac{pi}{2}} cos^4theta , mathrm{d}theta tag{$(u=tantheta)$} \
    &= frac{3pi}{16}
    approx 0.58904862254808623221 cdots.
    end{align*}







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 22 at 7:03









    Sangchul LeeSangchul Lee

    95.5k12171279




    95.5k12171279








    • 1




      $begingroup$
      +1. I knew there'd be a Glasser solution somewhere.
      $endgroup$
      – J.G.
      Jan 22 at 7:31














    • 1




      $begingroup$
      +1. I knew there'd be a Glasser solution somewhere.
      $endgroup$
      – J.G.
      Jan 22 at 7:31








    1




    1




    $begingroup$
    +1. I knew there'd be a Glasser solution somewhere.
    $endgroup$
    – J.G.
    Jan 22 at 7:31




    $begingroup$
    +1. I knew there'd be a Glasser solution somewhere.
    $endgroup$
    – J.G.
    Jan 22 at 7:31











    4












    $begingroup$

    Here is another way to get to the same point as what @Sangchul Lee gives.



    Let
    $$I = int_0^infty frac{x^4}{(x^4 - x^2 + 1)^4} , dx.$$
    Then
    $$I = int_0^1 frac{x^4}{(x^4 - x^2 + 1)^4} , dx + int_1^infty frac{x^4}{(x^4 - x^2 + 1)^4} , dx.$$
    Enforcing a substitution of $x mapsto 1/x$ in the right most integral leads to
    begin{align}
    I &= int_0^1 frac{x^4 (1 + x^6)}{(x^4 - x^2 + 1)^4} , dx\
    &= int_0^1 frac{x^4 (1 + x^2)(x^4 - x^2 + 1)}{(x^4 - x^2 + 1)^4} , dx\
    &= int_0^1 frac{x^4 (1 + x^2)}{(x^4 - x^2 + 1)^3} , dx\
    &= int_0^1 frac{1 + 1/x^2}{left [left (x - frac{1}{x} right )^2 + 1 right ]^3} , dx.
    end{align}
    On setting $-u = x - 1/x$, $-du = (1 + 1/x^2) , dx$ one has
    begin{align}
    I &= int_0^infty frac{du}{(u^2 + 1)^3}\
    &= int_0^{frac{pi}{2}} cos^4 theta , dtheta\
    &= int_0^{frac{pi}{2}} left (frac{1}{2} cos 2theta + frac{1}{8} cos 4theta + frac{3}{8} right ) , dtheta\
    &= frac{3pi}{16},
    end{align}
    as expected.






    share|cite|improve this answer









    $endgroup$


















      4












      $begingroup$

      Here is another way to get to the same point as what @Sangchul Lee gives.



      Let
      $$I = int_0^infty frac{x^4}{(x^4 - x^2 + 1)^4} , dx.$$
      Then
      $$I = int_0^1 frac{x^4}{(x^4 - x^2 + 1)^4} , dx + int_1^infty frac{x^4}{(x^4 - x^2 + 1)^4} , dx.$$
      Enforcing a substitution of $x mapsto 1/x$ in the right most integral leads to
      begin{align}
      I &= int_0^1 frac{x^4 (1 + x^6)}{(x^4 - x^2 + 1)^4} , dx\
      &= int_0^1 frac{x^4 (1 + x^2)(x^4 - x^2 + 1)}{(x^4 - x^2 + 1)^4} , dx\
      &= int_0^1 frac{x^4 (1 + x^2)}{(x^4 - x^2 + 1)^3} , dx\
      &= int_0^1 frac{1 + 1/x^2}{left [left (x - frac{1}{x} right )^2 + 1 right ]^3} , dx.
      end{align}
      On setting $-u = x - 1/x$, $-du = (1 + 1/x^2) , dx$ one has
      begin{align}
      I &= int_0^infty frac{du}{(u^2 + 1)^3}\
      &= int_0^{frac{pi}{2}} cos^4 theta , dtheta\
      &= int_0^{frac{pi}{2}} left (frac{1}{2} cos 2theta + frac{1}{8} cos 4theta + frac{3}{8} right ) , dtheta\
      &= frac{3pi}{16},
      end{align}
      as expected.






      share|cite|improve this answer









      $endgroup$
















        4












        4








        4





        $begingroup$

        Here is another way to get to the same point as what @Sangchul Lee gives.



        Let
        $$I = int_0^infty frac{x^4}{(x^4 - x^2 + 1)^4} , dx.$$
        Then
        $$I = int_0^1 frac{x^4}{(x^4 - x^2 + 1)^4} , dx + int_1^infty frac{x^4}{(x^4 - x^2 + 1)^4} , dx.$$
        Enforcing a substitution of $x mapsto 1/x$ in the right most integral leads to
        begin{align}
        I &= int_0^1 frac{x^4 (1 + x^6)}{(x^4 - x^2 + 1)^4} , dx\
        &= int_0^1 frac{x^4 (1 + x^2)(x^4 - x^2 + 1)}{(x^4 - x^2 + 1)^4} , dx\
        &= int_0^1 frac{x^4 (1 + x^2)}{(x^4 - x^2 + 1)^3} , dx\
        &= int_0^1 frac{1 + 1/x^2}{left [left (x - frac{1}{x} right )^2 + 1 right ]^3} , dx.
        end{align}
        On setting $-u = x - 1/x$, $-du = (1 + 1/x^2) , dx$ one has
        begin{align}
        I &= int_0^infty frac{du}{(u^2 + 1)^3}\
        &= int_0^{frac{pi}{2}} cos^4 theta , dtheta\
        &= int_0^{frac{pi}{2}} left (frac{1}{2} cos 2theta + frac{1}{8} cos 4theta + frac{3}{8} right ) , dtheta\
        &= frac{3pi}{16},
        end{align}
        as expected.






        share|cite|improve this answer









        $endgroup$



        Here is another way to get to the same point as what @Sangchul Lee gives.



        Let
        $$I = int_0^infty frac{x^4}{(x^4 - x^2 + 1)^4} , dx.$$
        Then
        $$I = int_0^1 frac{x^4}{(x^4 - x^2 + 1)^4} , dx + int_1^infty frac{x^4}{(x^4 - x^2 + 1)^4} , dx.$$
        Enforcing a substitution of $x mapsto 1/x$ in the right most integral leads to
        begin{align}
        I &= int_0^1 frac{x^4 (1 + x^6)}{(x^4 - x^2 + 1)^4} , dx\
        &= int_0^1 frac{x^4 (1 + x^2)(x^4 - x^2 + 1)}{(x^4 - x^2 + 1)^4} , dx\
        &= int_0^1 frac{x^4 (1 + x^2)}{(x^4 - x^2 + 1)^3} , dx\
        &= int_0^1 frac{1 + 1/x^2}{left [left (x - frac{1}{x} right )^2 + 1 right ]^3} , dx.
        end{align}
        On setting $-u = x - 1/x$, $-du = (1 + 1/x^2) , dx$ one has
        begin{align}
        I &= int_0^infty frac{du}{(u^2 + 1)^3}\
        &= int_0^{frac{pi}{2}} cos^4 theta , dtheta\
        &= int_0^{frac{pi}{2}} left (frac{1}{2} cos 2theta + frac{1}{8} cos 4theta + frac{3}{8} right ) , dtheta\
        &= frac{3pi}{16},
        end{align}
        as expected.







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        answered Jan 22 at 7:15









        omegadotomegadot

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