Integration of powers of trigonometric function with linear term












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I got stuck trying to find a general formula for the following integral
$$int_0^{pi} t cdotcos^{2n}{left(frac{t}{2}right)} , dt =
4 int_0^{pi/2} t cdotcos^{2n}t ,dt ; , ; text{ for } n in mathbb{N}.$$

It doesn't seem to be listed on Wikipedias Lists of integrals. After playing around for a while I found the following potentially useful identities$$begin{align}
int_0^{pi} t cdotcos^{2n}{left(frac{t}{2}right)} , dt &=
frac{(2n-1)!!}{(2n)!!} pi^2 - 4int_{0}^{pi/2}{int_0^{t} cos^{2n}{tau} , dtau} , dt \
&= frac{(2n-1)!!}{(2n)!!}cdot left( frac{pi^2}{2} - sum_{k=1}^n frac{(2k)!!}{(2k-1)!! , k^2} right) ,,
end{align}$$

where I would consider the last line a solution if one was able to find a closed formula for the sum that is not given in terms of another integral or another horrible function like Wolfram Alpha suggests.






Edits:




  • Moved the power closer to $cos$ to avoid confusion.

  • Corrected some constants in the identities.

  • Emphasized that $n in mathbb{N}$.










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  • $begingroup$
    Is $cos$ raised to the power of $2n$ or $t$?
    $endgroup$
    – DavidG
    Jan 3 at 3:16










  • $begingroup$
    If $n in mathbb{N}$ there are some nice closed solutions (I believe).
    $endgroup$
    – DavidG
    Jan 3 at 3:43










  • $begingroup$
    Numerical results of your identities don't hold up, I think you need $pi^{2}/2$ instead.
    $endgroup$
    – Ininterrompue
    Jan 3 at 6:34










  • $begingroup$
    @Ininterrompue, thank you! I fixed the constants now.
    $endgroup$
    – chickenNinja123
    Jan 3 at 9:18
















1












$begingroup$


I got stuck trying to find a general formula for the following integral
$$int_0^{pi} t cdotcos^{2n}{left(frac{t}{2}right)} , dt =
4 int_0^{pi/2} t cdotcos^{2n}t ,dt ; , ; text{ for } n in mathbb{N}.$$

It doesn't seem to be listed on Wikipedias Lists of integrals. After playing around for a while I found the following potentially useful identities$$begin{align}
int_0^{pi} t cdotcos^{2n}{left(frac{t}{2}right)} , dt &=
frac{(2n-1)!!}{(2n)!!} pi^2 - 4int_{0}^{pi/2}{int_0^{t} cos^{2n}{tau} , dtau} , dt \
&= frac{(2n-1)!!}{(2n)!!}cdot left( frac{pi^2}{2} - sum_{k=1}^n frac{(2k)!!}{(2k-1)!! , k^2} right) ,,
end{align}$$

where I would consider the last line a solution if one was able to find a closed formula for the sum that is not given in terms of another integral or another horrible function like Wolfram Alpha suggests.






Edits:




  • Moved the power closer to $cos$ to avoid confusion.

  • Corrected some constants in the identities.

  • Emphasized that $n in mathbb{N}$.










share|cite|improve this question











$endgroup$












  • $begingroup$
    Is $cos$ raised to the power of $2n$ or $t$?
    $endgroup$
    – DavidG
    Jan 3 at 3:16










  • $begingroup$
    If $n in mathbb{N}$ there are some nice closed solutions (I believe).
    $endgroup$
    – DavidG
    Jan 3 at 3:43










  • $begingroup$
    Numerical results of your identities don't hold up, I think you need $pi^{2}/2$ instead.
    $endgroup$
    – Ininterrompue
    Jan 3 at 6:34










  • $begingroup$
    @Ininterrompue, thank you! I fixed the constants now.
    $endgroup$
    – chickenNinja123
    Jan 3 at 9:18














1












1








1


0



$begingroup$


I got stuck trying to find a general formula for the following integral
$$int_0^{pi} t cdotcos^{2n}{left(frac{t}{2}right)} , dt =
4 int_0^{pi/2} t cdotcos^{2n}t ,dt ; , ; text{ for } n in mathbb{N}.$$

It doesn't seem to be listed on Wikipedias Lists of integrals. After playing around for a while I found the following potentially useful identities$$begin{align}
int_0^{pi} t cdotcos^{2n}{left(frac{t}{2}right)} , dt &=
frac{(2n-1)!!}{(2n)!!} pi^2 - 4int_{0}^{pi/2}{int_0^{t} cos^{2n}{tau} , dtau} , dt \
&= frac{(2n-1)!!}{(2n)!!}cdot left( frac{pi^2}{2} - sum_{k=1}^n frac{(2k)!!}{(2k-1)!! , k^2} right) ,,
end{align}$$

where I would consider the last line a solution if one was able to find a closed formula for the sum that is not given in terms of another integral or another horrible function like Wolfram Alpha suggests.






Edits:




  • Moved the power closer to $cos$ to avoid confusion.

  • Corrected some constants in the identities.

  • Emphasized that $n in mathbb{N}$.










share|cite|improve this question











$endgroup$




I got stuck trying to find a general formula for the following integral
$$int_0^{pi} t cdotcos^{2n}{left(frac{t}{2}right)} , dt =
4 int_0^{pi/2} t cdotcos^{2n}t ,dt ; , ; text{ for } n in mathbb{N}.$$

It doesn't seem to be listed on Wikipedias Lists of integrals. After playing around for a while I found the following potentially useful identities$$begin{align}
int_0^{pi} t cdotcos^{2n}{left(frac{t}{2}right)} , dt &=
frac{(2n-1)!!}{(2n)!!} pi^2 - 4int_{0}^{pi/2}{int_0^{t} cos^{2n}{tau} , dtau} , dt \
&= frac{(2n-1)!!}{(2n)!!}cdot left( frac{pi^2}{2} - sum_{k=1}^n frac{(2k)!!}{(2k-1)!! , k^2} right) ,,
end{align}$$

where I would consider the last line a solution if one was able to find a closed formula for the sum that is not given in terms of another integral or another horrible function like Wolfram Alpha suggests.






Edits:




  • Moved the power closer to $cos$ to avoid confusion.

  • Corrected some constants in the identities.

  • Emphasized that $n in mathbb{N}$.







integration indefinite-integrals trigonometric-integrals






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share|cite|improve this question













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edited Jan 3 at 17:38







chickenNinja123

















asked Jan 2 at 23:38









chickenNinja123chickenNinja123

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7713












  • $begingroup$
    Is $cos$ raised to the power of $2n$ or $t$?
    $endgroup$
    – DavidG
    Jan 3 at 3:16










  • $begingroup$
    If $n in mathbb{N}$ there are some nice closed solutions (I believe).
    $endgroup$
    – DavidG
    Jan 3 at 3:43










  • $begingroup$
    Numerical results of your identities don't hold up, I think you need $pi^{2}/2$ instead.
    $endgroup$
    – Ininterrompue
    Jan 3 at 6:34










  • $begingroup$
    @Ininterrompue, thank you! I fixed the constants now.
    $endgroup$
    – chickenNinja123
    Jan 3 at 9:18


















  • $begingroup$
    Is $cos$ raised to the power of $2n$ or $t$?
    $endgroup$
    – DavidG
    Jan 3 at 3:16










  • $begingroup$
    If $n in mathbb{N}$ there are some nice closed solutions (I believe).
    $endgroup$
    – DavidG
    Jan 3 at 3:43










  • $begingroup$
    Numerical results of your identities don't hold up, I think you need $pi^{2}/2$ instead.
    $endgroup$
    – Ininterrompue
    Jan 3 at 6:34










  • $begingroup$
    @Ininterrompue, thank you! I fixed the constants now.
    $endgroup$
    – chickenNinja123
    Jan 3 at 9:18
















$begingroup$
Is $cos$ raised to the power of $2n$ or $t$?
$endgroup$
– DavidG
Jan 3 at 3:16




$begingroup$
Is $cos$ raised to the power of $2n$ or $t$?
$endgroup$
– DavidG
Jan 3 at 3:16












$begingroup$
If $n in mathbb{N}$ there are some nice closed solutions (I believe).
$endgroup$
– DavidG
Jan 3 at 3:43




$begingroup$
If $n in mathbb{N}$ there are some nice closed solutions (I believe).
$endgroup$
– DavidG
Jan 3 at 3:43












$begingroup$
Numerical results of your identities don't hold up, I think you need $pi^{2}/2$ instead.
$endgroup$
– Ininterrompue
Jan 3 at 6:34




$begingroup$
Numerical results of your identities don't hold up, I think you need $pi^{2}/2$ instead.
$endgroup$
– Ininterrompue
Jan 3 at 6:34












$begingroup$
@Ininterrompue, thank you! I fixed the constants now.
$endgroup$
– chickenNinja123
Jan 3 at 9:18




$begingroup$
@Ininterrompue, thank you! I fixed the constants now.
$endgroup$
– chickenNinja123
Jan 3 at 9:18










3 Answers
3






active

oldest

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1












$begingroup$

The integral probably does not have a closed form in terms of something simpler than the hypergeometric ${}_{3}F_{2}$ unless something amazing happens with the series. I don't think you can do much better than



$$begin{aligned} I_{n} &= frac{(2n-1)!!}{(2n)!!}frac{sqrt{pi},Gamma^{2}(n+1)}{Gamma(n+3/2)Gamma(n+2)}{}_{3}F_{2}left(1, n+1, n+1; n+3/2, n+2; 1right) \
&= frac{{}_{3}F_{2}left(1, n+1, n+1; n+3/2, n+2; 1right)}{(n+1)(n+1/2)} end{aligned}$$



where I have used the double factorial identity



$$ frac{(2n-1)!!}{(2n)!!} = frac{(2n)!}{2^{2n}(n!)^{2}}$$



and Legendre's duplication formula to simplify the gamma functions and factorials. The hypergeometric function reduces to elementary functions in $x^{2}$ when $n$ is an integer; for example, Mathematica gives the $n=1$ and $n=2$ cases as



$$begin{aligned} {}_{3}F_{2}left(1, 2, 2; 5/2, 3; x^{2}right) &= frac{3}{x^{2}}left(frac{arcsin^{2}x}{x^{2}} - 1right) \
{}_{3}F_{2}left(1, 3, 3; 7/2, 4; x^{2}right) &= frac{15}{8x^{2}}left(frac{3arcsin^{2}x}{x^{4}} - frac{3}{x^{2}} - 1right). end{aligned}$$



The linear term makes things more difficult. By making the substitution $x mapsto pi/2 - x$, one can use the beta function and Legendre's duplication formula to show that the related integral has a nice form



$$begin{aligned} int_{0}^{pi/2}4xleft(cos^{2n}x + sin^{2n}xright)mathrm{d}x &= 2piint_{0}^{pi/2}sin^{2n}x,mathrm{d}x = pi,frac{Gamma(1/2)Gamma(1/2+n)}{Gamma(1+n)} \ &= frac{pi^{2}}{2^{2n}}frac{(2n)!}{(n!)^{2}} = pi^{2}frac{(2n-1)!!}{(2n)!!}. end{aligned}$$



Lastly, the tangent half-angle substitution can be used to write the integral in the form



$$ I_{n} = int_{0}^{pi/2}xcos^{2n}x,mathrm{d}x = int_{0}^{infty}frac{tan^{-1}x}{(1+x^{2})^{n+1}},mathrm{d}x $$



but I do not believe this is much easier than the original integral.






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  • $begingroup$
    If you have a closed formula for $_3F_2(1,n+1,n+1;n+3/2,n+2;1)$ for $n in mathbb{N}$ then that's exactly what I was looking for!
    $endgroup$
    – chickenNinja123
    Jan 3 at 9:29



















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NOT A FULL SOLUTION:



Here, complex analysis (if permitted) is your friend (sort of):



begin{equation}
Releft[tcdot e^{-t^n i} right] = t cdot cosleft(t^nright)
end{equation}



Thus,



begin{align}
I = int_{0}^{pi} t cdot cosleft(t^nright) = Releft[int_{0}^{pi} tcdot e^{-t^n i} :dtright]
end{align}



Here let $u = t^{n}i$ to yield:



begin{align}
I &= Releft[int_{0}^{pi} tcdot e^{-t^n i} :dtright] = Releft[int_{0}^{pi^ni} left(frac{u}{i} right)^{frac{1}{n}}cdot e^{-u} cdot frac{du}{left(frac{u}{i} right)^{frac{n - 1}{n}}}right] \
&= Releft[ i^{1 - frac{2}{n}} int_{0}^{pi^ni} u^{frac{2}{n} - 1}e^{-u}:duright] \&= Releft[i^{1 - frac{2}{n}} gammaleft(u^{frac{2}{n}}, pi^niright)right]
end{align}



Where $gamma(a,b)$ is the lower incomplete gamma function.






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  • $begingroup$
    I'm terribly sorry I didn't make it clear enough where I wanted the powers to be. :/ I do like your approach, though.
    $endgroup$
    – chickenNinja123
    Jan 3 at 9:22










  • $begingroup$
    @chickenNinja123 - No worries mate!
    $endgroup$
    – DavidG
    Jan 3 at 9:22










  • $begingroup$
    first time I see the lower gamma function defined for some non-real upper bound. I had to go to wikipedia to check it, and it is fine!
    $endgroup$
    – Masacroso
    Jan 3 at 22:41





















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One more version - in case it is useful for further simplifications.



Use
$$
cos^{2n}left(frac{t}{2}right)=frac1{2^{2 n}}binom{2 n}{n}+frac1{2^{2 n-1}}{sum _{j=1}^nbinom{2 n}{n-j} cos (jt)}
$$

and
$$
int_0^{pi} t cosleft(ntright) , dt=begin{cases}
frac{pi^2}2,&n=0,\
-frac2{(2k-1)^2},&n=2k-1,\
0,&n=2k>0
end{cases}
$$

to obtain
$$int_0^{pi} t cdotcos^{2n}{left(frac{t}{2}right)} , dt=
frac{pi^2}{2^{2 n+1}}binom{2 n}{n}-frac1{2^{2 n-2}}{sum _{k=1}^{left[frac{n+1}2right]} frac1{(2k-1)^2}binom{2 n}{n+2k-1}}.$$

Mathematica handles the last sum as
$$
frac{pi^2}{2^{2 n+1}}binom{2 n}{n}-frac1{2^{2 n-2}} binom{2 n}{n+1} , _5F_4left(frac{1}{2},frac{1}{2},1,frac{1}{2}-frac{n}{2},1-frac{n}{2};frac{3}{2},frac{3}{2},frac{n}{2}+1,frac{n}{2}+frac{3}{2};1right)
$$

(which is no better than the sum itself).






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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

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    active

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    active

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    1












    $begingroup$

    The integral probably does not have a closed form in terms of something simpler than the hypergeometric ${}_{3}F_{2}$ unless something amazing happens with the series. I don't think you can do much better than



    $$begin{aligned} I_{n} &= frac{(2n-1)!!}{(2n)!!}frac{sqrt{pi},Gamma^{2}(n+1)}{Gamma(n+3/2)Gamma(n+2)}{}_{3}F_{2}left(1, n+1, n+1; n+3/2, n+2; 1right) \
    &= frac{{}_{3}F_{2}left(1, n+1, n+1; n+3/2, n+2; 1right)}{(n+1)(n+1/2)} end{aligned}$$



    where I have used the double factorial identity



    $$ frac{(2n-1)!!}{(2n)!!} = frac{(2n)!}{2^{2n}(n!)^{2}}$$



    and Legendre's duplication formula to simplify the gamma functions and factorials. The hypergeometric function reduces to elementary functions in $x^{2}$ when $n$ is an integer; for example, Mathematica gives the $n=1$ and $n=2$ cases as



    $$begin{aligned} {}_{3}F_{2}left(1, 2, 2; 5/2, 3; x^{2}right) &= frac{3}{x^{2}}left(frac{arcsin^{2}x}{x^{2}} - 1right) \
    {}_{3}F_{2}left(1, 3, 3; 7/2, 4; x^{2}right) &= frac{15}{8x^{2}}left(frac{3arcsin^{2}x}{x^{4}} - frac{3}{x^{2}} - 1right). end{aligned}$$



    The linear term makes things more difficult. By making the substitution $x mapsto pi/2 - x$, one can use the beta function and Legendre's duplication formula to show that the related integral has a nice form



    $$begin{aligned} int_{0}^{pi/2}4xleft(cos^{2n}x + sin^{2n}xright)mathrm{d}x &= 2piint_{0}^{pi/2}sin^{2n}x,mathrm{d}x = pi,frac{Gamma(1/2)Gamma(1/2+n)}{Gamma(1+n)} \ &= frac{pi^{2}}{2^{2n}}frac{(2n)!}{(n!)^{2}} = pi^{2}frac{(2n-1)!!}{(2n)!!}. end{aligned}$$



    Lastly, the tangent half-angle substitution can be used to write the integral in the form



    $$ I_{n} = int_{0}^{pi/2}xcos^{2n}x,mathrm{d}x = int_{0}^{infty}frac{tan^{-1}x}{(1+x^{2})^{n+1}},mathrm{d}x $$



    but I do not believe this is much easier than the original integral.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      If you have a closed formula for $_3F_2(1,n+1,n+1;n+3/2,n+2;1)$ for $n in mathbb{N}$ then that's exactly what I was looking for!
      $endgroup$
      – chickenNinja123
      Jan 3 at 9:29
















    1












    $begingroup$

    The integral probably does not have a closed form in terms of something simpler than the hypergeometric ${}_{3}F_{2}$ unless something amazing happens with the series. I don't think you can do much better than



    $$begin{aligned} I_{n} &= frac{(2n-1)!!}{(2n)!!}frac{sqrt{pi},Gamma^{2}(n+1)}{Gamma(n+3/2)Gamma(n+2)}{}_{3}F_{2}left(1, n+1, n+1; n+3/2, n+2; 1right) \
    &= frac{{}_{3}F_{2}left(1, n+1, n+1; n+3/2, n+2; 1right)}{(n+1)(n+1/2)} end{aligned}$$



    where I have used the double factorial identity



    $$ frac{(2n-1)!!}{(2n)!!} = frac{(2n)!}{2^{2n}(n!)^{2}}$$



    and Legendre's duplication formula to simplify the gamma functions and factorials. The hypergeometric function reduces to elementary functions in $x^{2}$ when $n$ is an integer; for example, Mathematica gives the $n=1$ and $n=2$ cases as



    $$begin{aligned} {}_{3}F_{2}left(1, 2, 2; 5/2, 3; x^{2}right) &= frac{3}{x^{2}}left(frac{arcsin^{2}x}{x^{2}} - 1right) \
    {}_{3}F_{2}left(1, 3, 3; 7/2, 4; x^{2}right) &= frac{15}{8x^{2}}left(frac{3arcsin^{2}x}{x^{4}} - frac{3}{x^{2}} - 1right). end{aligned}$$



    The linear term makes things more difficult. By making the substitution $x mapsto pi/2 - x$, one can use the beta function and Legendre's duplication formula to show that the related integral has a nice form



    $$begin{aligned} int_{0}^{pi/2}4xleft(cos^{2n}x + sin^{2n}xright)mathrm{d}x &= 2piint_{0}^{pi/2}sin^{2n}x,mathrm{d}x = pi,frac{Gamma(1/2)Gamma(1/2+n)}{Gamma(1+n)} \ &= frac{pi^{2}}{2^{2n}}frac{(2n)!}{(n!)^{2}} = pi^{2}frac{(2n-1)!!}{(2n)!!}. end{aligned}$$



    Lastly, the tangent half-angle substitution can be used to write the integral in the form



    $$ I_{n} = int_{0}^{pi/2}xcos^{2n}x,mathrm{d}x = int_{0}^{infty}frac{tan^{-1}x}{(1+x^{2})^{n+1}},mathrm{d}x $$



    but I do not believe this is much easier than the original integral.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      If you have a closed formula for $_3F_2(1,n+1,n+1;n+3/2,n+2;1)$ for $n in mathbb{N}$ then that's exactly what I was looking for!
      $endgroup$
      – chickenNinja123
      Jan 3 at 9:29














    1












    1








    1





    $begingroup$

    The integral probably does not have a closed form in terms of something simpler than the hypergeometric ${}_{3}F_{2}$ unless something amazing happens with the series. I don't think you can do much better than



    $$begin{aligned} I_{n} &= frac{(2n-1)!!}{(2n)!!}frac{sqrt{pi},Gamma^{2}(n+1)}{Gamma(n+3/2)Gamma(n+2)}{}_{3}F_{2}left(1, n+1, n+1; n+3/2, n+2; 1right) \
    &= frac{{}_{3}F_{2}left(1, n+1, n+1; n+3/2, n+2; 1right)}{(n+1)(n+1/2)} end{aligned}$$



    where I have used the double factorial identity



    $$ frac{(2n-1)!!}{(2n)!!} = frac{(2n)!}{2^{2n}(n!)^{2}}$$



    and Legendre's duplication formula to simplify the gamma functions and factorials. The hypergeometric function reduces to elementary functions in $x^{2}$ when $n$ is an integer; for example, Mathematica gives the $n=1$ and $n=2$ cases as



    $$begin{aligned} {}_{3}F_{2}left(1, 2, 2; 5/2, 3; x^{2}right) &= frac{3}{x^{2}}left(frac{arcsin^{2}x}{x^{2}} - 1right) \
    {}_{3}F_{2}left(1, 3, 3; 7/2, 4; x^{2}right) &= frac{15}{8x^{2}}left(frac{3arcsin^{2}x}{x^{4}} - frac{3}{x^{2}} - 1right). end{aligned}$$



    The linear term makes things more difficult. By making the substitution $x mapsto pi/2 - x$, one can use the beta function and Legendre's duplication formula to show that the related integral has a nice form



    $$begin{aligned} int_{0}^{pi/2}4xleft(cos^{2n}x + sin^{2n}xright)mathrm{d}x &= 2piint_{0}^{pi/2}sin^{2n}x,mathrm{d}x = pi,frac{Gamma(1/2)Gamma(1/2+n)}{Gamma(1+n)} \ &= frac{pi^{2}}{2^{2n}}frac{(2n)!}{(n!)^{2}} = pi^{2}frac{(2n-1)!!}{(2n)!!}. end{aligned}$$



    Lastly, the tangent half-angle substitution can be used to write the integral in the form



    $$ I_{n} = int_{0}^{pi/2}xcos^{2n}x,mathrm{d}x = int_{0}^{infty}frac{tan^{-1}x}{(1+x^{2})^{n+1}},mathrm{d}x $$



    but I do not believe this is much easier than the original integral.






    share|cite|improve this answer











    $endgroup$



    The integral probably does not have a closed form in terms of something simpler than the hypergeometric ${}_{3}F_{2}$ unless something amazing happens with the series. I don't think you can do much better than



    $$begin{aligned} I_{n} &= frac{(2n-1)!!}{(2n)!!}frac{sqrt{pi},Gamma^{2}(n+1)}{Gamma(n+3/2)Gamma(n+2)}{}_{3}F_{2}left(1, n+1, n+1; n+3/2, n+2; 1right) \
    &= frac{{}_{3}F_{2}left(1, n+1, n+1; n+3/2, n+2; 1right)}{(n+1)(n+1/2)} end{aligned}$$



    where I have used the double factorial identity



    $$ frac{(2n-1)!!}{(2n)!!} = frac{(2n)!}{2^{2n}(n!)^{2}}$$



    and Legendre's duplication formula to simplify the gamma functions and factorials. The hypergeometric function reduces to elementary functions in $x^{2}$ when $n$ is an integer; for example, Mathematica gives the $n=1$ and $n=2$ cases as



    $$begin{aligned} {}_{3}F_{2}left(1, 2, 2; 5/2, 3; x^{2}right) &= frac{3}{x^{2}}left(frac{arcsin^{2}x}{x^{2}} - 1right) \
    {}_{3}F_{2}left(1, 3, 3; 7/2, 4; x^{2}right) &= frac{15}{8x^{2}}left(frac{3arcsin^{2}x}{x^{4}} - frac{3}{x^{2}} - 1right). end{aligned}$$



    The linear term makes things more difficult. By making the substitution $x mapsto pi/2 - x$, one can use the beta function and Legendre's duplication formula to show that the related integral has a nice form



    $$begin{aligned} int_{0}^{pi/2}4xleft(cos^{2n}x + sin^{2n}xright)mathrm{d}x &= 2piint_{0}^{pi/2}sin^{2n}x,mathrm{d}x = pi,frac{Gamma(1/2)Gamma(1/2+n)}{Gamma(1+n)} \ &= frac{pi^{2}}{2^{2n}}frac{(2n)!}{(n!)^{2}} = pi^{2}frac{(2n-1)!!}{(2n)!!}. end{aligned}$$



    Lastly, the tangent half-angle substitution can be used to write the integral in the form



    $$ I_{n} = int_{0}^{pi/2}xcos^{2n}x,mathrm{d}x = int_{0}^{infty}frac{tan^{-1}x}{(1+x^{2})^{n+1}},mathrm{d}x $$



    but I do not believe this is much easier than the original integral.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 3 at 7:39

























    answered Jan 3 at 7:21









    IninterrompueIninterrompue

    67519




    67519












    • $begingroup$
      If you have a closed formula for $_3F_2(1,n+1,n+1;n+3/2,n+2;1)$ for $n in mathbb{N}$ then that's exactly what I was looking for!
      $endgroup$
      – chickenNinja123
      Jan 3 at 9:29


















    • $begingroup$
      If you have a closed formula for $_3F_2(1,n+1,n+1;n+3/2,n+2;1)$ for $n in mathbb{N}$ then that's exactly what I was looking for!
      $endgroup$
      – chickenNinja123
      Jan 3 at 9:29
















    $begingroup$
    If you have a closed formula for $_3F_2(1,n+1,n+1;n+3/2,n+2;1)$ for $n in mathbb{N}$ then that's exactly what I was looking for!
    $endgroup$
    – chickenNinja123
    Jan 3 at 9:29




    $begingroup$
    If you have a closed formula for $_3F_2(1,n+1,n+1;n+3/2,n+2;1)$ for $n in mathbb{N}$ then that's exactly what I was looking for!
    $endgroup$
    – chickenNinja123
    Jan 3 at 9:29











    3












    $begingroup$

    NOT A FULL SOLUTION:



    Here, complex analysis (if permitted) is your friend (sort of):



    begin{equation}
    Releft[tcdot e^{-t^n i} right] = t cdot cosleft(t^nright)
    end{equation}



    Thus,



    begin{align}
    I = int_{0}^{pi} t cdot cosleft(t^nright) = Releft[int_{0}^{pi} tcdot e^{-t^n i} :dtright]
    end{align}



    Here let $u = t^{n}i$ to yield:



    begin{align}
    I &= Releft[int_{0}^{pi} tcdot e^{-t^n i} :dtright] = Releft[int_{0}^{pi^ni} left(frac{u}{i} right)^{frac{1}{n}}cdot e^{-u} cdot frac{du}{left(frac{u}{i} right)^{frac{n - 1}{n}}}right] \
    &= Releft[ i^{1 - frac{2}{n}} int_{0}^{pi^ni} u^{frac{2}{n} - 1}e^{-u}:duright] \&= Releft[i^{1 - frac{2}{n}} gammaleft(u^{frac{2}{n}}, pi^niright)right]
    end{align}



    Where $gamma(a,b)$ is the lower incomplete gamma function.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I'm terribly sorry I didn't make it clear enough where I wanted the powers to be. :/ I do like your approach, though.
      $endgroup$
      – chickenNinja123
      Jan 3 at 9:22










    • $begingroup$
      @chickenNinja123 - No worries mate!
      $endgroup$
      – DavidG
      Jan 3 at 9:22










    • $begingroup$
      first time I see the lower gamma function defined for some non-real upper bound. I had to go to wikipedia to check it, and it is fine!
      $endgroup$
      – Masacroso
      Jan 3 at 22:41


















    3












    $begingroup$

    NOT A FULL SOLUTION:



    Here, complex analysis (if permitted) is your friend (sort of):



    begin{equation}
    Releft[tcdot e^{-t^n i} right] = t cdot cosleft(t^nright)
    end{equation}



    Thus,



    begin{align}
    I = int_{0}^{pi} t cdot cosleft(t^nright) = Releft[int_{0}^{pi} tcdot e^{-t^n i} :dtright]
    end{align}



    Here let $u = t^{n}i$ to yield:



    begin{align}
    I &= Releft[int_{0}^{pi} tcdot e^{-t^n i} :dtright] = Releft[int_{0}^{pi^ni} left(frac{u}{i} right)^{frac{1}{n}}cdot e^{-u} cdot frac{du}{left(frac{u}{i} right)^{frac{n - 1}{n}}}right] \
    &= Releft[ i^{1 - frac{2}{n}} int_{0}^{pi^ni} u^{frac{2}{n} - 1}e^{-u}:duright] \&= Releft[i^{1 - frac{2}{n}} gammaleft(u^{frac{2}{n}}, pi^niright)right]
    end{align}



    Where $gamma(a,b)$ is the lower incomplete gamma function.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I'm terribly sorry I didn't make it clear enough where I wanted the powers to be. :/ I do like your approach, though.
      $endgroup$
      – chickenNinja123
      Jan 3 at 9:22










    • $begingroup$
      @chickenNinja123 - No worries mate!
      $endgroup$
      – DavidG
      Jan 3 at 9:22










    • $begingroup$
      first time I see the lower gamma function defined for some non-real upper bound. I had to go to wikipedia to check it, and it is fine!
      $endgroup$
      – Masacroso
      Jan 3 at 22:41
















    3












    3








    3





    $begingroup$

    NOT A FULL SOLUTION:



    Here, complex analysis (if permitted) is your friend (sort of):



    begin{equation}
    Releft[tcdot e^{-t^n i} right] = t cdot cosleft(t^nright)
    end{equation}



    Thus,



    begin{align}
    I = int_{0}^{pi} t cdot cosleft(t^nright) = Releft[int_{0}^{pi} tcdot e^{-t^n i} :dtright]
    end{align}



    Here let $u = t^{n}i$ to yield:



    begin{align}
    I &= Releft[int_{0}^{pi} tcdot e^{-t^n i} :dtright] = Releft[int_{0}^{pi^ni} left(frac{u}{i} right)^{frac{1}{n}}cdot e^{-u} cdot frac{du}{left(frac{u}{i} right)^{frac{n - 1}{n}}}right] \
    &= Releft[ i^{1 - frac{2}{n}} int_{0}^{pi^ni} u^{frac{2}{n} - 1}e^{-u}:duright] \&= Releft[i^{1 - frac{2}{n}} gammaleft(u^{frac{2}{n}}, pi^niright)right]
    end{align}



    Where $gamma(a,b)$ is the lower incomplete gamma function.






    share|cite|improve this answer









    $endgroup$



    NOT A FULL SOLUTION:



    Here, complex analysis (if permitted) is your friend (sort of):



    begin{equation}
    Releft[tcdot e^{-t^n i} right] = t cdot cosleft(t^nright)
    end{equation}



    Thus,



    begin{align}
    I = int_{0}^{pi} t cdot cosleft(t^nright) = Releft[int_{0}^{pi} tcdot e^{-t^n i} :dtright]
    end{align}



    Here let $u = t^{n}i$ to yield:



    begin{align}
    I &= Releft[int_{0}^{pi} tcdot e^{-t^n i} :dtright] = Releft[int_{0}^{pi^ni} left(frac{u}{i} right)^{frac{1}{n}}cdot e^{-u} cdot frac{du}{left(frac{u}{i} right)^{frac{n - 1}{n}}}right] \
    &= Releft[ i^{1 - frac{2}{n}} int_{0}^{pi^ni} u^{frac{2}{n} - 1}e^{-u}:duright] \&= Releft[i^{1 - frac{2}{n}} gammaleft(u^{frac{2}{n}}, pi^niright)right]
    end{align}



    Where $gamma(a,b)$ is the lower incomplete gamma function.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 3 at 3:39









    DavidGDavidG

    2,3811724




    2,3811724












    • $begingroup$
      I'm terribly sorry I didn't make it clear enough where I wanted the powers to be. :/ I do like your approach, though.
      $endgroup$
      – chickenNinja123
      Jan 3 at 9:22










    • $begingroup$
      @chickenNinja123 - No worries mate!
      $endgroup$
      – DavidG
      Jan 3 at 9:22










    • $begingroup$
      first time I see the lower gamma function defined for some non-real upper bound. I had to go to wikipedia to check it, and it is fine!
      $endgroup$
      – Masacroso
      Jan 3 at 22:41




















    • $begingroup$
      I'm terribly sorry I didn't make it clear enough where I wanted the powers to be. :/ I do like your approach, though.
      $endgroup$
      – chickenNinja123
      Jan 3 at 9:22










    • $begingroup$
      @chickenNinja123 - No worries mate!
      $endgroup$
      – DavidG
      Jan 3 at 9:22










    • $begingroup$
      first time I see the lower gamma function defined for some non-real upper bound. I had to go to wikipedia to check it, and it is fine!
      $endgroup$
      – Masacroso
      Jan 3 at 22:41


















    $begingroup$
    I'm terribly sorry I didn't make it clear enough where I wanted the powers to be. :/ I do like your approach, though.
    $endgroup$
    – chickenNinja123
    Jan 3 at 9:22




    $begingroup$
    I'm terribly sorry I didn't make it clear enough where I wanted the powers to be. :/ I do like your approach, though.
    $endgroup$
    – chickenNinja123
    Jan 3 at 9:22












    $begingroup$
    @chickenNinja123 - No worries mate!
    $endgroup$
    – DavidG
    Jan 3 at 9:22




    $begingroup$
    @chickenNinja123 - No worries mate!
    $endgroup$
    – DavidG
    Jan 3 at 9:22












    $begingroup$
    first time I see the lower gamma function defined for some non-real upper bound. I had to go to wikipedia to check it, and it is fine!
    $endgroup$
    – Masacroso
    Jan 3 at 22:41






    $begingroup$
    first time I see the lower gamma function defined for some non-real upper bound. I had to go to wikipedia to check it, and it is fine!
    $endgroup$
    – Masacroso
    Jan 3 at 22:41













    1












    $begingroup$

    One more version - in case it is useful for further simplifications.



    Use
    $$
    cos^{2n}left(frac{t}{2}right)=frac1{2^{2 n}}binom{2 n}{n}+frac1{2^{2 n-1}}{sum _{j=1}^nbinom{2 n}{n-j} cos (jt)}
    $$

    and
    $$
    int_0^{pi} t cosleft(ntright) , dt=begin{cases}
    frac{pi^2}2,&n=0,\
    -frac2{(2k-1)^2},&n=2k-1,\
    0,&n=2k>0
    end{cases}
    $$

    to obtain
    $$int_0^{pi} t cdotcos^{2n}{left(frac{t}{2}right)} , dt=
    frac{pi^2}{2^{2 n+1}}binom{2 n}{n}-frac1{2^{2 n-2}}{sum _{k=1}^{left[frac{n+1}2right]} frac1{(2k-1)^2}binom{2 n}{n+2k-1}}.$$

    Mathematica handles the last sum as
    $$
    frac{pi^2}{2^{2 n+1}}binom{2 n}{n}-frac1{2^{2 n-2}} binom{2 n}{n+1} , _5F_4left(frac{1}{2},frac{1}{2},1,frac{1}{2}-frac{n}{2},1-frac{n}{2};frac{3}{2},frac{3}{2},frac{n}{2}+1,frac{n}{2}+frac{3}{2};1right)
    $$

    (which is no better than the sum itself).






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      One more version - in case it is useful for further simplifications.



      Use
      $$
      cos^{2n}left(frac{t}{2}right)=frac1{2^{2 n}}binom{2 n}{n}+frac1{2^{2 n-1}}{sum _{j=1}^nbinom{2 n}{n-j} cos (jt)}
      $$

      and
      $$
      int_0^{pi} t cosleft(ntright) , dt=begin{cases}
      frac{pi^2}2,&n=0,\
      -frac2{(2k-1)^2},&n=2k-1,\
      0,&n=2k>0
      end{cases}
      $$

      to obtain
      $$int_0^{pi} t cdotcos^{2n}{left(frac{t}{2}right)} , dt=
      frac{pi^2}{2^{2 n+1}}binom{2 n}{n}-frac1{2^{2 n-2}}{sum _{k=1}^{left[frac{n+1}2right]} frac1{(2k-1)^2}binom{2 n}{n+2k-1}}.$$

      Mathematica handles the last sum as
      $$
      frac{pi^2}{2^{2 n+1}}binom{2 n}{n}-frac1{2^{2 n-2}} binom{2 n}{n+1} , _5F_4left(frac{1}{2},frac{1}{2},1,frac{1}{2}-frac{n}{2},1-frac{n}{2};frac{3}{2},frac{3}{2},frac{n}{2}+1,frac{n}{2}+frac{3}{2};1right)
      $$

      (which is no better than the sum itself).






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        One more version - in case it is useful for further simplifications.



        Use
        $$
        cos^{2n}left(frac{t}{2}right)=frac1{2^{2 n}}binom{2 n}{n}+frac1{2^{2 n-1}}{sum _{j=1}^nbinom{2 n}{n-j} cos (jt)}
        $$

        and
        $$
        int_0^{pi} t cosleft(ntright) , dt=begin{cases}
        frac{pi^2}2,&n=0,\
        -frac2{(2k-1)^2},&n=2k-1,\
        0,&n=2k>0
        end{cases}
        $$

        to obtain
        $$int_0^{pi} t cdotcos^{2n}{left(frac{t}{2}right)} , dt=
        frac{pi^2}{2^{2 n+1}}binom{2 n}{n}-frac1{2^{2 n-2}}{sum _{k=1}^{left[frac{n+1}2right]} frac1{(2k-1)^2}binom{2 n}{n+2k-1}}.$$

        Mathematica handles the last sum as
        $$
        frac{pi^2}{2^{2 n+1}}binom{2 n}{n}-frac1{2^{2 n-2}} binom{2 n}{n+1} , _5F_4left(frac{1}{2},frac{1}{2},1,frac{1}{2}-frac{n}{2},1-frac{n}{2};frac{3}{2},frac{3}{2},frac{n}{2}+1,frac{n}{2}+frac{3}{2};1right)
        $$

        (which is no better than the sum itself).






        share|cite|improve this answer











        $endgroup$



        One more version - in case it is useful for further simplifications.



        Use
        $$
        cos^{2n}left(frac{t}{2}right)=frac1{2^{2 n}}binom{2 n}{n}+frac1{2^{2 n-1}}{sum _{j=1}^nbinom{2 n}{n-j} cos (jt)}
        $$

        and
        $$
        int_0^{pi} t cosleft(ntright) , dt=begin{cases}
        frac{pi^2}2,&n=0,\
        -frac2{(2k-1)^2},&n=2k-1,\
        0,&n=2k>0
        end{cases}
        $$

        to obtain
        $$int_0^{pi} t cdotcos^{2n}{left(frac{t}{2}right)} , dt=
        frac{pi^2}{2^{2 n+1}}binom{2 n}{n}-frac1{2^{2 n-2}}{sum _{k=1}^{left[frac{n+1}2right]} frac1{(2k-1)^2}binom{2 n}{n+2k-1}}.$$

        Mathematica handles the last sum as
        $$
        frac{pi^2}{2^{2 n+1}}binom{2 n}{n}-frac1{2^{2 n-2}} binom{2 n}{n+1} , _5F_4left(frac{1}{2},frac{1}{2},1,frac{1}{2}-frac{n}{2},1-frac{n}{2};frac{3}{2},frac{3}{2},frac{n}{2}+1,frac{n}{2}+frac{3}{2};1right)
        $$

        (which is no better than the sum itself).







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 3 at 22:08

























        answered Jan 3 at 21:34









        მამუკა ჯიბლაძემამუკა ჯიბლაძე

        747817




        747817






























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