I got stuck trying to find a general formula for the following integral $$\int_0^{\pi} t \cdot\cos^{2n}{\left(\frac{t}{2}\right)} \, dt = 4 \int_0^{\pi/2} t \cdot\cos^{2n}t \,dt \; , \; \text{ for } n \in \mathbb{N}.$$ 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$$\begin{align} \int_0^{\pi} t \cdot\cos^{2n}{\left(\frac{t}{2}\right)} \, dt &= \frac{(2n-1)!!}{(2n)!!} \pi^2 - 4\int_{0}^{\pi/2}{\int_0^{t} \cos^{2n}{\tau} \, d\tau} \, dt \\ &= \frac{(2n-1)!!}{(2n)!!}\cdot \left( \frac{\pi^2}{2} - \sum_{k=1}^n \frac{(2k)!!}{(2k-1)!! \, k^2} \right) \,, \end{align}$$ 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.


  • Moved the power closer to $\cos$ to avoid confusion.
  • Corrected some constants in the identities.
  • Emphasized that $n \in \mathbb{N}$.
  • $\begingroup$ Is $\cos$ raised to the power of $2n$ or $t$? $\endgroup$
    – user150203
    Jan 3, 2019 at 3:16
  • $\begingroup$ If $n \in \mathbb{N}$ there are some nice closed solutions (I believe). $\endgroup$
    – user150203
    Jan 3, 2019 at 3:43
  • $\begingroup$ Numerical results of your identities don't hold up, I think you need $\pi^{2}/2$ instead. $\endgroup$ Jan 3, 2019 at 6:34
  • $\begingroup$ @Ininterrompue, thank you! I fixed the constants now. $\endgroup$ Jan 3, 2019 at 9:18

3 Answers 3



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

\begin{equation} \Re\left[t\cdot e^{-t^n i} \right] = t \cdot \cos\left(t^n\right) \end{equation}


\begin{align} I = \int_{0}^{\pi} t \cdot \cos\left(t^n\right) = \Re\left[\int_{0}^{\pi} t\cdot e^{-t^n i} \:dt\right] \end{align}

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

\begin{align} I &= \Re\left[\int_{0}^{\pi} t\cdot e^{-t^n i} \:dt\right] = \Re\left[\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] \\ &= \Re\left[ i^{1 - \frac{2}{n}} \int_{0}^{\pi^ni} u^{\frac{2}{n} - 1}e^{-u}\:du\right] \\&= \Re\left[i^{1 - \frac{2}{n}} \gamma\left(u^{\frac{2}{n}}, \pi^ni\right)\right] \end{align}

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

  • $\begingroup$ I'm terribly sorry I didn't make it clear enough where I wanted the powers to be. :/ I do like your approach, though. $\endgroup$ Jan 3, 2019 at 9:22
  • $\begingroup$ @chickenNinja123 - No worries mate! $\endgroup$
    – user150203
    Jan 3, 2019 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, 2019 at 22:41

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

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

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

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

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

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

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

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

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

  • $\begingroup$ 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! $\endgroup$ Jan 3, 2019 at 9:29

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

Use $$ \cos^{2n}\left(\frac{t}{2}\right)=\frac1{2^{2 n}}\binom{2 n}{n}+\frac1{2^{2 n-1}}{\sum _{j=1}^n\binom{2 n}{n-j} \cos (jt)} $$ and $$ \int_0^{\pi} t \cos\left(nt\right) \, dt=\begin{cases} \frac{\pi^2}2,&n=0,\\ -\frac2{(2k-1)^2},&n=2k-1,\\ 0,&n=2k>0 \end{cases} $$ to obtain $$\int_0^{\pi} t \cdot\cos^{2n}{\left(\frac{t}{2}\right)} \, dt= \frac{\pi^2}{2^{2 n+1}}\binom{2 n}{n}-\frac1{2^{2 n-2}}{\sum _{k=1}^{\left[\frac{n+1}2\right]} \frac1{(2k-1)^2}\binom{2 n}{n+2k-1}}.$$ Mathematica handles the last sum as $$ \frac{\pi^2}{2^{2 n+1}}\binom{2 n}{n}-\frac1{2^{2 n-2}} \binom{2 n}{n+1} \, _5F_4\left(\frac{1}{2},\frac{1}{2},1,\frac{1}{2}-\frac{n}{2},1-\frac{n}{2};\frac{3}{2},\frac{3}{2},\frac{n}{2}+1,\frac{n}{2}+\frac{3}{2};1\right) $$ (which is no better than the sum itself).


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