Evaluate $\int_{0}^{\pi }\theta ^{3}\log^{3}\left ( 2\sin\frac{\theta }{2} \right )\mathrm{d}\theta $

Question

Evaluate

$$\int_{0}^{\pi }\theta ^{3}\log^{3}\left ( 2\sin\frac{\theta }{2} \right )\,\mathrm{d}\theta $$

Several days ago,I found this interesting integral from a paper about generalized log-sine integrals,but I can't remember the title of it. The answer of the integral is

\begin{align*} -\mathrm{Ls}_{7}^{\left ( 3 \right )}\left ( \pi \right)&=\frac{9}{35}\log^72+\frac{4}{5}\pi ^{2} \log^52+9\zeta \left ( 3 \right )\log^42-\frac{31}{30}\pi ^{4}\log^32\\ &-\left [ 72\mathrm{Li}_5\left ( \frac{1}{2} \right )-\frac{9}{8}\zeta \left ( 5 \right )-\frac{51}{4}\pi ^{2}\zeta \left ( 3 \right ) \right ]\log^22\\ &+\left [ 72\mathrm{Li}_{5,1}\left ( \frac{1}{2} \right )-216\mathrm{Li}_6\left ( \frac{1}{2} \right )+36\pi ^{2}\mathrm{Li}_4\left ( \frac{1}{2} \right ) \right ]\log2+72\mathrm{Li}_{6,1}\left ( \frac{1}{2} \right )\\ &-216\mathrm{Li}_7\left ( \frac{1}{2} \right )+36\pi ^{2}\mathrm{Li}_5\left ( \frac{1}{2} \right )-\frac{1161}{32}\zeta \left ( 7 \right )-\frac{375}{32}\pi ^{2}\zeta \left ( 5 \right )+\frac{1}{10}\pi ^{4}\zeta \left ( 3 \right ) \end{align*} where $$\mathrm{Ls}_n^{\left ( k \right )}\left ( \alpha \right ):=-\int_{0}^{\alpha }\theta ^{k}\log^{n-1-k}\left | 2\sin\frac{\theta }{2} \right |\mathrm{d}\theta $$ is the generalized log-sine integral and $$\mathrm{Li}_{\lambda ,1}\left ( z \right )=\sum_{k=1}^{\infty }\frac{z^{k}}{k^{\lambda }}\sum_{j=1}^{k-1}\frac{1}{j}$$ is the multiple polylogarithm.

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I found a beautiful way to solve the integrals below $$\int_{0}^{\frac{\pi }{2}}t^{2n}\log^{m}\left ( 2\cos t \right )\mathrm{d}t $$ Let's consider $$\mathcal{I}\left ( x,y \right )=\int_{0}^{\frac{\pi }{2}}\cos\left ( xt \right )\left ( 2\cos t \right )^{y}\mathrm{d}t$$ By using Gamma function,the integral become $$\mathcal{I}\left ( x,y \right )=\frac{\pi \, \Gamma \left ( y+1 \right )}{2\Gamma \left ( \dfrac{x+y+2}{2} \right )\Gamma \left ( \dfrac{y-x+2}{2} \right )}$$ Then we can get $$\mathcal{I}\left ( x,y \right )=\frac{\pi }{2}\exp\left ( \sum_{k=2}^{\infty }\frac{\left ( -1 \right )^{k}}{k\cdot 2^{k}}\zeta \left ( k \right )\left [ \left ( 2y \right )^{k}-\left ( y-x \right )^{k}-\left ( x+y \right )^{k} \right ] \right )$$ On the other hand,using taylor series $$\mathcal{I}\left ( x,y \right )=\sum_{n=0}^{\infty }\frac{\left ( -1 \right )^{n}}{\left ( 2n \right )!}x^{2n}\sum_{m=0}^{\infty }\frac{y^{m}}{m!}\int_{0}^{\frac{\pi }{2}}t^{2n}\log^m\left ( 2\cos t \right )\mathrm{d}t$$ So,the comparison of coefficient shows the answer.For example $$\int_{0}^{\frac{\pi }{2}}t^{2}\log^2\left ( 2\cos t \right )\mathrm{d}t=4\cdot \frac{\pi }{2}\left [ \frac{12}{4\cdot 16} \zeta \left ( 4 \right )+\frac{1}{2}\frac{8}{2^{2}\cdot 4^{2}}\zeta \left ( 2 \right )^{2}\right ]=\frac{11}{1440}\pi ^{5}$$

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I wonder can we use the same way to prove the integral in the beginning,if not,is there another way to handle it?

Answer 1

First I want to define with the Stirling numbers of the first kind $\left[ \begin{array}{c} n \\ k \end{array} \right]$ a special generalization of the Riemann Zeta function :

$$\zeta_n(m):=\sum\limits_{k=1}^\infty \frac{1}{k^m}\left(\frac{n!}{(k-1)!}\left[\begin{array}{c} k \\ n+1 \end{array} \right]\right)$$

and

$$\eta_n(m):=\sum\limits_{k=1}^\infty \frac{(-1)^{k-1}}{k^m}\left(\frac{n!}{(k-1)!}\left[\begin{array}{c} k \\ n+1 \end{array} \right]\right)$$

which are convergent for the integer values $\enspace m\geq 2$ .

For $\enspace n=0\enspace$ we have $\enspace\zeta_0(m)=\zeta(m)\enspace$ and $\enspace\eta_0(m)=\eta(m)\enspace$ .

Note: Obviously (because of the other results) these series can be expressed by sums of the polylogarithm function and modifications of that.

Please also see here, part Expansion by harmonic numbers, with $\enspace\displaystyle w(n,m):=\frac{m!}{(n-1)!}\left[ \begin{array}{c} n \\ {m+1} \end{array} \right]\enspace$ and it's recursion formula.

Secondly, an extension of an integral as a series, $n\in\mathbb{N}_0$ and $z\in\mathbb{R}\setminus \{2\mathbb{N}\}$ and $nz>-1$:

$ \displaystyle \int\limits_0^\pi x^n \left(2\sin\frac{x}{2}\right)^z dx=i^{-z} \int\limits_0^\pi x^n e^{i\frac{xz}{2}}(1- e^{-ix})^z dx= e^{-i\frac{\pi z}{2}} \int\limits_0^\pi x^n \sum\limits_{k=0}^\infty\binom{z}{k}(-1)^k e^{-ix(\frac{z}{2}-k)} dx$

$\displaystyle =\int\limits_0^\pi x^n e^{i(x-\pi)\frac{z}{2}} dx+ \sum\limits_{v=0}^n \frac{(-1)^v\pi^{n-v} n!}{i^{v+1}(n-v)!} \sum\limits_{k=1}^\infty \binom{z}{k}\frac{1}{(\frac{z}{2}-k)^{v+1}} $

$\displaystyle \hspace{3.5cm} -i^{n-1}n!e^{-i\frac{\pi z}{2}} \sum\limits_{k=1}^\infty \binom{z}{k}\frac{ (-1)^k}{(\frac{z}{2}-k)^{n+1}}$

using the main branch of the logarithm and therefore $\displaystyle i=e^{i\frac{\pi}{2}}$ .

The Stirling numbers of the first kind are usually defined by $\enspace \displaystyle \sum\limits_{k=0}^n \left[ \begin{array}{c} n \\ k \end{array} \right] x^k := x(x+1)…(x+n-1) $ .

Because of $\enspace \displaystyle (\sum\limits_{v=0}^\infty x^v \frac{d^k}{dz^k}\binom{z}{v}) |_{z=0} =\frac{d^k}{dz^k}(1+x)^z |_{z=0} =(\ln(1+x))^k=k!\sum\limits_{v=k}^\infty (-1)^{v-k} \left[\begin{array}{c} v \\ k \end{array} \right] \frac{x^v}{v!}$

we get $\enspace \displaystyle \binom{z}{k}|_{z=0}=0^k\enspace$ , $\enspace \displaystyle \frac{d}{dz} \binom{z}{k} |_{z=0} = (-1)^{k-1} \left[\begin{array}{c} k \\ 1 \end{array} \right] \frac{1}{k!}= \frac{(-1)^{k-1}}{k} \enspace$ , $\enspace \displaystyle \frac{d^2}{dz^2} \binom{z}{k} |_{z=0} = (-1)^{k-2} \left[\begin{array}{c} k \\ 2 \end{array} \right] \frac{2!}{k!}= \frac{(-1)^k 2}{k}\sum\limits_{j=1}^{k-1}\frac{1}{j} \enspace$ and $\enspace \displaystyle \frac{d^3}{dz^3} \binom{z}{k} |_{z=0} = (-1)^{k-3} \left[\begin{array}{c} k \\ 3 \end{array} \right] \frac{3!}{k!}= \frac{(-1)^{k-1} 3}{k}( (\sum\limits_{j=1}^{k-1}\frac{1}{j})^2 - \sum\limits_{j=1}^{k-1}\frac{1}{j^2} ) $ .

For $(n;k):=(3;3)$ follows

$\displaystyle \int\limits_0^\pi x^3 \left(\ln\left(2\sin\frac{x}{2} \right)\right)^3 dx =$

$\hspace{2cm}\displaystyle =\frac{9\pi^2}{2}\left(\zeta(5)+3\eta(5)-4\eta_1(4)+2\eta_2(3)\right) $

$\hspace{2.5cm}\displaystyle - 90\left(\zeta(7)+\eta(7)\right) +72\left(\zeta_1(6)+\eta_1(6)\right) - 18\left(\zeta_2(5)+\eta_2(5)\right) $

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Note:

For the calculations I have used $\enspace\displaystyle\int\limits_0^\pi x^n e^{iax}dx = \frac{(-1)^{n+1} n!}{(ia)^{n+1}}+e^{i\pi a}\sum\limits_{v=0}^n\frac{(-1)^v \pi^{n-v}n!}{(ia)^{v+1}(n-v)!}$

with $\enspace\displaystyle a=-(\frac{z}{2}-k)$ .

And it was necessary to calculate $\enspace\displaystyle\frac{d^m}{dz^m} \binom{z}{k}\frac{1}{(\frac{z}{2}-k)^{v+1}}|_{z=0}\enspace$ and $\enspace\displaystyle\frac{d^m}{dz^m} e^{-i\frac{\pi z}{2}}\binom{z}{k}\frac{1}{(\frac{z}{2}-k)^{n+1}}|_{z=0}\enspace$ for $\enspace m\in\{0,1,2,3\}$ .

Answer 2

I think you can apply the method only partially for the integral

\begin{align} \int_{0}^{\pi }\theta ^{3}\log^{3}\left ( 2\sin\frac{\theta }{2} \right )\,\mathrm{d}\theta &= 2^4\int_{0}^{\pi /2 }\left(\frac{\pi}{2}-\theta \right)^{3}\log^{3}\left ( 2\cos\theta\right )\,\mathrm{d}\theta\\ &=2π^3\int_{0}^{\pi /2 }\log^{3}\left ( 2\cos\theta\right )\,\mathrm{d}\theta - 12 π^2\int_{0}^{\pi /2 } θ\log^{3}\left ( 2\cos\theta\right )\,\mathrm{d}\theta \\&+ 24 π\int_{0}^{\pi /2 } θ^2\log^{3}\left ( 2\cos\theta\right )\,\mathrm{d}\theta - 16\int_{0}^{\pi /2 }θ^3\log^{3}\left ( 2\cos\theta\right )\,\mathrm{d}\theta \end{align}

Where for even powers of $\theta$ you can use your formula. The other integrals are not trivial.

Note that the approach you suggested is driven by the fact that if

$$\mathcal{I}\left ( x,y \right )=\int_{0}^{\frac{\pi }{2}}\cos\left ( x \theta \right )\left ( 2\cos \theta \right )^{y}\mathrm{d}\theta$$

Then we can solve the integral by differentiation with respect to both $x$ and $y$ but since we cannot get rid of $\sin(x \theta)$ we can apply the derivative even number of times.

$$\frac{\partial^{2n}\partial ^m}{\partial x^{2n}\partial y^m} \mathcal{I}\left ( 0,0 \right )=(-1)^n\int_{0}^{\frac{\pi }{2}}\theta^{2n} \log^m\left ( 2\cos \theta \right )\mathrm{d}\theta$$

Answer 3

I think you can get from the paper( Jonathan M. Borwein and
and Armin 2013） reslut Log-sine evaluations of Mahler measures Theorem 2.6 use this identity $$-\sum_{n,k\ge 0}Ls^{(k)}_{n+k+1}(\pi)\dfrac{\lambda ^n}{n!}\cdot\dfrac{i\mu)^k}{k!}=i\sum_{n\ge 0}(-1)^n\binom{\lambda}{n}\dfrac{e^{i\pi\frac{\lambda}{2}}-(-1)^ne^{i\pi\mu}}{\mu-\dfrac{\lambda}{2}+n}$$ then $$\int_{0}^{\pi}\theta^3\log^3{\left(2\sin{\dfrac{\theta}{2}}\right)}d\theta=-Ls_{7}^{(3)}=\dfrac{d^3}{d\mu^3}\dfrac{d^3}{d\lambda^3}\sum_{n\ge 0}\binom{n}{\lambda}\dfrac{(-1)^ne^{i\pi\frac{\lambda}{2}}-e^{i\pi\mu}}{\mu-\dfrac{\lambda}{2}+n}=6\pi^2\lambda_{5}\left(\dfrac{1}{2}\right)+36Li_{5,1,1}(-1)-\pi^4\zeta{(3)}-\dfrac{759}{32}\pi^2\zeta{(5)}-\dfrac{45}{32}\zeta{(7)}$$

where $$\lambda_{n}(x)=(n-2)!\sum_{k=0}^{n-2}\dfrac{(-1)^k}{k!}Li_{n-k}(x)\log^k|x|+\dfrac{(-1)^n}{n}\log^{n}|x|$$

Answer 4

In general, the log sine integral for general $a,b$,

$$\int_0^{\pi}x^a\ln^b\left(2\sin\tfrac{x}2\right) dx$$

and at the special upper limit $\sigma = \pi$, can be concisely expressed in terms of the Nielsen generalized polylogarithm,

$$S_{n,p}(z) = \frac{(-1)^{n+p-1}}{(n-1)!\,p!}\int_0^1\frac{(\ln t)^{n-1}\big(\ln(1-z\,t)\big)^p}{t}dt$$

This reduces to the usual polylogarithm when $p=1$,

$$S_{n-1,1}(z)=\mathrm{Li}_n(z)$$

For brevity, since we will use only $\color{blue}{z=-1}$, let,

$$S_{n,p}(-1) = S_{n,p}$$

The OP's integral then has the compact form of only $6$ terms,

$$\large{\frac1{18}\int_0^{\pi}x^3\ln^3\left(2\sin\tfrac{x}2\right)dx \\=-10S_{5,2}+14S_{4,3}-8S_{3,4}+\frac{\pi^2}6\Big(4S_{3,2}-9S_{2,3}+6S_{1,4}\Big)\\ =\, 0.3341049\dots}$$

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P.S. Note also that,

$$32S_{3,2}(-1) = 16\zeta(2)\zeta(3)-29\zeta(5)$$

$$32S_{2,3}(-1) = 16\zeta(2)\zeta(3)-31\zeta(5)+64S_{1,4}(-1)$$

$$128S_{5,2}(-1) = 64\zeta(2)\zeta(5)+112\zeta(3)\zeta(4)-251\zeta(7)$$

$$30S_{1,4}(-1) = -\ln^4(2)\, \rm{Li}_1(\tfrac12)-5\ln^3(2)\, \rm{Li}_2(\tfrac12)-15\ln^2(2) \,\rm{Li}_3(\tfrac12)\\ -30 \ln(2)\, \rm{Li}_4(\tfrac12)-30\rm{Li}_5(\tfrac12)+30\zeta(5)$$

Perhaps $S_{3,4}(-1)$ and $S_{4,3}(-1)$ can also be similarly expressed though it is not sure. If so, then the OP's integral can be evaluated without exotic functions by using only the polylogarithm $\mathrm{Li}_n(z)$, since $\ln(z)$ and $\zeta(n)$ are just special cases of it.