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

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.

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 $$\int_{0}^{\frac{\pi }{2}}t^{2}\log^2\left ( 2\cos t \right )\mathrm{d}t=4\cdot \frac{\pi }{2}\left [ \frac{12}{4\cdot 16} \zeta \left ( 4 \right )+\frac{1}{2}\frac{8}{2^{2}\cdot 4^{2}}\zeta \left ( 2 \right )^{2}\right ]=\frac{11}{1440}\pi ^{5}$$

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

• I recommend $\int_{0}^{\frac{\pi}{2} }$, because interior of log become minus, and I just found only that it can be transformed $\theta ^{3}\log^{3}\left ( 1+\cos\theta \right )\mathrm{d}\theta$. – Takahiro Waki Jan 7 '17 at 10:44
• Wolfy agrees with your $\int t^2...$ expression. – marty cohen Jan 9 '17 at 5:35
• When you edit lots of questions with only minor changes like "d" to "mathrm{d}", it pushes all those old questions to the front page. People who come to the site looking to see what the new questions are, have to wade through all these old questions. 3 or 4 edits per day would be about right. And changing "d" to "mathrm{d}" is only a stylistic quirk; it doesn't improve the post. – B. Goddard Jan 9 '17 at 12:41
• @B. Goddard sorry about that ,i won't do that again.. – Renascence_5. Jan 9 '17 at 12:44
• For your example see: On an Intriguing Integral and Some Series Related to ζ(4), DAVID BORWEIN AND JONATHAN M. BORWEIN and, An elementary evaluation of an intriguing integral via Fourier series TUO YEONG LEE and WANDI LIN – FDP May 21 at 8:14

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

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

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

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

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 $$\enspace m\in\{0,1,2,3\}$$ .

• @Renascence_5. : Thanks! - But to get your formula we have to know something about the relation between $\zeta_1$, $\zeta_2$ and the polylogarithm (which is an additional problem). – user90369 Jan 12 '17 at 8:54
• What is $\zeta_n$ ? – Zaid Alyafeai Jan 12 '17 at 10:39
• @Zaid Alyafeai : That's defined in my answer above for $n=1$ and $n=2$ . – user90369 Jan 12 '17 at 10:49
• What are they called ? It seems like related to Euler sums also what is the general formula ? – Zaid Alyafeai Jan 12 '17 at 11:01
• The relation to polylogarithms could be done using $$\sum_{k=1}^\infty H^{(p)}_k x^k = \frac{\mathrm{Li}_p(x)}{1-x}$$ – Zaid Alyafeai Jan 12 '17 at 11:07

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.

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

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

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,

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

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.