Htykuut

Evaluate

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

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+9zeta left ( 3 right )log^42-frac{31}{30}pi ^{4}log^32\
&-left [ 72mathrm{Li}_5left ( frac{1}{2} right )-frac{9}{8}zeta left ( 5 right )-frac{51}{4}pi ^{2}zeta left ( 3 right ) right ]log^22\
&+left [ 72mathrm{Li}_{5,1}left ( frac{1}{2} right )-216mathrm{Li}_6left ( frac{1}{2} right )+36pi ^{2}mathrm{Li}_4left ( frac{1}{2} right ) right ]log2+72mathrm{Li}_{6,1}left ( frac{1}{2} right )\
&-216mathrm{Li}_7left ( frac{1}{2} right )+36pi ^{2}mathrm{Li}_5left ( 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
$$mathrm{Ls}_n^{left ( k right )}left ( alpha right ):=-int_{0}^{alpha }theta ^{k}log^{n-1-k}left | 2sinfrac{theta }{2} right |mathrm{d}theta $$
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
$$int_{0}^{frac{pi }{2}}t^{2n}log^{m}left ( 2cos t right )mathrm{d}t $$
Let's consider
$$mathcal{I}left ( x,y right )=int_{0}^{frac{pi }{2}}cosleft ( xt right )left ( 2cos t right )^{y}mathrm{d}t$$
By using Gamma function,the integral become
$$mathcal{I}left ( x,y right )=frac{pi , Gamma left ( y+1 right )}{2Gamma left ( dfrac{x+y+2}{2} right )Gamma left ( dfrac{y-x+2}{2} right )}$$
Then we can get
$$mathcal{I}left ( x,y right )=frac{pi }{2}expleft ( sum_{k=2}^{infty }frac{left ( -1 right )^{k}}{kcdot 2^{k}}zeta left ( k right )left [ left ( 2y right )^{k}-left ( y-x right )^{k}-left ( x+y right )^{k} right ] right )$$
On the other hand,using taylor series
$$mathcal{I}left ( x,y right )=sum_{n=0}^{infty }frac{left ( -1 right )^{n}}{left ( 2n right )!}x^{2n}sum_{m=0}^{infty }frac{y^{m}}{m!}int_{0}^{frac{pi }{2}}t^{2n}log^mleft ( 2cos t right )mathrm{d}t$$
So,the comparison of coefficient shows the answer.For example
$$int_{0}^{frac{pi }{2}}t^{2}log^2left ( 2cos t right )mathrm{d}t=4cdot frac{pi }{2}left [ frac{12}{4cdot 16} zeta left ( 4 right )+frac{1}{2}frac{8}{2^{2}cdot 4^{2}}zeta left ( 2 right )^{2}right ]=frac{11}{1440}pi ^{5}$$

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

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 :

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

and

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

which are convergent for the integer values $enspace mgeq 2$ .

For $enspace n=0enspace$ we have $enspacezeta_0(m)=zeta(m)enspace$ and $enspaceeta_0(m)=eta(m)enspace$ .

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 $enspacedisplaystyle w(n,m):=frac{m!}{(n-1)!}left[ begin{array}{c} n \ {m+1} end{array} right]enspace$ and it's recursion formula.

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

$ displaystyle intlimits_0^pi x^n left(2sinfrac{x}{2}right)^z dx=i^{-z} intlimits_0^pi x^n e^{ifrac{xz}{2}}(1- e^{-ix})^z dx= e^{-ifrac{pi z}{2}} intlimits_0^pi x^n sumlimits_{k=0}^inftybinom{z}{k}(-1)^k e^{-ix(frac{z}{2}-k)} dx$

$displaystyle =intlimits_0^pi x^n e^{i(x-pi)frac{z}{2}} dx+ sumlimits_{v=0}^n frac{(-1)^vpi^{n-v} n!}{i^{v+1}(n-v)!} sumlimits_{k=1}^infty binom{z}{k}frac{1}{(frac{z}{2}-k)^{v+1}} $

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

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

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

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

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

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

$displaystyle intlimits_0^pi x^3 left(lnleft(2sinfrac{x}{2} right)right)^3 dx =$

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

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

Note:

For the calculations I have used $enspacedisplaystyleintlimits_0^pi x^n e^{iax}dx = frac{(-1)^{n+1} n!}{(ia)^{n+1}}+e^{ipi a}sumlimits_{v=0}^nfrac{(-1)^v pi^{n-v}n!}{(ia)^{v+1}(n-v)!}$

with $enspacedisplaystyle a=-(frac{z}{2}-k)$ .

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

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
$$-sum_{n,kge 0}Ls^{(k)}_{n+k+1}(pi)dfrac{lambda ^n}{n!}cdotdfrac{imu)^k}{k!}=isum_{nge 0}(-1)^nbinom{lambda}{n}dfrac{e^{ipifrac{lambda}{2}}-(-1)^ne^{ipimu}}{mu-dfrac{lambda}{2}+n}$$
then
$$int_{0}^{pi}theta^3log^3{left(2sin{dfrac{theta}{2}}right)}dtheta=-Ls_{7}^{(3)}=dfrac{d^3}{dmu^3}dfrac{d^3}{dlambda^3}sum_{nge 0}binom{n}{lambda}dfrac{(-1)^ne^{ipifrac{lambda}{2}}-e^{ipimu}}{mu-dfrac{lambda}{2}+n}=6pi^2lambda_{5}left(dfrac{1}{2}right)+36Li_{5,1,1}(-1)-pi^4zeta{(3)}-dfrac{759}{32}pi^2zeta{(5)}-dfrac{45}{32}zeta{(7)}$$

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

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

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

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

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.

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