I am trying to evaluate the following integral to get a function in terms of $n$
$$ \int_0^{1/2} x^{2n}\ln(\sin(\pi x))dx $$
where $n\in \mathbb{N}_0$.
I have tried integration by parts, series expansions and complex number identities but nothing seems to help. Trying with small values of $n$ on wolfram has given me that the answer should be a summation involving the zeta function but not much more.
Substitute $t=\pi x$ and integrate by parts twice
\begin{align} \int_0^{1/2} x^{2n}\ln(\sin \pi x)\>dx = &\>-\frac1{\pi^{2n+1}(2n+1)(2n+2)}\int_0^{\pi/2} t^{2n+2}\csc^2 t\>dt\\ \end{align} Then, integrate
\begin{align} I_m =&\> \int_0^{\pi/2} t^{2m}\csc^2 t\>dt \overset{y=e^{i t}}= -\frac4{(-1)^{m}i }\int_1^i \frac{y\ln^{2m}y}{(1-y^2)^2} dy \\ =&\> -\frac4{(-1)^{m} i}\left( \int_0^1 \frac{iy\ln^{2m}(iy)}{(1+y^2)^2} d(iy) + \int_1^0 \frac{y\ln^{2m}y}{(1-y^2)^2} dy\right)\\ \overset{y^2\to y}=&\> \frac2{(-1)^{m} i }\left( \int_0^1 \frac{(\frac{i\pi}2+\frac12\ln y)^{2m}}{(1+y)^2} dy - \int_0^1 \frac{(\frac1{2}\ln y)^{2m}}{(1-y)^2} dy\right)\\ \end{align} and, for its real value, retain only the odd terms of the binomial expansion of $(\frac{i\pi}2+\frac12\ln y)^{2m}$, i.e. \begin{align} I_m = & \frac2{(-1)^{m} }\sum_{j=1}^m \binom{2m}{2j-1} (\frac{\pi}2)^{2(m-j)+1}\frac{ (-1)^{m-j}}{2^{2j-1} }\int_0^1 \frac{\ln^{2j-1} y}{(1+y)^2}dy\\ \end{align} Recognize $\int_0^1 \frac{\ln y}{(1+y)^2}dy=-\ln2$ and $ \int_0^1 \frac{\ln^{s} y}{(1+y)^2} dy \overset{s>1}=(-1)^s(1-\frac1{2^{s-1}})s!\zeta(s)$ to arrive at
\begin{align} I_m =&\frac{m\pi^{2m-1}}{2^{2m-2}}\ln2 -\frac{(2m)!}{2^{2m-1} }\sum_{j=2}^m \frac{ (-1)^{j} \pi^{2m-2j+1}}{(2m-2j+1)!}\left(1-\frac1{2^{2j-2}}\right)\zeta(2j-1) \end{align}
Then
\begin{align} \int_0^{1/2} x^{2n}\ln(\sin \pi x)\>dx = &\>-\frac1{\pi^{2n+1}(2n+1)(2n+2)}I_{n+1} \end{align}
Listed below are some sample results based on the general form derived above
\begin{align} & \int_0^{1/2} \ln(\sin \pi x)\>dx =-\frac1{2\pi}I_1 = -\frac12\ln2\\ & \int_0^{1/2} x^2 \ln(\sin \pi x)\>dx =-\frac1{12\pi^3}I_2 = -\frac1{24}\ln2+\frac3{16\pi^2}\zeta(3)\\ & \int_0^{1/2} x^4 \ln(\sin \pi x)\>dx =-\frac1{30\pi^5}I_3 = -\frac1{160}\ln2+\frac3{32\pi^2}\zeta(3)- \frac{45}{64\pi^4}\zeta(5)\\ & \int_0^{1/2} x^6 \ln(\sin \pi x)\>dx =-\frac1{56\pi^7}I_4 = \>\cdots \cdots \end{align}
$$\mathrm{I=\int_0^\frac\pi 2 x^{2k} ln(sin(x))dx=\int_0^\frac \pi 2\sum_{n=1}^\infty\frac{x^{2k}(-1)^n(sin(x)-1)^n}{n}dx=\sum_{n=1}^\infty\sum_{m=0}^n\binom nm \frac{(-1)^m}{n}\int_0^\frac12 x^{2k} sin^m(x\pi)=\sum_{n=1}^\infty\sum_{m=0}^n\binom nm \frac{(-1)^m}{n (2i)^m} \sum_{y=0}^m\binom my \int_0^\frac12 x^{2k} e^{i\pi x(2y-m)}=\quad \sum_{n=1}^\infty\sum_{m=0}^n \sum_{y=0}^m\binom nm \binom my \left(\frac i2\right)^m\frac{γ\left(2k+1,\frac12\pi i (m-2y)\right)}{n\big ((m-2y)\pi i\big)^{2k+1}}}$$
I will work on this more as I am human and can make a small typo. Here is a nice series I will maybe use to find the solution. This will unfortunately use a large summation expansion for the answer using the binomial theorem and the complex sine definition. Finally, one can rewrite these in terms of a lower incomplete gamma function. Please tell me if you have feedback and correct me!
Hint
$$I_n=\int_0^{\frac12} x^{2n}\log\big[\sin(\pi x)\big]\,dx$$ Using one integration by parts leads to $$I_n=-\int_0^{\frac12}\frac{\pi x^{2 n+1} \cot (\pi x)}{2 n+1}\,dx=-\int_0^{\frac12}\frac{ x^{2 n}} {2 n+1}\, \big[\pi x\cot (\pi x)\big]$$
Using the infinite product representation of the sine function (have a look here) $$\big[\pi x\cot (\pi x)\big]=1-2\sum^\infty_{k=1}\zeta(2k)\,x^{2k}$$
