TITLE: Evaluate the closed form $\int_{0}^{\pi/2}\sin^k(x)\ln\left[\sin(x)\sin^2\left({x\over 2}\right)\right]\mathrm dx=F(k)$
QUESTION [7 upvotes]: Proposed:

$$\int_{0}^{\pi/2}\sin^k(x)\ln\left[\sin(x)\sin^2\left({x\over 2}\right)\right]\mathrm dx=F(k)\tag1$$

Where $k=1,3,5,...$
Where $F(1)=-2$, $F(3)=-{17\over 18}$, $F(5)=-{587\over 900}$, $... $
It looks that $F(k)$ only yield rational numbers.
My try: 
Split out the integral
$$\int_{0}^{\pi/2}\sin^k(x)\ln\sin(x)+2\int_{0}^{\pi/2}\sin^k(x)\ln\left[\sin\left({x\over 2}\right)\right]\mathrm dx\tag2$$
How may one find the closed form for $F(k)?$

REPLY [4 votes]: Let us try to achieve the required result, if possible using elementary approaches.
At first,
$$F(k) = \int\limits_{0}^{\pi/2}\sin^kx\ln\left({1\over2}\sin x(1-\cos x)\right)\,\mathrm dx = \int\limits_{0}^{\pi/2}\sin^kx\ln\left({\sin x\over2} - {\sin 2x\over 4}\right)\,\mathrm dx.$$
$\mathbf{k=1}$
$$F(1) = \int\limits_{0}^{\pi/2}\sin x\ln\left({1\over2}\sin x(1-\cos x)\right)\,\mathrm dx = \int\limits_{0}^{\pi/2}\ln\left({\sin x\over2} - {\sin 2x\over 4}\right)\,\mathrm d(1-\cos x).$$
By parts:
$$ F(1) = (1-\cos x)\left.\ln\left({1\over2}\sin x(1-\cos x)\right)\right|_{0}^{\pi/2} - \int\limits_{0}^{\pi/2}(1-\cos x)\,{{1\over2}(\cos x - \cos2x)\over{{1\over2}\sin x(1-\cos x)}}\,\mathrm d x$$
$$ = -\ln2 - \int\limits_{0}^{\pi/2}{\cos x - 1 + 2\sin^2 x\over{\sin x}}\mathrm dx = -\ln2 - 2\int\limits_{0}^{\pi/2}\sin x\mathrm dx + \int\limits_{0}^{\pi/2}{1-\cos x\over1-\cos^2x}\sin x\,\mathrm dx$$
$$ = -\ln2 + 2\cos x\biggr|_0^{\pi/2} - \int\limits_{0}^{\pi/2}{d(1+\cos x) \over1+\cos x}\,\mathrm dx = -\ln2 - 2 - \ln(1+\cos x)\biggr|_0^{\pi/2} = -2,$$
$$\boxed{F(1) = -2}.$$
$\mathbf{k=3}$
$$\int\limits_0^x\sin^3y\,\mathrm dy = \int\limits_0^x(1-\cos^2y)\sin y\,\mathrm dy$$
$$ = \int\limits_0^x\left(2(1-\cos y)-(1-\cos y)^2\right)\,\mathrm d(1-\cos y)  = \int\limits_0^{1-\cos x}(2t-t^2)\,\mathrm dy = J_3(1-\cos x),$$
where
$$J_3(t) = t^2-{1\over3}t^3.$$
Then
$$F(3) = \int\limits_{0}^{\pi/2}\sin^3x\ln\left({1\over2}\sin x(1-\cos x)\right)\,\mathrm dx = \int\limits_{0}^{\pi/2}\ln\left({\sin x\over2} - {\sin 2x\over 4}\right)\,\mathrm dJ_3(1-\cos x).$$
By parts:
$$ F(3) = J_3(1-\cos x)\left.\ln\left({1\over2}\sin x(1-\cos x)\right)\right|_{0}^{\pi/2} - \int\limits_{0}^{\pi/2}J_3(1-\cos x)\,{{1\over2}(\cos x - \cos2x)\over{{1\over2}\sin x(1-\cos x)}}\,\mathrm dx$$
$$ = -J_3(1)\ln2 - \int\limits_{0}^{\pi/2}J_3(1-\cos x){\cos x - 1 + 2\sin^2 x\over{\sin x(1-\cos x)}}\mathrm dx$$
$$ = -J_3(1)\ln2 + \int\limits_{0}^{\pi/2}{J_3(1-\cos x)\over\sin x}\mathrm dx - 2\int\limits_{0}^{\pi/2}{J_3(1-\cos x)\over1-\cos x}\sin x\,\mathrm dx$$
$$ = -J_3(1)\ln2 + \int\limits_{0}^{\pi/2}{J_3(1-\cos x)\over(1-\cos x)(1+\cos x)}\mathrm d(1-\cos x) - 2\int\limits_{0}^{\pi/2}{J_3(1-\cos x)\over1-\cos x}\,\mathrm d(1-\cos x)$$
$$ = -J_3(1)\ln2 + \int\limits_{0}^{1}{J_3(t)\over t(2-t)}\mathrm dt - 2\int\limits_{t}^{\pi/2}{J_3(t)\over t}\,\mathrm dt$$
$$ = -J_3(1)\ln2 + {1\over2}\int\limits_{0}^{1}{J_3(t)\over t}\mathrm dt + {1\over2}\int\limits_{0}^{1}{J_3(t)\over2-t}\mathrm dt - 2\int\limits_{t}^{\pi/2}{J_3(t)\over t}\,\mathrm dt$$
$$ = -J_3(1)\ln2 + {1\over2}J_3(2)\int\limits_{0}^{1}{\mathrm dt\over2-t} - {1\over2}\int\limits_{0}^{1}{J_3(t)-J_3(2)\over t-2}\mathrm dt - {3\over2}\int\limits_{t}^{\pi/2}{J_3(t)\over t}\,\mathrm dt$$
$$ = -J_3(1)\ln2 + {1\over2}J_3(2)\ln(2-t)\biggr|_{0}^{1} - {1\over2}\int\limits_{0}^{1}{J_3(t)-J_3(2)\over t-2}\mathrm dt - {3\over2}\int\limits_{t}^{\pi/2}{J_3(t)\over t}\,\mathrm dt$$
$$ = \left(-J_3(1)+{1\over2}J_3(2)\right)\ln2 - {1\over2}\int\limits_{0}^{1}{J_3(t)-J_3(2)\over t-2}\mathrm dt - {3\over2}\int\limits_{t}^{\pi/2}{J_3(t)\over t}\,\mathrm dt$$
$$ = \left(-{2\over3}+{1\over2}\cdot{4\over3}\right)\ln2 - {17\over18} = -{17\over18}$$
(see also Wolfram Alfa),
$$\boxed{F(3) = -{17\over18}}.$$
$\mathbf{k=2n+1}$
Similarly,
$$\int\limits_0^x\sin^{2n+1}y\,\mathrm dy = \int\limits_0^x(1-\cos^2y)^n\sin y\,\mathrm dy = \int\limits_0^{1-\cos x}(2z-z^2)^n\,\mathrm dz = J_{2n+1}(1-\cos x),$$
where
$$J_{2n+1}(t) = \int\limits_0^{t}(2z-z^2)^n\,\mathrm dz.$$
Then
$$F(2n+1) = \left(-J_{2n+1}(1)+{1\over2}J_{2n+1}(2)\right)\ln2 $$
$$- {1\over2}\int\limits_0^1{J_{2n+1}(t)-J_{2n+1}(2)\over t-2}\,\mathrm dt- {3\over2}\int\limits_0^1{J_{2n+1}(t)\over t}\,\mathrm dt.$$
Note that
$$J_{2n+1}(1) = \int\limits_0^1(2t-t^2)^n\,\mathrm dt = \int\limits_0^1(1-(1-t)^2)^n\,\mathrm dt = \int\limits_0^1(1-t^2)^n\,\mathrm dt = {\sqrt{\pi}\,\Gamma(n + 1)\over2\Gamma(n + 3/2)}$$
(see also Wolfram Alfa),
$$J_{2n+1}(2) = \int\limits_0^2(2t-t^2)^n\,\mathrm dt = {\sqrt{\pi}\,\Gamma(n + 1)\over\Gamma(n + 3/2)} = {n!\cdot2^{n+1}\over(2n+1)!!}$$
(see also Wolfram Alfa), therefore
$$F(2n+1) = - {1\over2}\int\limits_0^1{J_{2n+1}(t)-J_{2n+1}(2)\over t-2}\,\mathrm dt- {3\over2}\int\limits_0^1{J_{2n+1}(t)\over t}\,\mathrm dt,$$
and the integrands are polynomials with rational coefficients.
Besides,
$$J_{2n+1}(t) - J_{2n+1}(2) = \int\limits_2^t(2y-y^2)^n\,\mathrm dy = -\int\limits_2^t((2-y)(2-(2-y))^n\,\mathrm d(2-y) = -\int\limits_0^{2-t}(z(2-z))^n\,\mathrm dz = -J_{2n+1}(2-t).$$
That gives
$$\boxed{F(2n+1) = - {1\over2}\int\limits_0^1\left(R_n(2-t)+3R_n(t)\right)\,\mathrm dt},$$
where
$$\boxed{R_n(t) = {1\over t}\int\limits_0^t(2z-z^2)^n\,\mathrm dz}.$$
This allows one to obtain closed form for the required integrals (as shown below) and explains the rational form of the results.
The general formula
Let us use Newton binomial formula
$$(2z-z^2)^n = z^n(2-z)^n = z^n\sum\limits_{i=0}^n\genfrac{(}{)}{0}{0}{n}{i}(-1)^i\cdot2^{n-i}z^i,$$
then
$$R_n(t) = {1\over t}\int\limits_0^t(2z-z^2)^n\,\mathrm dz$$
$$ = {1\over t}\sum\limits_{i=0}^n\genfrac{(}{)}{0}{0}{n}{i}(-1)^i\cdot2^{n-i}\int\limits_0^t z^{n+i}\,\mathrm dz = \sum\limits_{i=0}^n{(-1)^i\over n+i+1}\genfrac{(}{)}{0}{0}{n}{i}\cdot2^{n-i}t^{n+i},$$
$$F(2n+1) = - {1\over2}\int\limits_0^1\left(R_n(2-t)+3R_n(t)\right)\,\mathrm dt$$
$$ = - {1\over2}\sum\limits_{i=0}^n{(-1)^i\over n+i+1}\genfrac{(}{)}{0}{0}{n}{i}\cdot2^{n-i}\int\limits_0^1\left((2-t)^{n+i}+3t^{n+i}\right)\,\mathrm dt $$
$$ = - {1\over2}\sum\limits_{i=0}^n{(-1)^i\over (n+i+1)^2}\genfrac{(}{)}{0}{0}{n}{i}\cdot2^{n-i}\left(-(2-t)^{n+i+1}+3t^{n+i+1}\right)\biggr|_0^1 $$
$$ = - {1\over2}\sum\limits_{i=0}^n{(-1)^i\over (n+i+1)^2}\genfrac{(}{)}{0}{0}{n}{i}\cdot2^{n-i}\left(2+2^{n+i+1}\right),$$
$$\boxed{\boxed{F(2n+1) = -4^n\sum\limits_{i=0}^n{(-1)^i\over (n+i+1)^2}\genfrac{(}{)}{0}{0}{n}{i}(1+2^{-n-i})}}.\tag1$$
Results
Calculation results by $(1)$ using the Wolfram Alpha program (сliсk to the right symbols "="):
$\mathbf{n=0}\quad F(1)\ $=$\ -2$
$\mathbf{n=1}\quad F(3)\ $=$\ -\dfrac{17}{18}$
$\mathbf{n=2}\quad F(5)\ $=$\ -\dfrac{587}{900}$
$\mathbf{n=3}\quad F(7)\ $=$\ -\dfrac{629}{1225}$
$\mathbf{n=4}\quad F(9)\ $=$\ -\dfrac{342319}{793800}$
$\mathbf{n=5}\quad F(11)\ $=$\ -\dfrac{3613679}{38419920}$
Done!