Proving a proposition which leads the irrationality of $\frac{\zeta(5)}{\zeta(2)\zeta(3)}$

Question : Is the following $(\star)$ true for $a,b,c\in\mathbb Z$ ?

$$\begin{align}\int_{0}^{\frac{\pi}{2}}(ax^4+b\pi x^3+c{\pi}^{2}x^2)\log(\sin x)dx=0\Rightarrow a=b=c=0\qquad(\star)\end{align}$$

Motivation : I've been asking the following question on MSE :

Is $\frac{\zeta (m+n)}{\zeta (m)\zeta (n)}$ a rational number for $m,n\in\mathbb Z$ where $\zeta (s)=\sum_{n=1}^{\infty}\frac{1}{n^s}$?

Then, I've just been to able to prove the following theorem :

Theorem : If $(\star)$ is true, then $\frac{\zeta (5)}{\zeta (2)\zeta (3)}$ is an irrational number.

However, I can't prove that $(\star)$ is true. Can anyone help?

In the following, I'm going to prove the theorem. Here, note that $(\star)$ is saying that $$aI_{4}+bI_{3}+cI_2=0\ (a,b,c\in\mathbb Z)\Rightarrow a=b=c=0$$ where $$I_m=\frac{1}{{\pi}^{m+1}}\int_{0}^{\frac{\pi}{2}}x^m\log(\sin x)dx.$$

Lemma : $$I_2=-\frac{1}{24}\log 2+\frac{3}{16}\frac{\zeta (3)}{{\pi}^2}.$$ $$I_3=-\frac{1}{64}\log 2+\frac{9}{64}\frac{\zeta (3)}{{\pi}^2}-\frac{93}{128}\frac{\zeta(5)}{{\pi}^4}.$$ $$I_4=-\frac{1}{160}\log 2+\frac{3}{32}\frac{\zeta (3)}{{\pi}^2}-\frac{45}{64}\frac{\zeta(5)}{{\pi}^4}.$$

Proof for lemma : By $$\log(\sin x)=-\sum_{n=1}^{\infty}\frac 1n\cos(2n\pi)-\log 2$$ for $0\lt x\lt \frac{\pi}{2}$, we get $${\pi}^{m+1}I_m=-\sum_{n=1}^{\infty}\frac 1n\int_{0}^{\frac{\pi}{2}}x^m\cos(2n\pi)dx-\frac{1}{m+1}\left(\frac{\pi}{2}\right)^{m+1}\log 2.$$ Here, by utilizing partial integration, we get $$\int_{0}^{\frac{\pi}{2}}x^2\cos(2nx)dx=(-1)^n\frac{\pi}{4n^2},$$ $$\int_{0}^{\frac{\pi}{2}}x^3\cos(2nx)dx=\frac{3}{8n^4}-(-1)^n\frac{3}{8n^4}+(-1)^n\frac{3{\pi}^2}{16n^2},$$ $$\int_{0}^{\frac{\pi}{2}}x^4\cos(2nx)dx=-(-1)^n\frac{3\pi}{4n^4}+(-1)^n\frac{{\pi}^3}{8n^2}.$$ Hence, since we get $$\sum_{n=1}^{\infty}\frac{(-1)^n}{n^3}=-\left(1-2\cdot\frac{1}{2^3}\right)\zeta(3)=-\frac 34\zeta(3)$$ $$\sum_{n=1}^{\infty}\frac{(-1)^n}{n^5}=-\left(1-2\cdot\frac{1}{2^5}\right)\zeta(5)=-\frac{15}{16}\zeta(5),$$ we get $I_2,I_3,I_4$ as desired. Now the proof for lemma is completed.

Proof for theorem : By the above lemma, we can write $$\left(\begin{matrix} I_2 \\ I_3 \\ I_4 \\ \end{matrix} \right)=\left(\begin{matrix} -\frac{1}{24} & \frac{3}{16} & 0 \\ -\frac{1}{64} & \frac{9}{64} & -\frac{93}{128} \\ -\frac{1}{160} & \frac{3}{32} & -\frac{45}{64} \\ \end{matrix} \right)\left(\begin{matrix} \log 2 \\ {\zeta (3)}/{{\pi}^2} \\ {\zeta(5)}/{{\pi}^4} \\ \end{matrix} \right)$$ Hence, we get $$\frac{\zeta(3)}{{\pi}^2}=-\frac 13(1240I_4-1200I_3+264I_2),$$ $$\frac{\zeta(5)}{{\pi}^4}=-\frac 13(120I_4-112I_3+24I_2).$$ Hence, since $\zeta(2)={{\pi}^2}/{6}$, we get $$\frac{\zeta(5)}{\zeta(2)\zeta(3)}=6\frac{{\zeta(5)}/{{\pi}^4}}{{\zeta(3)}/{{\pi}^2}}=6\frac{120I_4-112I_3+24I_2}{1240I_4-1200I_3+264I_2}.$$ Now if $\frac{\zeta(5)}{\zeta(2)\zeta(3)}$ is a rational number, then there exist natural numbers $r,s$ such that $$\frac{120I_4-112I_3+24I_2}{1240I_4-1200I_3+264I_2}=\frac sr.$$ Hence, we get $$(120r-1240s)I_4-(112r-1200s)I_3+(24r-264s)I_2=0$$ $$\Rightarrow 120r-1240s=112r-1200s=24r-264s=0$$ $$\Rightarrow 3r=31s,\ 7r=75s,\ r=11s.$$ However, there don't exist such $r,s$. Hence, we know that $\frac{\zeta(5)}{\zeta(2)\zeta(3)}$ is an irrational number. Now the proof for theorem is completed.

Update : I crossposted to MO.

Let $M$ be the $3 \times 3$ matrix of numerical coefficients at the beginning of the proof of main theorem. I've just verified that its determinant is $3/40960$, hence non-zero. This shows that $M$ has an inverse (also with a non-null determinant) and then the $3 \times 3$ system can be `inverted' to furnish $\ln{2}$ and the $\zeta{(n)}/\pi^{n-1}$ values in terms of the integrals $I_k$.

F. M. S. Lima, University of Brasilia (fabio-at-fis.unb.br)