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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 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?

Remark : This question has been asked previously on math.SE without receiving any answers, where you can see the proof of the above theorem.

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  • $\begingroup$ It may be helpful to mention that you know how to evaluate the integral, but it doesn't help in proving the desired property. $\endgroup$ Nov 7, 2013 at 15:57
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    $\begingroup$ I like the question, but unfortunately I doubt this is progress towards proving the irrationality. (Of course I'd be very happy to be proved wrong.) By splitting (*) apart into the $a$, $b$, and $c$ pieces, you are trying to show that these three integrals are linearly independent over $\mathbb{Q}$. As you note on math.SE, they are $\mathbb{Q}$-linear combinations of $\log 2$, $\zeta(3)/\pi^2$, and $\zeta(5)/\pi^4$. In particular, if $\zeta(5)/(\zeta(2)\zeta(3))$ were irrational then the three integrals wouldn't be $\mathbb{Q}$-linearly independent. $\endgroup$
    – Henry Cohn
    Nov 7, 2013 at 16:07
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    $\begingroup$ In order to prove (*), you'd need to have some way of detecting when an integral vanishes. I imagine the reason why there have been no answers on math.SE is that nobody has any idea how to do it: it's hard to prove an integral is nonzero without some special structure to help, and it's not clear to me what to use here. But I could just be missing something... $\endgroup$
    – Henry Cohn
    Nov 7, 2013 at 16:09
  • $\begingroup$ After substituting $x=\pi u$, the question concerns $$\int_0^{1/2}(au^4+bu^3+cu^2)\log\sin\pi u\,du$$ which seems a little more pleasant to me. $\endgroup$ Nov 7, 2013 at 23:12

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