# How to solve this equation for $y$? Is it an irrational number or rational?

How to solve this equation for $y$? and is it an irrational number or a rational?

$$\frac{2}{7}\pi^2\log2+\frac{16}{7}\int_{0}^{\frac{\pi}{2}}x\log(\sin x)\,\mathrm{d}x-\sqrt{y}\pi^3=0.$$

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what's the base of the logarithm? – Caran-d'Ache Mar 31 '13 at 14:47
@Stahl Do you think $\zeta(3)=\sqrt{y}\pi^3,y\in\mathbb{Q}$ is right? – daoyi Mar 31 '13 at 15:17
You can isolate $y$ without too much trouble, but an explicit value for $\zeta(3)$ is not known. In particular, we don't know if $\zeta(3) = q\pi^3$ for some $q\in\Bbb{Q}$, so it's not clear from your equation whether or not $y\in\Bbb{Q}$. – Stahl Mar 31 '13 at 15:20
@daoyi If that is your real question, why didn't you ask directly? – Erick Wong Mar 31 '13 at 15:21

Just "isolate" $\,y\,$ :

$$\sqrt y=\frac{1}{\pi^3}\left(\frac{2}{7}\pi^2\log 2+\frac{16}{7}\int\limits_0^{\pi/2}x\log\sin x\,dx\right)\ldots$$

You may want now to square the whole right hand side, or evaluate explicitly the integral, or...

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Well, assuming that the logarithm has base $b$, the integral will be equal to: $$\frac{16}{7}\int_{0}^{\frac{\pi}{2}}x\log(\sin x)\,\mathrm{d}x=-\frac{\pi ^2 \ln(4)-7 \zeta (3)}{7 \ln(b)},$$ where $\zeta (3)$ - is the Riemann zeta function. So $$\frac{2}{7}\pi^2\log2+\frac{16}{7}\int_{0}^{\frac{\pi}{2}}x\log(\sin x)\,\mathrm{d}x=\frac{\zeta(3)}{\ln(b)}$$ And $$y={\bigg(\!\!\frac{\zeta(3)}{\ln(b)\pi^3}\!\!\bigg)}^{2}$$ Buy the way $\zeta (3)$ is proved to be irrational.

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Thank you a lot! In fact I know that $b=\mathrm{e}=\lim\limits_{n\to\infty}(1+1/n)^n$. I want to know if $y$ is a rational? – daoyi Mar 31 '13 at 15:08
@daoyi I think it's still open whether $\zeta(3)/\pi^3$ is rational. We know from numerical evidence that if it is rational, it can't be a very nice one (probably millions of digits long at least). Presumably no one knows whether $(\zeta(3)/\pi^3)^2$ is rational either. – Erick Wong Mar 31 '13 at 15:16
@daoyi You have an irrational number (see the link in the answer) divided by an irrational number $\pi$. It's an irrational number. – Caran-d'Ache Mar 31 '13 at 15:16
@Caran-d'Ache $\sqrt{12}/\sqrt{3}$? – Erick Wong Mar 31 '13 at 15:17
@ErickWong Well, that's an open problem and guess with Apéry's constant it's "the million-dollar question". – Caran-d'Ache Mar 31 '13 at 15:19