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Let $\zeta(s)$ the Riemann Zeta function, where $s=x+iy$ is the complex variable. In this post we consider the special values $s=x+i\cdot 0=x$, thus our specialization is for real numbers. I was considering a family of integrals, easier than this particular $$-\int_0^1\frac{\zeta(x(x-1))}{\zeta(x)\zeta(1-x)}\,dx.\tag{1}$$

My intention is ask basics about this (that is more complicated than those in which I was interested). See the approximation and graph that provide us Wolfram Alpha online calculator

int -zeta(x(x-1))/(zeta(1-x)zeta(x))dx, from x=0 to x=1.

Question. A) How do you justify rigurously $$-\int_0^1\frac{\zeta(x(x-1))}{\zeta(x)\zeta(1-x)}\,dx=-2\int_{1/2}^{1}\frac{\zeta(x(x-1))}{\zeta(x)\zeta(1-x)}\,dx?\tag{2}$$ B) How one can calculate a good approximation (justify your words, please; I am interested in the justification to set it, not about a very good approximation) of $$-\int_{1/2}^{1}\frac{\zeta(x(x-1))}{\zeta(x)\zeta(1-x)}\,dx?\tag{3}$$ Many thanks.

About $A)$ in my simple cases (different integrals than $(2)$) I know how do a change of variable but I don't know if it justify riguriously the similar relationship. About $B)$ I don't know if there is a special issue to get an approximation of this integral (or this kind of integrals, I don't know if these integrals were studied).

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    $\begingroup$ A) let $x=1-t$ in the right integral and add it to itself to get the left integral $\endgroup$ – Simply Beautiful Art Jul 15 '17 at 11:07
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    $\begingroup$ As per an approximation, as $x\to1$, $\zeta(x)=\frac1{x-1}+\gamma+o(1)$. $\endgroup$ – Simply Beautiful Art Jul 15 '17 at 11:09
  • $\begingroup$ Many thanks for your contribution @SimplyBeautifulArt $\endgroup$ – user243301 Jul 15 '17 at 11:15
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Define $f(x)=\dfrac{\zeta ((x-1) x)}{\zeta (1-x) \zeta (x)}$

We have on $[0,1]$ that $f(\frac12+x)=f(\frac12-x)$ for any $x\in[0,\frac12]$

Indeed $$f\left(\frac12+x\right)=\frac{\zeta \left(\left(x-\frac{1}{2}\right) \left(x+\frac{1}{2}\right)\right)}{\zeta \left(\frac{1}{2}-x\right) \zeta \left(x+\frac{1}{2}\right)}$$ and $$f\left(\frac{1}{2}-x\right)=\frac{\zeta \left(\left(-x-\frac{1}{2}\right) \left(\frac{1}{2}-x\right)\right)}{\zeta \left(\frac{1}{2}-x\right) \zeta \left(x+\frac{1}{2}\right)}$$ denominators are the same and in the numerators the arguments of $\zeta$ are equal

$\left(-x-\dfrac{1}{2}\right) \left(\dfrac{1}{2}-x\right)=\left(x-\dfrac{1}{2}\right) \left(x+\dfrac{1}{2}\right)=x^2-\dfrac{1}{4}$

Thus the function $f$ has a symmetry axis $x=\dfrac{1}{2}$ which explains the reason why the integrals on $\left[0,\frac12\right]$ and $\left[\frac12,1\right]$ are equal

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  • $\begingroup$ Many thanks, this afternoon I am going to study your answer. $\endgroup$ – user243301 Jul 15 '17 at 11:15

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