Connection between the area of a n-sphere and the Riemann zeta function?

The Riemann Xi-Function is defined as $$\xi(s) = \tfrac{1}{2} s(s-1) \pi^{-s/2} \Gamma\left(\tfrac{1}{2} s\right) \zeta(s)$$ and it satisfies the reflection formula $$\xi(s) = \xi(1-s).$$ But the area $A$ of a $s$-dimensional sphere is $$A(s) = \frac{2 \pi^{s/2}}{\Gamma\left(\tfrac{1}{2} s\right)}$$ so that we can write the Xi-Function like $$\xi(s) = s(s-1) \frac{\zeta(s)}{A(s)}.$$ If we insert this into the reflection formula we get the following relation between the area of a n-sphere and the Riemann zeta function $$\frac{\zeta(s)}{\zeta(1-s)} = \frac{A(s)}{A(1-s)}$$ Has this relation been noted in the literature? Why should there be such a connection between the area of a n-sphere and the zeta function? Can this relation be explained geometrically?

UPDATE: I agree with the comments that negative dimensional spheres are difficult to interpret geometrically. However if we insert the formula for the area $A$ into Eulers reflection formula $$\Gamma(z)\Gamma(1-z)= \frac{\pi}{\sin(\pi z)}$$ we get $$A(1-s) = \frac{4\sin\left(\pi \frac{s+1}{2}\right)}{A(s+1)}$$ Inserting this into our relation with the Riemann zeta function yields $$\frac{\zeta(s)}{\zeta(1-s)} = \frac{A(s)A(s+1)}{4\sin\left(\pi \frac{s+1}{2}\right)}$$ which avoids the negative dimensions for $s>0$. Can this formula be interpreted geometrically?

• Why should there be such a connection between the area of a n-sphere and the zeta function ? - What is a sphere ? A bounded sum of powers: $x^2+y^2+z^2=r^2.~$ And what is the $\zeta$ function ? A bounded sum of powers: recall its famous infinite series expression for $\Re(s)>1.~$ And by expanding the power of a sum into a sum of powers through Newton's binomial series we get binomial coefficients, aka beta functions, which can be expressed in terms of factorials, aka $\Gamma$ functions. In general, all $\Gamma$ functions of argument $\frac1n$ are connected to superellipses $x^n+y^n=r^n$. – Lucian May 20 '16 at 10:14
• I can't follow you. You are saying e.g. $\zeta(3) = \frac{1}{1^3} + \frac{1}{2^3} + \frac{1}{3^3} + ...$ is a bounded sum of powers? It is a bounded sum, but it is a sum with infinite terms to the power 3. Whereas a 3d-sphere is a bounded sum of three terms to the power two. Can you elaborate again on what kind of connection you see between these sum of powers? – asmaier May 21 '16 at 20:49
• The superellipse $x^n+y^n=r^n$ has an area which is directly expressible in terms of beta$/\Gamma$ functions, whose arguments are rational numbers of denominator n. At the same time, $\zeta(n)=\displaystyle\sum_{k=1}^\infty\frac1{k^n}.~$ Both are bounded sums of powers. For instance, the fact that $\Gamma\Big(\tfrac12\Big)=\sqrt\pi$ and $\zeta(2n)=a_n\cdot\pi^n$ are both related to the the same fundamental constant is not a coincidence. Since $A(s)$ is expressible in terms of $\Gamma(s),$ the conclusion follows. – Lucian May 27 '16 at 7:43
• What geometrical meaning do you give to $A(-\frac 1 2)$ and $A(\textrm i)$? – Alex M. Jun 5 '16 at 17:05
• I don't think you can hope for a simple geometrical interpretation of this. Saying that $A(s)$ is the area of a sphere only makes sense when $s$ is an integer (the dimension of the space). But if $s$ is an integer then $1-s$ is negative (or zero) so there are no choices where both $A(s)$ and $A(1-s)$ can be given a classical geometrical interpretation. – Winther Jun 9 '16 at 2:51

The relation $$\frac{\zeta(s)}{A(s)}=\frac{\zeta(1-s)}{A(1-s)}$$ stems from the fact that $$\frac{\zeta(s)}{A(s)}=\frac{1}{2}\mathcal{M}\left(\psi\right)\left(\frac{s}{2}\right),$$ where $\mathcal{M}$ denotes the Mellin transform and $$\psi(x)=\sum_{n=1}^\infty e^{-\pi n^2 x}.$$ This function is symmetric due to the functional equation for $\psi(x)$, and that yields the functional equation for the zeta function. The reason for the connection to the surface area appears in my previous answer https://math.stackexchange.com/a/1494471/6075, which provides a geometric explanation for integer dimensions - the case where we can actually interpret the surface area. In that answer, I showed that the factor of $A(s)$ appears due to the spherical symmetry of a naturally arising $s$-dimensional integral expression for $\zeta(s)$.
Remark 2: Note that the surface area here, $A(s)$, corresponds to the $s-1$-dimensional sphere, that is the surface of the $s$-dimensional ball, whereas in the linked answer, $A_{k-1}$ is the surface area of the $k-1$-dimensional sphere, that is the surface of the $k$ dimensional ball.