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Does the expression $$\frac{\pi^{\frac{n}{2}}}{\Gamma(\frac{n}{2}+1)}R^n,$$ which gives the volume of an $n$-dimensional ball of radius $R$ when $n$ is a nonnegative integer, have any known significance when $n$ is an odd negative integer?

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    $\begingroup$ You should explain the context why you are asking this question to make it more relevant. $\endgroup$ Mar 4 '20 at 14:22
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    $\begingroup$ Pure curiosity. $\endgroup$ Mar 4 '20 at 15:28
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    $\begingroup$ It can occur in the dimensional regularization of Feynman diagram integrals in quantum field theory, but I don't think it has too much significance there. $\endgroup$
    – gmvh
    Mar 6 '20 at 10:12
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    $\begingroup$ Here is a funny formula: with $\nu !=\Gamma (\nu+1)$, we have $$ \sum_{\nu\in \mathbb N\cup (\mathbb N+\frac12)}\frac{π^\nu}{\nu !} R^{2\nu}=e^{π R^2} +\sum_{k\in \mathbb N}\frac{π^{k+\frac12}}{(k+\frac12)!} R^{2k+1} =\sum_{\nu\in \mathbb N\cup (\mathbb N+\frac12)}\vert \mathbb B^{2\nu}\vert R^{2\nu} =\sum_{n\in \mathbb N} \vert \mathbb B^{n}\vert R^{n}. $$ $\endgroup$
    – Bazin
    Mar 6 '20 at 20:46
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    $\begingroup$ Maybe en.wikipedia.org/wiki/Reflection_formula is relevant here? $\endgroup$ Mar 7 '20 at 2:43
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I will answer regarding the dimension $-1$.

An example of such space is a set of periodic lattices on a real line.

Indeed, you can see that the Hausdorff dimension of a periodic lattice is $-1$: If we scale down the lattice twice, it would be able to include two original lattices. Since scaling the fractal down twice makes it twice as big, its dimension is $\frac{\ln 2}{\ln (1/2)}=-1$.

The formula for the volume of a ball gives $\frac1{\pi R}$ for $n=-1$. This means the unit ball ($R=1$) includes only one lattice: the $\pi$-periodic one.

If we reduce the radius (the step of the lattice divided by $\pi$), its volume increases: a lattice with step $1/2$ can be represented as two lattices of step $1$. Thus the volume of a lattice of step $1/2$ consists of two "points" ($1$-periodic lattices), and so has volume twice the volume of $1$-periodic lattice.

Alternatively, we can consider it being 1 point with "weight" 2 (the weight being proportional to the density of a lattice, so that our space has fuzzy membership function).

Similarly, if we increase the lattice step ("radius" of the $-1$-sphere), we can consider it a lattice with weight below $1$, the same as its volume.

So, the volume of the $-1$-ball is the density of the lattice.

Interesting observation: in positive-dimentional space the ball becomes a point with reduction of its radius to zero. In zero-dimensional Euclidean space the ball is always a point, disregarding the "radius" and in this $-1$-dimensional sphere the ball becomes a zero-volume point when we increase the radius infinitely.

You can find some further ideas here.

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    $\begingroup$ This proposed value of the “volume” of the (-1)-ball does not agree with the formula given in the original post. $\endgroup$ Apr 19 '20 at 16:31
  • $\begingroup$ @JamesPropp oh yes, i missed a mistake $\endgroup$
    – Anixx
    Apr 19 '20 at 16:33
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    $\begingroup$ I do not see why “the unit sphere is a pi-periodic lattice” (and I doubt that it is a “known fact”). Also, Anixx’s style of answer is unlikely to discriminate between even and odd (negative) dimensionalities, so I think it’s on the wrong track. $\endgroup$ Apr 19 '20 at 17:03
  • $\begingroup$ @JamesPropp This answer is specifically about the dimension -1, and the generalization to other dimensions may require efforts. Particularly, because in dimension -1 the periodic lattice is both a sphere and a cube. $\endgroup$
    – Anixx
    Apr 19 '20 at 17:10
  • $\begingroup$ @JamesPropp And, in general, when we are talking about negative dimensions, the bodies in them are littices, spectrums and other (infinite but countable) collections of dots, but their volumes are determined based on their overall density (or the infinite part of numerocity if you prefer), not on the number of dots they include. $\endgroup$
    – Anixx
    Apr 19 '20 at 17:25
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I want to make another answer to this question, although it is is not really an answer, and I do not know whether it is relevant or not, but it may be useful.

Let's take

$$V(n,R)=\frac{\pi^{\frac{n}{2}}}{\Gamma(\frac{n}{2}+1)}R^n$$

Also, using the reflection formula for the Zeta function we can see:

$$n\zeta(1-n)\frac{\pi^{\frac{n}{2}}}{\Gamma(\frac{n}{2}+1)}=(1-n)\zeta(n)\frac{\pi^{\frac{1-n}{2}}}{\Gamma(\frac{1-n}{2}+1)}$$

Interestingly enough, in the algebra of divergent integrals described here, there is a rule $\operatorname{reg}\omega_+^n=-n\zeta(1-n)$, (where $\omega_+=\int_{-1/2}^\infty dx$ is an infinite divergent integral and $\operatorname{reg}$ denotes regularization), the overall formula becomes

$$\operatorname{reg} V(n, \omega_+)=\operatorname{reg} V(1-n, \omega_+)$$

This is a very nice-looking relation but its deep meaning is unclear. Particularly because it is not evident what meaning should have the balls with infinite radius, and especially the balls of infinite radius in negative dimension :-). Still formally this relation looks beautiful.

See also this question: What is the connection between the Riemann Xi-function and n-sphere?

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