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Let $d \in \mathbb N$ and $E_{d}:=\{x \in \mathbb R^{d}:|x|\leq 1\}$

Prove that $$ \lambda^{d}(E_{d})=\frac{\pi^{\frac{d}{2}}}{\Gamma(\frac{d}{2}+1)} $$

and determine $\lambda^{d}(E_{d})$ as $d \to \infty$

I struggle with d-dimensional volume, so I will try the behavior for $d \to \infty$

Note:

$$\frac{\pi^{\frac{d}{2}}}{\Gamma(\frac{d}{2}+1)}=\frac{\pi^{\frac{d}{2}}}{\int_{0}^{\infty}x^{\frac{d}{2}}e^{-x}dx}$$

Looking particularly at:

$\int_{0}^{\infty}x^{\frac{d}{2}}e^{-x}dx$ it looks like partial integration, but I wouldn't as $d \in \mathbb N$. I would use substitution, namely $y = x^{\frac{d}{2}}\Rightarrow \frac{2}{d}dy=x^{\frac{d}{2}-1}dx$. But this is a dead end, as $x$ does not disappear

Any ideas?

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Let $t=d/2$. Observe that $$0 \leq \frac{\pi^t}{\Gamma(t+1)} \leq \frac{\pi^{\lfloor t \rfloor+1}}{\Gamma(\lfloor t \rfloor+1)} $$ and recall that the series $$\sum_{n=0}^\infty \frac{\pi^n}{n!} $$ converges (to $e^\pi$). Convergent series have their terms approaching zero, and the squeeze law shows that the same is true for $\lambda^d (E_d)$.

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Hint for the derivation: compute $\int_{\mathbb{R}^d}{e^{-|x|^2}}$ by separating integrals and by polar coordinates.

For the asymptotic behavior: use the functional equation of $\Gamma$ to have a non-integral formula for $d=2p$ and $d=2p-1$. Then let $p$ go to infinity in both formulas and use Stirling’s theorem.

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  • $\begingroup$ Why would I use the integral $\int_{\mathbb R^{d}}e^{-|x|^2}$ ? $\endgroup$ – MinaThuma Jan 8 '19 at 16:42
  • $\begingroup$ It is the shortest way I know to derive the “volume” of the unit sphere thus the volume of the unit ball. $\endgroup$ – Mindlack Jan 8 '19 at 17:13

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