# Show that $\lambda^{d}(E_{d})=\frac{\pi^{\frac{d}{2}}}{\Gamma(\frac{d}{2}+1)}$ and determine behavior for $d \to \infty$

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?

• Are you looking for the derivation of the volume or the limiting behavior? – user1337 Jan 8 '19 at 12:12
• Rather the deviation as $d$ gets larger – MinaThuma Jan 8 '19 at 12:21
• I think you are looking for volume of n-ball. – StubbornAtom Jan 8 '19 at 12:59
• – StubbornAtom Jan 8 '19 at 13:05

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 $$\lambda^d (E_d)$$.

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.

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