Q: Find the volumes enclosed by $\ln^2(x)+\ln^2(y)+\ln^2(z)+\cdots=1$
I would like to find the volumes enclosed and then plot them to see their distribution as the dimension increases.
I've seen the plots of the volumes of hyper-spheres as the dimension increases and want to do the same for the equation above.
You might notice that the equation looks similar to the equation of a hyper-sphere but with logarithms. This is because this is the equation of a hyper-sphere in exponential space instead of regular $x,y,...$ space.
I think the area enclosed by $\ln^2(x)+\ln^2(y)=1$ can be written using a Bessel function.
Let $B_n$ be the interior of $$ \sum_{k=1}^n\log(x_k)^2=1\tag1 $$ Let $x_k=e^{u_k}$, then $$ \begin{align} \int_{B_n}\mathrm{d}x &=\int_{\|u\|\le1}e^{(1,1,\dots,1)\cdot u}\,\mathrm{d}u\tag{2a}\\ &=\int_{\|u\|\le1}e^{\sqrt{n}\,u_1}\,\mathrm{d}u\tag{2b}\\ &=\frac{\pi^{\frac{n-1}2}}{\Gamma\!\left(\frac{n+1}2\right)}\int_{-1}^1e^{\sqrt{n}\,t}\left(1-t^2\right)^{\frac{n-1}2}\,\mathrm{d}t\tag{2c}\\ &=\frac{2\pi^{\frac{n-1}2}}{\Gamma\!\left(\frac{n+1}2\right)}\int_0^1\cosh\left(\sqrt{n}\,t\right)\left(1-t^2\right)^{\frac{n-1}2}\,\mathrm{d}t\tag{2d}\\ &=\frac{2\pi^{\frac{n-1}2}}{\Gamma\!\left(\frac{n+1}2\right)}\sum_{k=0}^\infty\int_0^1\frac{n^kt^{2k}}{(2k)!}\left(1-t^2\right)^{\frac{n-1}2}\,\mathrm{d}t\tag{2e}\\ &=\frac{\pi^{\frac{n-1}2}}{\Gamma\!\left(\frac{n+1}2\right)}\sum_{k=0}^\infty\frac{n^k}{(2k)!}\int_0^1t^{k-\frac12}(1-t)^{\frac{n-1}2}\,\mathrm{d}t\tag{2f}\\ &=\frac{\pi^{\frac{n-1}2}}{\Gamma\!\left(\frac{n+1}2\right)}\sum_{k=0}^\infty\frac{n^k}{(2k)!}\frac{\Gamma\!\left(k+\frac12\right)\Gamma\!\left(\frac{n+1}2\right)}{\Gamma\!\left(k+\frac{n}2+1\right)}\tag{2g}\\ &=\frac{\pi^{\frac{n}2}}{\Gamma\!\left(\frac{n}2+1\right)}\sum_{k=0}^\infty\frac{n^k}{(2k)!}\frac{\Gamma\!\left(k+\frac12\right)\Gamma\!\left(\frac{n}2+1\right)}{\Gamma\!\left(\frac12\right)\Gamma\!\left(k+\frac{n}2+1\right)}\tag{2h}\\ &=\frac{\pi^{\frac{n}2}}{\Gamma\!\left(\frac{n}2+1\right)}\sum_{k=0}^\infty\frac{n^k}{(2k)!}\frac{(2k-1)!!\,\Gamma\!\left(\frac{n}2+1\right)}{2^k\,\Gamma\!\left(k+\frac{n}2+1\right)}\tag{2i}\\ &=\frac{\pi^{\frac{n}2}}{\Gamma\!\left(\frac{n}2+1\right)}\sum_{k=0}^\infty\frac{n^k}{4^kk!}\frac{\Gamma\!\left(\frac{n}2+1\right)}{\Gamma\!\left(k+\frac{n}2+1\right)}\tag{2j}\\ &=\frac{\pi^{\frac{n}2}}{\Gamma\!\left(\frac{n}2+1\right)}\,\vphantom{F}_0F_1\!\left(;\frac n2+1;\frac n4\right)\tag{2k} \end{align} $$ Explanation:
$\text{(2a):}$ $x_k=e^{u_k}$
$\text{(2b):}$ rotate coordinates
$\text{(2c):}$ integrate in slices where the area of a slice is $\dfrac{\pi^{\frac{n-1}2}}{\Gamma\!\left(\frac{n+1}2\right)}\left(1-t^2\right)^{\frac{n-1}2}$
$\text{(2d):}$ $e^x=\cosh(x)+\sinh(x)$
$\phantom{\text{(2d):}}$ the integral of $\sinh(x)$ over $[-1,0]$ is the the negative of that over $[0,1]$
$\phantom{\text{(2d):}}$ the integral of $\cosh(x)$ over $[-1,0]$ is the same as over $[0,1]$
$\text{(2e):}$ apply the series for $\cosh(x)$
$\text{(2f):}$ substitute $t\mapsto t^{1/2}$
$\text{(2g):}$ apply the Beta integral
$\text{(2h):}$ cancel $\Gamma\!\left(\frac{n}2+1\right)$ and multiply by $\frac{\sqrt\pi}{\Gamma\left(\frac12\right)}\frac{\Gamma\left(\frac{n}2+1\right)}{\Gamma\left(\frac{n}2+1\right)}$
$\text{(2i):}$ $\frac{\Gamma\left(k+\frac12\right)}{\Gamma\left(\frac12\right)}=\frac{(2k-1)!!}{2^k}$
$\text{(2j):}$ $(2k)!=2^kk!(2k-1)!!$
$\text{(2k):}$ write as a generalized hypergeometric function
Jack D'Aurizio's answer matches $\text{(2c)}$ except that $\dfrac{\pi^{\frac{n-1}2}}{\Gamma\!\left(\frac{n+1}2\right)}$ in $\text{(2c)}$ is replaced by $\dfrac{\pi^{\frac{n}2}}{\Gamma\!\left(\frac{n}2+1\right)}$.
The maximum occurs at $n=5$, but it is only $0.003717818$ greater than at $n=6$.
Here is the surface for $n=3$:
Its volume is $$ \frac{4\pi}3\left(\cosh\left(\sqrt3\right)+\frac{\sinh\left(\sqrt3\right)}{\sqrt3}\right)\tag3 $$