Question

Triple integral

Answer 1

\[\large \int_0^1\int_0^1\int_0^1\frac{dxdydz}{1-xyz}\]

Answer 2

One may wish to examine the convergence of \(\large \int_0^1\dfrac{dx}{1-x}\). Seems unlikely that if it converges we'll be able to stay in Real Numbers.

Answer 3

Formula (10) gives an immediate closed form: http://mathworld.wolfram.com/RiemannZetaFunction.html

Answer 4

Let \(\mathcal B=\{(x,y,z):0<x,y,z<1\}\). So the integral is
\[\iiint_\mathcal{B}\frac{\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z}{1-xyz}\]
Since \(0<x,y,z<1\) (meaning each of \(x\), \(y\) and \(z\) are between \(0\) and \(1\), respectively - forgive the abuse of notation), we have
\[\frac{1}{1-xyz}=\sum_{n\ge0}(xyz)^n\]Replacing accordingly and exchanging integration and summation, you get
\[\begin{align*}
\iiint_\mathcal{B}\sum_{n\ge0}(xyz)^n\,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z&=\sum_{n\ge0}\iiint_\mathcal{B} x^ny^nz^n\,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z\\[1ex]
&=\sum_{n\ge0}\frac{1}{(n+1)^3}\\[1ex]
&=\sum_{n\ge1}\frac{1}{n^3}\\[1ex]
&=\zeta(3)
\end{align*}\]where \(\displaystyle\zeta(s)=\sum_{n\ge1}\frac{1}{n^s}\).

Answer 5

Awesome, thanks! I can't believe I didn't think to look at it as a geometric series, neat.