What is the relation between Zeta Function and nth Integral?

I was reading the article in Wikipedia about Apery Constant and I saw the triple integral

$$\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\frac{1}{1-xyz}dx~dy~dz$$

By curiosity I removed the z variable and the third integral leaving only the double integral

$$\int_{0}^{1}\int_{0}^{1}\frac{1}{1-xy}dx~dy$$

calculated it with Desmos and got the result of $$\frac{\pi^{2}}{6}$$

I then tried with one integral but unfortunately that was undefined

I throw even more curiosity and wanted to know what will happen with 4 integrals and I got the value $$\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\frac{1}{1-xyzw}dx~dy~dz~dw=\zeta(4) = \frac{\pi^{4}}{90}$$

So far I have reached till $$\zeta(4)$$ does this pattern continue and what is the thing that makes this true?

• Yes, it continues. To see why expand $\frac{1}{1-x_1x_2\dots x_n}$ as a geometric series and interchange summation and integrations.
– Zima
Jul 24 at 18:02
• The non-convergence of the integral $\int\limits_0^1 \frac{1}{1-x} \, dx$ is consistent with the pole of $\zeta(s)$ at $s=1$. Jul 24 at 18:08

1 Answer

Yes the pattern exists. Indeed,

\begin{align} \int_{[0,1]^n} \frac1{1 - \prod_{\ell=1}^n x_\ell}\mathrm dx_1\ldots\mathrm dx_n &= \sum_{k=0}^\infty \int_{[0,1]^{n}} \left(\prod_{\ell=1}^n x_\ell\right)^k\mathrm dx_1\ldots\mathrm dx_n\\ &= \sum_{k=0}^{\infty} \left(\int_0^1 x^k\mathrm d x\right)^n\\ &= \sum_{k=0}^{\infty} \frac1{(k+1)^n} = \zeta(n) \end{align}