Cauchy integral formula on a polydisc

Remarks

The proof applies the one-variable Cauchy integral formula in $z_1$, then in $z_2$ inside the integral, and so on, $n$ times. Continuity of $f$ on $T$ allows the iterated integrals to be combined into one torus integral. The formula holds under the weaker hypothesis that $f$ is continuous on $\overline{P}$ and holomorphic on $P$. Expanding $1/(\zeta_j - z_j)$ in a geometric series in $(z_j - a_j)/(\zeta_j - a_j)$ gives the power series of $f$ at $a$, with coefficients $$c_\alpha = \frac{1}{(2\pi i)^n} \int_T \frac{f(\zeta)}{(\zeta - a)^{\alpha + \mathbf{1}}} \, d\zeta, \qquad (\zeta - a)^{\alpha + \mathbf{1}} = \prod_{j} (\zeta_j - a_j)^{\alpha_j + 1}.$$ Example: for $f(z) = z_1 z_2$ and $P$ the unit polydisc of $\C^2$, at $z = 0$ the right side is $(2\pi i)^{-2} \int_T d\zeta_1 \, d\zeta_2 = 0 = f(0)$. Only the values of $f$ on the $n$-dimensional torus $T$ are used, not on the whole $(2n-1)$-dimensional boundary of $P$.

Going deeper

For a student

This is the basic tool of local theory in several variables. From it one gets that holomorphic functions are locally sums of convergent power series, the Cauchy estimates $\abs{c_\alpha} \le \max_T \abs{f} / r^\alpha$, and the infinite differentiability of holomorphic functions. Since the kernel is a product of one-variable kernels, it is tied to the product shape of the polydisc. On other domains one uses integral formulas with other kernels, such as the Bochner–Martinelli formula.