Evaluation of Harmonic Sums with Integrals

TL;DR

This paper uses multiple integration techniques to evaluate harmonic sums and Riemann zeta values, revealing connections with probability distributions and geometric volumes, and generalizing classical identities.

Contribution

It introduces novel integral representations and probabilistic interpretations for harmonic sums and zeta values, extending known results to higher dimensions.

Findings

Derived new closed-form formulas for S(k) and ζ(2k).

Connected harmonic sums to volumes of convex polytopes.

Linked integrals to probabilities involving Cauchy and uniform random variables.

Abstract

We consider the sums $S(k)=\sum_{n=0}^{\infty}\frac{(-1)^{nk}}{(2n+1)^k}$ and $\zeta(2k)=\sum_{n=1}^{\infty}\frac{1}{n^{2k}}$ with $k$ being a positive integer. We evaluate these sums with multiple integration, a modern technique. First, we start with three different double integrals that have been previously used in the literature to show $S(2)=\pi^2/8,$ which implies Euler's identity $\zeta(2)=\pi^2/6.$ Then, we generalize each integral in order to find the considered sums. The $k$ dimensional analogue of the first integral is the density function of the quotient of $k$ independent, nonnegative Cauchy random variables. In seeking this function, we encounter a special logarithmic integral that we can directly relate to $S(k).$ The $k$ dimensional analogue of the second integral, upon a change of variables, is the volume of a convex polytope, which can be expressed as a probability involving certain pairwise sums of $k$ independent uniform random variables. We use combinatorial arguments to find the volume, which in turn gives new closed formulas for $S(k)$ and $\zeta(2k).$ The $k$ dimensional analogue of the last integral, upon another change of variables, is an integral of the joint density function of $k$ Cauchy random variables over a hyperbolic polytope. This integral can be expressed as a probability involving certain pairwise products of these random variables, and it is equal to the probability from the second generalization. Thus, we specifically highlight the similarities in the combinatorial arguments between the second and third generalizations.