Two contributions to the absolute-value-Hardy-Hilbert-type integral inequalities
Received: 04 Sep 2025 | Accepted: 02 Dec 2025 | Final Version: 31 Dec 2025
Abstract
This article presents two new Hardy-Hilbert-type integral inequalities involving the absolute value function. The first result can be viewed as an analog of a well-known inequality of this type, while the second is distinguished by its originality, incorporating the sine function into the integrand. Complete and detailed proofs are provided for both results.
- Hardy-Hilbert integral inequality, absolute value function, trigonometric functions, change of variables technique.
1. Introduction
The Hardy-Hilbert integral inequality is one of the most celebrated results in mathematical analysis. Its precise formulation is given below. Let \(p>1, q=p/(p-1)\) such that \(1/p+1/q=1\), and \(f,g:(0,+\infty)\to (0,+\infty)\) be two (measurable) functions such that \(\int_{0}^{+\infty}f^p(x)dx < +\infty, \int_{0}^{+\infty}g^q(x)dx < +\infty.\) Then we have \begin{equation} \int_{0}^{+\infty}\int_{0}^{+\infty} \frac{1}{x+y}f(x)g(y)dxdy\le \frac{\pi}{\sin(\pi/p)}\left[\int_{0}^{+\infty}f^p(x)dx\right]^{1/p}\left[\int_{0}^{+\infty}g^q(x)dx\right]^{1/q}... \end{equation}
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