[Paper Review] The integrals in Gradshteyn and Ryzhik. Part 10: the digamma function

Luis A. Medina, Victor H. Moll|ArXiv.org|Sep 21, 2007

Mathematical functions and polynomials4 references19 citations

TL;DR

This paper provides analytical evaluations of definite integrals from Gradshteyn and Ryzhik involving the digamma function ψ(x), using integral representations and properties of the gamma and psi functions. It derives key results such as ∫₀^∞[e⁻ˣ − (1+x)⁻ᵃ]/x dx = ψ(a), and extends these to families of integrals involving powers and logarithmic terms, linking them to special functions and constants like γ and π.

ABSTRACT

Many integrals in the classical table by Gradshteyn and Ryzhik can be evaluated in terms of the digamma function (= the logarithmic derivative of the gamma function). Some of them are presented here.

Motivation & Objective

• To systematically evaluate definite integrals from Gradshteyn and Ryzhik that involve the digamma function ψ(x).
• To establish new integral representations of ψ(a) using exponential and rational functions.
• To connect these integrals to known special functions and constants such as the Euler-Mascheroni constant γ and π.
• To extend results to families of integrals involving powers, logarithms, and beta functions.
• To provide a foundation for future evaluations of integrals with (1+x)⁻¹ terms using the incomplete beta function.

Proposed method

• Derives ∫₀^∞[e⁻ˣ − (1+x)⁻ᵃ]/x dx = ψ(a) using the Frullani-type integral ∫₀^∞(e⁻ᶻ − e⁻ˢᶻ)/z dz = ln s.
• Applies the integral representation Γ′(a) = ∫₀^∞ e⁻ˢ sᵃ⁻¹ ln s ds and swaps integration order to relate to ψ(a).
• Uses the identity ∫₀^∞(e⁻ᶻ − e⁻ˢᶻ)/z dz = ln s to evaluate differences of exponential and rational terms.
• Applies changes of variables such as w = −ln x, t = 1/(x+1), and u = xᵖ to transform integrals into known forms.
• Employs the beta function and its logarithmic derivative to evaluate integrals with (1−x)ᵇ⁻¹ or (1−x²)ⁿ⁻¹ terms.
• Uses reflection and duplication formulas for ψ(x) and Γ(x), including ψ(1/2) = −γ − 2ln 2 and ψ(1) = −γ.

Experimental results

Research questions

• RQ1 How can the digamma function ψ(a) be represented as a definite integral involving e⁻ˣ and (1+x)⁻ᵃ?
• RQ2 What is the analytical evaluation of ∫₀^∞(e⁻ˣᵖ − e⁻ˣʲ)/x dx in terms of the Euler-Mascheroni constant γ?
• RQ3 How do changes of variables transform integrals involving (1+x)⁻ᵃ into forms related to the beta function?
• RQ4 What is the connection between integrals of the form ∫₀¹ xᵃ⁻¹ ln(1−x) dx and the digamma function?
• RQ5 Can a two-parameter family of integrals ∫₀^∞(e⁻ˣᵃ − 1/(1+xᵇ))/x dx be evaluated independently of b, and what is its value?

Key findings

• The integral ∫₀^∞[e⁻ˣ − (1+x)⁻ᵃ]/x dx = ψ(a) holds for a > 0, providing a fundamental representation of the digamma function.
• The special case a = 1 yields ∫₀^∞(e⁻ˣ − 1/(1+x))/x dx = −γ, corresponding to formula 3.435.3.
• The identity ∫₀^∞(e⁻ˣᵖ − e⁻ˣʲ)/x dx = (p−q)/(pq) γ is derived, with p = q = 2 giving ∫₀^∞(e⁻ˣ² − e⁻ˣ²)/x dx = 0.
• For ∫₀^∞(e⁻ˣᵃ − 1/(1+xᵇ))/x dx, the result is −γ/a, independent of b, as shown in Proposition 11.1.
• The evaluation ∫₀¹ ln x / √[ⁿ](1−x²ⁿ) dx = −π/8 ⋅ B(1/(2n), 1/(2n)) / (n² sin(π/(2n))) is derived, matching formula 4.247.1.
• The integral ∫₀^∞ x e⁻ˣ (1−e⁻²ˣ)ⁿ⁻¹ᐟ² dx evaluates to (π/2²ⁿ⁺²) ⋅ (2ⁿ choose n) ⋅ (2ln 2 + ∑ₖ₌₁ⁿ 1/k), corresponding to 3.457.1.