[Paper Review] The integrals in Gradshteyn and Ryzhik. Part 7: Elementary examples

Tewodros Amdeberhan, Victor H. Moll|ArXiv.org|Jul 14, 2007

Mathematical functions and polynomials10 references3 citations

TL;DR

This paper provides rigorous derivations and contextual proofs for a selection of elementary definite integrals listed in Gradshteyn and Ryzhik’s renowned table of integrals. Using elementary techniques such as substitution, beta and gamma functions, trigonometric transformations, and recursive integration, the authors establish closed-form evaluations for integrals involving algebraic, logarithmic, exponential, and rational functions, with key results including evaluations in terms of the beta function, inverse trigonometric and logarithmic functions, and generalized hypergeometric structures.

ABSTRACT

The classical table of integrals by I. S. Gradshteyn and I. M. Ryzhik contains some elementary integrals. We discuss their evaluations.

Motivation & Objective

• To provide rigorous proofs for selected elementary definite integrals listed in Gradshteyn and Ryzhik’s table, particularly those lacking detailed derivations.
• To establish a systematic approach to evaluating integrals that combine elementary functions, such as rational, algebraic, logarithmic, and exponential forms.
• To generalize known integrals using substitutions and transformations, extending results to broader parameter ranges.
• To connect the evaluations to special functions like the beta and gamma functions, and to derive new closed-form expressions.
• To validate and extend entries in the table, including new generalizations such as the integral family involving $ (ax^2 + 2bx + c)^{-n-3/2} $.

Proposed method

• Employing the substitution $ y = 1 - \ oot{2}x $ to transform $ \int_0^1 (1 - \sqrt{x})^{p-1} dx $ into a sum of power integrals.
• Using the change of variables $ t = x^a $ to express $ \int_0^1 (1 - x^a)^{p-1} dx $ in terms of the beta function $ B(p, a^{-1}) $.
• Applying trigonometric substitutions $ t = \tan\varphi $ and $ u = \sin\varphi $ to reduce oscillatory integrals to rational or inverse trigonometric forms.
• Deriving recursive relations via differentiation under the integral sign and solving differential-difference equations for parameter-dependent integrals.
• Using partial fraction decomposition and series expansions to evaluate integrals involving $ (1 - s^2)^{-1} $ and $ (1 + z^2 x^2)^{-n-1} $.
• Applying integration by parts and parameter differentiation to derive closed forms for integrals involving $ \ln(a^s + x^s) $ and rational functions raised to fractional powers.

Experimental results

Research questions

• RQ1 How can the elementary integral $ \int_0^1 (1 - \sqrt{x})^{p-1} dx = \frac{2}{p(p+1)} $ be rigorously derived using elementary substitutions?
• RQ2 What generalization of $ \int_0^1 (1 - x^a)^{p-1} dx $ yields a closed-form expression in terms of the beta function?
• RQ3 How can the integral $ \int_{-\infty}^\infty \frac{dx}{(1+x^2)\sqrt{b + a x^2}} $ be evaluated for general $ a, b > 0 $, and what are the three cases based on $ a $ and $ b $?
• RQ4 What is the closed-form evaluation of $ \int_0^\infty \frac{dt}{(a + b t^2)^n \sqrt{1 + t^2}} $, and how does it relate to the beta function and trigonometric substitutions?
• RQ5 How can the integral $ \int_0^\infty e^{-px}(e^{-x} - 1)^n \frac{dx}{x^j} $ be evaluated recursively, and what is the explicit formula for $ T_j $ in terms of logarithmic and polynomial terms?

Key findings

• The integral $ \int_0^1 (1 - \sqrt{x})^{p-1} dx $ evaluates to $ \frac{2}{p(p+1)} $, derived via substitution $ y = 1 - \sqrt{x} $ and direct integration.
• The generalized form $ \int_0^1 (1 - x^a)^{p-1} dx $ equals $ a^{-1} B(p, a^{-1}) $, where $ B $ is the beta function, obtained through the substitution $ t = x^a $.
• The integral $ \int_{-\infty}^\infty \frac{dx}{(1+x^2)\sqrt{b + a x^2}} $ evaluates to $ \frac{2}{\sqrt{a}} \tan^{-1}\left( \frac{\sqrt{b-a}}{\sqrt{a}} \right) $ when $ a < b $, and similar expressions for $ a = b $ and $ a > b $, with the special case $ a=3, b=4 $ yielding $ \pi/3 $.
• The integral $ \int_0^\infty \frac{dt}{(a + b t^2)^n \sqrt{1 + t^2}} $ is expressed as $ \int_0^1 \frac{(1 - v^2)^{n-1} dv}{(a + (b-a)v^2)^n} $ via the substitution $ v = t / \sqrt{1 + t^2} $, enabling recursive evaluation.
• The integral $ \int_0^\infty e^{-px}(e^{-x} - 1)^n \frac{dx}{x^j} $ is evaluated as $ \frac{(-1)^j}{(j-1)!} \sum_{k=0}^n (-1)^k \binom{n}{k} (p + n - k)^{j-1} \ln(p + n - k) $, derived via iterative integration and parameter differentiation.
• The logarithmic integral $ \int_0^\infty \ln\left( \frac{a^s + x^s}{b^s + x^s} \right) dx = (a - b) \frac{\pi}{\sin(\pi/s)} $ is proven using integration by parts and the known result $ \int_0^\infty \frac{dx}{1 + x^s} = \frac{\pi}{s \sin(\pi/s)} $.