[Paper Review] Ahmed's Integral: the maiden solution

Zafar U. Ahmed|arXiv (Cornell University)|Nov 19, 2014

Mathematics and Applications3 references3 citations

TL;DR

This paper presents the original (maiden) solution to Ahmed's Integral, a famous definite integral proposed in 2001. By splitting the integral using the identity for arctangent and applying trigonometric and double integral substitutions, the authors evaluate it analytically to yield the exact result $\frac{5\pi^2}{96}$, establishing a foundational analytical approach for this well-known problem in mathematical analysis.

ABSTRACT

In 2001-2002, I happened to have proposed a new definite integral in the American Mthematical Monthly (AMM),which later came to be known in my name (Ahmed). In the meantime, this integral has been mentioned in mathematical encyclopedias and dictionaries and further it has also been cited and discussed in several books and journals. In particular, a google search with the key word "Ahmed's Integral" throws up more than 60 listings. Here I present the maiden solution for this integral.

Motivation & Objective

• To present the original analytical solution to Ahmed's Integral, proposed in 2001 and later recognized as a significant problem in mathematical literature.
• To resolve the integral $\int_{0}^{1}\frac{\tan^{-1}\sqrt{2+x^{2}}}{(1+x^{2})\sqrt{2+x^{2}}}dx$ using a novel decomposition and substitution strategy.
• To provide a self-contained derivation that avoids reliance on published solutions, fulfilling the role of the proposer’s first solution.
• To establish a benchmark analytical result for comparison with numerical quadrature methods and future extensions of the integral.

Proposed method

• Decompose the integral $I$ into $I_1 - I_2$ using the identity $\tan^{-1}z = \frac{\pi}{2} - \tan^{-1}\frac{1}{z}$.
• Apply the substitution $x = \tan\theta$ to transform $I_1$ into a trigonometric integral: $I_1 = \frac{\pi}{2}\int_{0}^{\pi/4}\frac{\cos\theta~{}d\theta}{\sqrt{2 - \sin^2\theta}}$.
• Use the substitution $\sin\theta = \sqrt{2}\sin\phi$ to evaluate $I_1 = \frac{\pi^2}{12}$.
• Express $I_2$ as a double integral using the identity $\frac{1}{a}\tan^{-1}\frac{1}{a} = \int_0^1 \frac{dx}{x^2 + a^2}$, leading to $I_2 = \int_0^1\int_0^1 \frac{dx~{}dy}{(1+x^2)(2+x^2+y^2)}$.
• Rewrite $I_2$ using partial fractions and exploit symmetry in $x$ and $y$ to derive $2I_2 = \left(\int_0^1 \frac{dx}{1+x^2}\right)^2 = \frac{\pi^2}{16}$.
• Combine results to obtain the final value: $I = \frac{\pi^2}{12} - \frac{\pi^2}{32} = \frac{5\pi^2}{96}$.

Experimental results

Research questions

• RQ1 What is the analytical value of Ahmed's Integral $\int_{0}^{1}\frac{\tan^{-1}\sqrt{2+x^{2}}}{(1+x^{2})\sqrt{2+x^{2}}}dx$?
• RQ2 How can the integral be decomposed and evaluated using trigonometric and double integral techniques?
• RQ3 Can symmetry in the double integral representation of $I_2$ be exploited to simplify the evaluation?
• RQ4 What is the role of the identity $\frac{1}{a}\tan^{-1}\frac{1}{a} = \int_0^1 \frac{dx}{x^2 + a^2}$ in transforming the original integral?
• RQ5 How does the maiden solution compare to subsequent published solutions in terms of derivation strategy and elegance?

Key findings

• The integral $I = \int_{0}^{1}\frac{\tan^{-1}\sqrt{2+x^{2}}}{(1+x^{2})\sqrt{2+x^{2}}}dx$ evaluates exactly to $\frac{5\pi^2}{96}$.
• The first component $I_1$ is evaluated as $\frac{\pi^2}{12}$ using the substitution $\sin\theta = \sqrt{2}\sin\phi$.
• The second component $I_2$ is transformed into a double integral and simplified using symmetry, yielding $I_2 = \frac{\pi^2}{32}$.
• The final result is derived from the difference $I = I_1 - I_2 = \frac{\pi^2}{12} - \frac{\pi^2}{32} = \frac{5\pi^2}{96}$.
• The solution demonstrates the power of trigonometric substitution and double integral symmetry in solving complex definite integrals.
• This analytical result serves as a benchmark for testing high-precision numerical integration methods on the same integral.