Basel problem
The Basel problem is a landmark question in mathematical analysis that seeks the exact value of the infinite series \sum_{n=1}^{\infty} \frac{1}{n^2}.[1] First posed by the Italian mathematician and priest Pietro Mengoli in his 1644 work Novae quadraturae arithmeticae, seu Geometriae pars plana, the problem challenged mathematicians for nearly 90 years.[2] It gained widespread attention in the late 17th century through the Bernoulli brothers—Jakob and Johann—who approximated the sum numerically but could not find a closed form, despite their prominence in the development of calculus.[2] Named after the Swiss city of Basel, where the Bernoullis taught at the University of Basel, the problem became a touchstone for advances in infinite series and analytic techniques.[2] In 1734, at the age of 27, Leonhard Euler, then a professor in St. Petersburg, provided the first complete solution, demonstrating that the sum equals \frac{\pi^2}{6}.[1] Euler's approach equated the infinite product representation of \frac{\sin x}{x} (derived from its roots at integer multiples of \pi) with its Taylor series expansion, then compared coefficients of x^2 to isolate the desired sum.[2] Although initially controversial due to its reliance on non-rigorous manipulations of infinite series, Euler's result was later justified through more formal methods and remains a cornerstone of analytic number theory.[2] This value corresponds to \zeta(2), the Riemann zeta function evaluated at 2, which Euler extended to other even integers, revealing further connections to \pi and Bernoulli numbers.[3]Introduction and Historical Context
Problem Statement
The Basel problem concerns the evaluation of the infinite series \sum_{n=1}^\infty \frac{1}{n^2}.[1] This series, known as a p-series with p=2, converges to the finite value \frac{\pi^2}{6}.[4][1] The problem derives its name from Basel, Switzerland, the city where the Bernoulli brothers taught at the University of Basel.[5] Although the series converges by the p-series test since p=2>1, determining its exact sum proved non-trivial, as the rapid decrease of terms alone does not reveal the closed form.[4] The significance of the Basel problem lies in providing the first exact evaluation of the Riemann zeta function \zeta(2) beyond the divergent \zeta(1), thereby establishing a profound link between the transcendental number \pi and sums over reciprocals of integer squares.[6] Euler's solution in 1734 marked a breakthrough in the analysis of infinite series and analytic number theory.[7]Early Attempts and Challenges
The Basel problem, concerning the infinite sum \sum_{n=1}^{\infty} \frac{1}{n^2}, was first explicitly posed by the Italian mathematician Pietro Mengoli in 1650 within his treatise Novae quadraturae arithmeticae, seu de additione fractionum. Mengoli included the series as part of a broader discussion on quadrature and the addition of fractions but offered no method for determining its exact value, leaving it as an open challenge.[8] His formulation highlighted the problem's connection to geometric series and partial fractions, yet it garnered little immediate attention and was largely overlooked for decades. Interest revived in the late 17th century with efforts by the Bernoulli family. Jakob Bernoulli, in his 1689 dissertation Tractatus de seriebus infinitis, popularized the problem—earning it the name "Basel problem" due to his affiliation with the University of Basel—and demonstrated its convergence by comparing the series to the integral \int_1^{\infty} \frac{1}{x^2} \, dx = 1 and using the inequality \frac{1}{n^2} < \frac{1}{(n-1)n} for telescoping bounds, establishing that the sum lies between 1 and 2. However, he could not derive a closed-form expression despite extensive numerical computations.[2] His brother Johann Bernoulli, along with contemporaries like Gottfried Wilhelm Leibniz, pursued similar approaches in the early 18th century, employing integral representations and partial fraction decompositions to obtain tighter bounds, such as showing the sum exceeds \frac{\pi^2}{12}, but these yielded only approximations without an exact result. English mathematician John Wallis contributed an early numerical estimate around 1.644 in 1655 via interpolation methods related to his work on \pi, further illustrating the series' slow convergence but not its precise value.[2] These attempts underscored significant mathematical obstacles of the era. The divergent nature of the related harmonic series \sum \frac{1}{n}, which grows logarithmically without bound, contrasted sharply with the p-series convergence for p=2, complicating efforts to identify a simple closed form.[8] Moreover, the absence of rigorous frameworks for infinite products—later key to the solution—and the undeveloped state of Fourier series analysis limited analytical progress, as mathematicians relied primarily on finite approximations, integrals, and inequalities that proved insufficient for exact evaluation.Euler's Solution and Its Foundations
Euler's Infinite Product Approach
In 1735, Leonhard Euler devised an ingenious method to evaluate the sum \sum_{n=1}^\infty \frac{1}{n^2} by linking it to the analytic properties of the sine function, drawing an analogy between polynomials and their roots to infinite products. He began by expressing \sin(\pi x) in terms of its zeros at all nonzero integers, proposing the factorization \sin(\pi x) = \pi x \prod_{n=1}^\infty \left(1 - \frac{x^2}{n^2}\right). This representation posits that the sine function behaves like an "infinite-degree polynomial" with roots at x = \pm 1, \pm 2, \dots, scaled appropriately by the leading coefficient \pi to match the behavior near x=0. Euler justified this form by comparing it to the finite product for \sin(\pi x) over the first N roots, which approximates the full sine as N \to \infty, though without modern notions of uniform convergence. Dividing both sides by \pi x yields \frac{\sin(\pi x)}{\pi x} = \prod_{n=1}^\infty \left(1 - \frac{x^2}{n^2}\right). Euler then expanded the left side using the known Taylor series for sine: \sin(\pi x) = \pi x - \frac{(\pi x)^3}{3!} + \frac{(\pi x)^5}{5!} - \frac{(\pi x)^7}{7!} + \cdots, so \frac{\sin(\pi x)}{\pi x} = 1 - \frac{\pi^2 x^2}{6} + \frac{\pi^4 x^4}{120} - \cdots. The coefficient of x^2 here is -\pi^2/6. On the right side, Euler expanded the infinite product as a power series in x. The constant term is 1. For the x^2 term, it arises solely from selecting the -x^2/n^2 contribution from one factor (for each n) and the constant 1 from all others, yielding -\left( \sum_{n=1}^\infty \frac{1}{n^2} \right) x^2, since cross terms involving two or more factors contribute to x^4 or higher powers. Equating the coefficients of x^2 from both sides gives \sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}. Euler's derivation relied on this coefficient comparison, assuming the product converges appropriately and that the series expansion interchanges validly with the infinite product, steps that were intuitive but lacked the rigorous analytic foundations later provided by Weierstrass and others. Despite these informalities, the argument successfully resolved the Basel problem and inspired subsequent developments in complex analysis.[2]Generalizations via Symmetric Polynomials
Euler's heuristic approach to the Basel problem, based on the infinite product representation of the sine function, admits a natural generalization to evaluate the Riemann zeta function at all positive even integers using elementary symmetric polynomials and Newton's identities. The infinite product \frac{\sin \pi x}{\pi x} = \prod_{n=1}^\infty \left(1 - \frac{x^2}{n^2}\right) expresses the normalized sine as a generating function whose power series coefficients in powers of x^2 are determined by the elementary symmetric sums in the infinite set of variables \{1/n^2 \mid n \in \mathbb{N}\}. Specifically, expanding the product yields \prod_{n=1}^\infty \left(1 - \frac{x^2}{n^2}\right) = \sum_{k=0}^\infty (-1)^k e_k x^{2k}, where e_k is the k-th elementary symmetric polynomial evaluated at those variables, i.e., e_k = \sum_{1 \leq n_1 < \cdots < n_k} \prod_{j=1}^k \frac{1}{n_j^2}.[8] Equating this to the known Taylor series expansion \frac{\sin \pi x}{\pi x} = \sum_{k=0}^\infty \frac{(-1)^k \pi^{2k} x^{2k}}{(2k+1)!} allows comparison of coefficients: e_k = \frac{\pi^{2k}}{(2k+1)!}. To relate these elementary symmetric sums e_k to the power sums p_m = \sum_{n=1}^\infty \frac{1}{n^{2m}} = \zeta(2m), Newton's identities provide a recursive bridge. These identities express power sums in terms of elementary symmetric polynomials (or vice versa) via the relation k e_k = \sum_{j=1}^k (-1)^{j-1} e_{k-j} p_j for k \geq 1, with e_0 = 1. Starting from the Basel case (k=1), where e_1 = \zeta(2) = \pi^2 / 6, the recursion successively determines \zeta(4), \zeta(6), \dots by solving for higher p_m using the known e_k from the sine series.[8][9] This recursive process leads to an explicit closed-form expression involving Bernoulli numbers B_{2k}, which Euler introduced to encapsulate the coefficients in the expansions. The general formula is \zeta(2k) = (-1)^{k+1} \frac{B_{2k} (2\pi)^{2k}}{2 (2k)!} for positive integers k, where the Bernoulli numbers satisfy the generating function \frac{t}{e^t - 1} = \sum_{m=0}^\infty B_m \frac{t^m}{m!} and B_{2k} alternates in sign with |B_{2k}| growing rapidly. For example, applying the recursion for k=2 yields \zeta(4) = \pi^4 / 90, confirming the pattern. This generalization extends Euler's original evaluation of \zeta(2) systematically to all even arguments, revealing the deep connection between the arithmetic of zeta values and the analytic properties of the sine function.[10][9] The rigorization of this approach relies on the Weierstrass factorization theorem, which justifies the infinite product form for entire functions of finite order like sine by constructing products over zeros with convergence factors. This theorem ensures the product's validity in the complex plane, transforming Euler's formal manipulation into a convergent series expansion suitable for coefficient comparison. Without this analytic foundation, established later in the 19th century, the symmetric polynomial relations would remain heuristic.[11]Immediate Consequences and Extensions
One immediate consequence of Euler's evaluation of the Basel sum \sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6} was the derivation of the sum over even indices. By separating the terms where n is even, \sum_{n=1}^\infty \frac{1}{(2n)^2} = \frac{1}{4} \sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{24}.[2] Similarly, the sum over odd indices follows as \sum_{n=1}^\infty \frac{1}{(2n-1)^2} = \frac{\pi^2}{6} - \frac{\pi^2}{24} = \frac{\pi^2}{8}.[2] These results enabled further extensions to alternating series through simple manipulation. The Dirichlet eta function at 2, \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^2}, is obtained by \sum_{n=1}^\infty \frac{1}{(2n-1)^2} - \sum_{n=1}^\infty \frac{1}{(2n)^2} = \frac{\pi^2}{8} - \frac{\pi^2}{24} = \frac{\pi^2}{12}.[2] Euler also explored finite sums using partial fraction decompositions related to trigonometric functions, building on his infinite product for sine to approximate partial series sums.[12] Historically, Euler's success with the Basel problem inspired his development of the Euler-Maclaurin summation formula around 1738, which provided a systematic way to relate sums to integrals and approximate infinite series more broadly.[13] This formula emerged from efforts to refine the Basel sum's approximation and influenced early evaluations of related p-series. In later work, such as his 1740 paper E393, Euler connected the Basel sum's coefficients to Bernoulli numbers without a complete proof, noting patterns like \zeta(2) = \frac{\pi^2}{6} = \frac{|B_2| (2\pi)^2}{2 \cdot 2!} where B_2 = \frac{1}{6}, laying groundwork for general even zeta values.[12]Connections to Analytic Number Theory
Relation to the Riemann Zeta Function
The Riemann zeta function \zeta(s) is defined for complex numbers s with real part greater than 1 by the Dirichlet series \zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s}. This series converges absolutely in that half-plane, and the function admits a meromorphic continuation to the entire complex plane, with a single simple pole at s=1. Euler's solution to the Basel problem provides the value \zeta(2) = \frac{\pi^2}{6}, marking the first explicit evaluation of the zeta function at a positive even integer beyond the trivial cases. This result initiated further computations by Euler for \zeta(2k) at higher even positives, expressed via Bernoulli numbers as \zeta(2k) = (-1)^{k+1} \frac{B_{2k} (2\pi)^{2k}}{2 (2k)!}, for positive integers k, where B_{2k} denotes the $2k-th Bernoulli number.[14] These closed-form expressions highlight the zeta function's role in linking sums of reciprocals to transcendental constants like \pi. A key structural property is Euler's product formula, valid for \operatorname{Re}(s) > 1, \zeta(s) = \prod_p \left(1 - p^{-s}\right)^{-1}, where the product runs over all prime numbers p. Riemann later established the functional equation \zeta(s) = 2^s \pi^{s-1} \sin\left(\frac{\pi s}{2}\right) \Gamma(1-s) \zeta(1-s), which relates values at s and $1-s, enabling the analytic continuation and revealing symmetries in the function's behavior. In analytic number theory, the zeta function underpins the Prime Number Theorem, which asserts that the number of primes up to x is asymptotically \sim \frac{x}{\log x} as x \to \infty; this was proved by showing that \zeta(s) \neq 0 for \operatorname{Re}(s) = 1, s \neq 1, using the non-vanishing properties derived from the functional equation and product formula.[15] The even-integer evaluations, starting from the Basel sum, further illustrate the zeta function's connections to classical constants and its foundational importance in understanding prime distributions and series summations.Advanced Connections via Weil's Conjecture
Weil's conjecture on Tamagawa numbers, formulated in the 1950s, posits that for a simply connected semisimple algebraic group G defined over a number field k, the Tamagawa number \tau(G) = 1. This number represents the volume of the adelic quotient G(k) \backslash G(\mathbb{A}_k) with respect to a canonical Tamagawa measure constructed from local measures at each place of k.[16] A modern proof of the Basel problem emerges from applying this conjecture to G = \mathrm{SL}(2) over k = \mathbb{Q}, where the conjecture implies that the volume of \mathrm{SL}(2, \mathbb{Q}) \backslash \mathrm{SL}(2, \mathbb{A}_\mathbb{Q}) is 1. The proof sketch decomposes this global volume into a product of local volumes at the archimedean place (infinity) and non-archimedean places (primes p). The archimedean volume is computed geometrically as the volume of a fundamental domain for \mathrm{SL}(2, \mathbb{Z}) in \mathrm{SL}(2, \mathbb{R}), yielding \pi^2/6 via integration in Iwasawa coordinates with respect to the Tamagawa measure. The non-archimedean local volumes, evaluated using p-adic Haar measures and Iwasawa decompositions, produce an Euler product \prod_p (1 - p^{-2}) after normalization. Equating the product of local volumes to the global volume of 1 gives (\pi^2/6) \times \prod_p (1 - p^{-2}) = 1, so \zeta(2) = \pi^2 / 6 since \prod_p (1 - p^{-2}) = 1/\zeta(2). For \mathrm{GL}(2), the simply connected cover \mathrm{SL}(2) links the calculation to special values of associated L-functions via adelic integrals over the group, relating the determinant character to the Riemann zeta function.[17] This approach is conditional on the conjecture, which was verified for \mathrm{SL}(2) in the 1980s through explicit computations aligning with broader proofs for simply connected groups using trace formulas and endoscopic methods. The result connects the Basel problem to class number problems, as Tamagawa numbers for related groups appear in formulas for the order of the Tate-Shafarevich group and special L-values, such as in the Birch and Swinnerton-Dyer conjecture. Post-2000 confirmations, including resolutions for exceptional groups via geometric methods over function fields, have extended these links to more general motivic L-functions.[18]Elementary Analytic Proofs
Proof Using Euler's Formula and L'Hôpital's Rule
The infinite product representation of the sine function plays a central role in this elementary proof of the Basel problem, connecting the sum \sum_{n=1}^\infty \frac{1}{n^2} to the Taylor expansion behavior of \sin x near zero. Euler's formula, e^{ix} = \cos x + i \sin x, provides the exponential form \sin x = \frac{e^{ix} - e^{-ix}}{2i}, which facilitates analysis of the function's zeros and factorization properties through complex exponentials. This representation underpins the derivation of the infinite product for \sin x, expressed as \sin x = x \prod_{n=1}^\infty \left(1 - \frac{x^2}{n^2 \pi^2}\right), or equivalently, \frac{\sin x}{x} = \prod_{n=1}^\infty \left(1 - \frac{x^2}{n^2 \pi^2}\right). To evaluate the sum without relying on explicit Taylor series coefficients, consider the limit \lim_{x \to 0} \frac{\frac{\sin x}{x} - 1}{x^2}. This limit captures the quadratic term in the expansion of \frac{\sin x}{x} around zero. Substituting the infinite product yields \lim_{x \to 0} \frac{\prod_{n=1}^\infty \left(1 - \frac{x^2}{n^2 \pi^2}\right) - 1}{x^2}. For the product side, the limit simplifies directly: expanding each factor $1 - \frac{x^2}{n^2 \pi^2} shows that the linear terms vanish, and the quadratic contribution is -\sum_{n=1}^\infty \frac{1}{n^2 \pi^2}, with higher-order terms vanishing as x \to 0. Thus, \lim_{x \to 0} \frac{\prod_{n=1}^\infty \left(1 - \frac{x^2}{n^2 \pi^2}\right) - 1}{x^2} = -\frac{1}{\pi^2} \sum_{n=1}^\infty \frac{1}{n^2}. For the left side, the limit \lim_{x \to 0} \frac{\frac{\sin x}{x} - 1}{x^2} = \lim_{x \to 0} \frac{\sin x - x}{x^3} is an indeterminate form $0/0. Applying L'Hôpital's rule gives \lim_{x \to 0} \frac{\cos x - 1}{3x^2}, still $0/0. Applying it again yields \lim_{x \to 0} \frac{-\sin x}{6x}, again $0/0. A third application results in \lim_{x \to 0} \frac{-\cos x}{6} = -\frac{1}{6}. Therefore, \lim_{x \to 0} \frac{\frac{\sin x}{x} - 1}{x^2} = -\frac{1}{6}. Equating the two sides of the limit, -\frac{1}{6} = -\frac{1}{\pi^2} \sum_{n=1}^\infty \frac{1}{n^2}, so \sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}. This derivation relies solely on the product formula, limits, and repeated applications of L'Hôpital's rule, avoiding Fourier analysis or advanced complex methods.Proof via Fourier Series Expansion
One approach to solving the Basel problem utilizes the Fourier series expansion of the function f(x) = x^2 on the interval [-\pi, \pi]. Since f(x) is even, its Fourier series contains only cosine terms: f(x) = \frac{a_0}{2} + \sum_{n=1}^\infty a_n \cos(nx), where the coefficients are given by a_0 = \frac{1}{\pi} \int_{-\pi}^\pi x^2 \, dx, \quad a_n = \frac{1}{\pi} \int_{-\pi}^\pi x^2 \cos(nx) \, dx \quad (n \geq 1). These integrals leverage the orthogonality of the cosine functions over [-\pi, \pi], specifically the relations \int_{-\pi}^\pi \cos(mx) \cos(nx) \, dx = 0 for m \neq n and \int_{-\pi}^\pi \cos^2(nx) \, dx = \pi for n \geq 1, with the constant term adjusted accordingly.[19] Computing a_0: a_0 = \frac{2}{\pi} \int_0^\pi x^2 \, dx = \frac{2}{\pi} \left[ \frac{x^3}{3} \right]_0^\pi = \frac{2}{\pi} \cdot \frac{\pi^3}{3} = \frac{2\pi^2}{3}. Thus, \frac{a_0}{2} = \frac{\pi^2}{3}. For n \geq 1, integration by parts yields a_n = \frac{2}{\pi} \int_0^\pi x^2 \cos(nx) \, dx = \frac{4 (-1)^n}{n^2}, and the sine coefficients b_n = 0. Therefore, the Fourier series is x^2 = \frac{\pi^2}{3} + \sum_{n=1}^\infty \frac{4 (-1)^n}{n^2} \cos(nx), \quad x \in [-\pi, \pi]. [20] Evaluating the series at x = \pi, where f(\pi) = \pi^2 and \cos(n\pi) = (-1)^n, gives \pi^2 = \frac{\pi^2}{3} + \sum_{n=1}^\infty \frac{4 (-1)^n}{n^2} \cdot (-1)^n = \frac{\pi^2}{3} + \sum_{n=1}^\infty \frac{4}{n^2}. Rearranging terms, \pi^2 - \frac{\pi^2}{3} = \sum_{n=1}^\infty \frac{4}{n^2} \implies \frac{2\pi^2}{3} = 4 \sum_{n=1}^\infty \frac{1}{n^2} \implies \sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}. This establishes the value of the Riemann zeta function at 2, \zeta(2) = \frac{\pi^2}{6}.[19]Integral and Differentiation-Based Proofs
Proof by Differentiation Under the Integral Sign
One approach to proving the Basel problem using differentiation under the integral sign involves a parametric double integral representation. Consider the parametric integral I(a) = \int_0^1 \int_0^1 \frac{(xy)^{a-1}}{1 - xy} \, dx \, dy for \Re(a) > 0. The geometric series expansion $1/(1 - xy) = \sum_{k=0}^\infty (xy)^k allows us to write I(a) = \sum_{k=0}^\infty \int_0^1 x^{a+k-1} \, dx \int_0^1 y^{a+k-1} \, dy = \sum_{k=0}^\infty \frac{1}{(a+k)^2}, provided the interchange is justified. For a > 1, the series converges absolutely, and the integrals are positive, so the monotone convergence theorem permits the interchange. For $0 < a \leq 1, the Leibniz rule for differentiation under the integral sign justifies analytic continuation. At a = 1, I(1) = \sum_{k=1}^\infty 1/k^2 = \zeta(2). To evaluate I(a), the double integral can be computed using iterated integration. The inner integral over x gives -\frac{1}{y} \log(1 - y) for a=1, leading to I(1) = \int_0^1 \frac{-\log(1 - y)}{y} \, dy. This integral equals \zeta(2), but to obtain the value \pi^2/6 independently, it can be evaluated using methods such as the Fourier series expansion of x^2 on [-\pi, \pi], which yields \int_0^1 \frac{-\log(1 - y)}{y} \, dy = \pi^2/6. Differentiating I(a) with respect to a gives I'(a) = \int_0^1 \int_0^1 \frac{(xy)^{a-1} \log (xy)}{1 - xy} \, dx \, dy = -2 \sum_{k=0}^\infty \frac{1}{(a + k)^3} = -2 \zeta(3, a). The interchange of derivative and integral is justified by the Leibniz rule. This relates to the Apéry constant \zeta(3) at a=1, providing a connection to higher zeta values but serving as a consistency check rather than a direct proof for \zeta(2). An alternative method uses the known integral representation \int_0^{\pi/2} \log (\sin x) \, dx = -\frac{\pi}{2} \log 2, derived from the infinite product for \sin x or Fourier series. Consider the parametric integral J(b) = \int_0^{\pi/2} (\sin x)^b \, dx = \frac{\sqrt{\pi}}{2} \frac{\Gamma\left(\frac{b+1}{2}\right)}{\Gamma\left(\frac{b+2}{2}\right)}, the Wallis integral, related to the beta function via t = \sin^2 x, yielding J(b) = \frac{1}{2} B\left(\frac{b+1}{2}, \frac{1}{2}\right). Differentiating twice with respect to b under the integral sign (justified by the Leibniz rule) gives J''(b) = \int_0^{\pi/2} (\sin x)^b (\log \sin x)^2 \, dx. At b = 0, J(0) = \pi/2, J'(0) = -\frac{\pi}{2} \log 2, and J''(0) = \int_0^{\pi/2} (\log \sin x)^2 \, dx. From the beta/gamma expression, the second derivative of \log J(b) at b=0 involves the trigamma function: \frac{J''(b)}{J(b)} - \left( \frac{J'(b)}{J(b)} \right)^2 = \frac{d^2}{db^2} \log J(b) = \frac{1}{4} \psi_1\left( \frac{b+1}{2} \right) - \frac{1}{4} \psi_1\left( \frac{b+2}{2} \right) + \frac{1}{4} \left(\psi\left( \frac{b+1}{2} \right) - \psi\left( \frac{b+2}{2} \right)\right)^2. At b=0, using \psi_1(1/2) = \pi^2/2 from the trigamma reflection formula \psi_1(x) + \psi_1(1-x) = \pi^2 / \sin^2(\pi x), this yields \psi_1(1) = \zeta(2) = \pi^2/6.Cauchy's Integral-Based Proof
Cauchy's proof employs the residue theorem from complex analysis to evaluate the infinite sum \sum_{n=1}^\infty \frac{1}{n^2}. The key function considered is f(z) = \frac{\pi \cot(\pi z)}{z^2}, which is meromorphic in the complex plane. This function has simple poles at all nonzero integers z = n \in \mathbb{Z} \setminus \{0\}, where the residue at each such pole is \operatorname{Res}_{z=n} f(z) = \frac{1}{n^2}, since \pi \cot(\pi z) has simple poles with residue 1 at the integers. At z = 0, f(z) has a pole of order 3.[21] To apply the residue theorem, integrate f(z) over a sequence of expanding square contours C_N in the complex plane, where N is a positive integer and the contour has vertices at \pm (N + 1/2) \pm i (N + 1/2). This choice ensures that the contour avoids the poles at the integers by passing midway between them. For large N, the enclosed poles are at z = 0 and all integers n with |n| \leq N, n \neq 0. By the residue theorem, \oint_{C_N} f(z) \, dz = 2\pi i \left( \operatorname{Res}_{z=0} f(z) + \sum_{\substack{|n| \leq N \\ n \neq 0}} \frac{1}{n^2} \right). As N \to \infty, the integral over C_N vanishes. On the horizontal sides of the square, where \operatorname{Im}(z) is fixed at \pm (N + 1/2), |\cot(\pi z)| is bounded, and the length of each side is $2(N + 1/2), but the $1/z^2 term ensures the contribution tends to 0. Similarly, on the vertical sides, where \operatorname{Re}(z) = \pm (N + 1/2), \cot(\pi z) behaves periodically and remains bounded away from integers, again making the integral tend to 0 due to the $1/z^2 decay. Thus, $0 = 2\pi i \left( \operatorname{Res}_{z=0} f(z) + \sum_{n \neq 0} \frac{1}{n^2} \right), which implies \operatorname{Res}_{z=0} f(z) = - \sum_{n \neq 0} \frac{1}{n^2} = -2 \sum_{n=1}^\infty \frac{1}{n^2}. To compute the residue at z = 0, expand f(z) in its Laurent series around 0. The known series expansion of the cotangent function is \cot u = \frac{1}{u} - \frac{u}{3} - \frac{u^3}{45} - \cdots = \frac{1}{u} + \sum_{k=1}^\infty (-1)^k \frac{2^{2k} B_{2k}}{(2k)!} u^{2k-1}, where B_{2k} are the Bernoulli numbers. Substituting u = \pi z yields \pi \cot(\pi z) = \frac{1}{z} - \frac{\pi^2 z}{3} - \frac{\pi^4 z^3}{45} - \cdots. Dividing by z^2 gives the Laurent series for f(z): f(z) = \frac{1}{z^3} - \frac{\pi^2}{3 z} - \frac{\pi^4 z}{45} - \cdots. The residue is the coefficient of $1/z, which is -\pi^2 / 3. Therefore, -\frac{\pi^2}{3} = -2 \sum_{n=1}^\infty \frac{1}{n^2}, so \sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}. This result aligns with real-variable methods, such as differentiation under the integral sign applied to suitable parameter-dependent integrals.[21]Historical Development of Cauchy's Method
Augustin-Louis Cauchy introduced a novel approach to solving the Basel problem in 1821 through his Cours d'analyse de l'École Royale Polytechnique (Note VIII), where he employed contour integrals in the complex plane to evaluate the sum \sum_{n=1}^\infty \frac{1}{n^2}. This work predated the complete formulation of his residue theorem, relying instead on an intuitive manipulation of integrals around polygonal contours enclosing poles, inspired by partial fraction expansions similar to those used by Euler but extended via complex analysis. Cauchy's method marked a significant step in applying nascent complex variable techniques to infinite series, though it was presented as an ad hoc calculation rather than a general theory.[22][23] Cauchy's approach drew influences from Joseph-Louis Lagrange's earlier efforts in the 1770s and 1790s to sum series using integral representations, such as those involving the beta function for related sums, which emphasized analytic continuation over purely real methods. Additionally, Joseph Fourier's contemporaneous work on heat conduction and series expansions (circa 1822) provided conceptual groundwork for handling periodic functions and their singularities, elements Cauchy adapted to his contour setups. These influences helped Cauchy bridge real analysis with emerging complex methods, though his proof assumed uniform convergence without full justification.[22][23] In the 1850s, Bernhard Riemann refined Cauchy's technique, particularly in his 1859 habilitation thesis on the Riemann zeta function, where he generalized the contour integral involving \pi \cot(\pi z) to derive sums for \zeta(2k) systematically. Riemann's refinements incorporated more precise handling of residues at infinity and along infinite contours, transforming Cauchy's specific calculation into a versatile tool for analytic number theory. This evolution shifted from isolated proofs to a framework for residue-based summation formulas applicable to the zeta function at even positive integers.[24] Nineteenth-century mathematicians, including Niels Henrik Abel and Carl Gustav Jacob Jacobi, engaged in debates over the rigor of Cauchy's method, criticizing its reliance on unproven assumptions about integral convergence and the analytic continuation of meromorphic functions outside established limits. These discussions, spanning the 1820s to 1840s, spurred developments in complex analysis that addressed such gaps, culminating in the full residue theorem by the 1880s. In contemporary textbooks, Cauchy's method is adapted for pedagogical clarity, often presented with the cotangent integral setup verified via the modern residue calculus to ensure accessibility while preserving its historical elegance.[25][26]Advanced and Geometric Proofs
Proof Using Parseval's Identity
One approach to solving the Basel problem utilizes Parseval's identity from Fourier analysis, which relates the L² norm of a function to the sum of the squares of its Fourier coefficients. This identity stems from the orthogonality of the trigonometric basis functions {1, \cos(nx), \sin(nx)}{n=1}^\infty over the interval [-\pi, \pi] with respect to the inner product \langle f, g \rangle = \frac{1}{\pi} \int{-\pi}^\pi f(x) g(x) , dx. For a square-integrable function f with Fourier series \frac{a_0}{2} + \sum_{n=1}^\infty (a_n \cos(nx) + b_n \sin(nx)), where a_n = \frac{1}{\pi} \int_{-\pi}^\pi f(x) \cos(nx) , dx for n \geq 0 and b_n = \frac{1}{\pi} \int_{-\pi}^\pi f(x) \sin(nx) , dx for n \geq 1, Parseval's identity asserts that \frac{1}{\pi} \int_{-\pi}^\pi [f(x)]^2 \, dx = \frac{a_0^2}{2} + \sum_{n=1}^\infty (a_n^2 + b_n^2). This equality follows directly from the completeness of the trigonometric system in L²[-\pi, \pi] and the Pythagorean theorem in Hilbert space.[27] To apply this to the Basel problem, consider the function f(x) = x on [-\pi, \pi], which is an odd function, so a_0 = 0 and a_n = 0 for all n \geq 1. The sine coefficients are b_n = \frac{1}{\pi} \int_{-\pi}^\pi x \sin(nx) \, dx = \frac{2 (-1)^{n+1}}{n}, \quad n \geq 1. These are computed via integration by parts: \int_0^\pi x \sin(nx) , dx = -\frac{\pi (-1)^n}{n}, yielding the doubled value after normalization by 1/\pi. Thus, Parseval's identity simplifies to \frac{1}{\pi} \int_{-\pi}^\pi x^2 \, dx = \sum_{n=1}^\infty b_n^2 = \sum_{n=1}^\infty \left( \frac{2 (-1)^{n+1}}{n} \right)^2 = 4 \sum_{n=1}^\infty \frac{1}{n^2}. The left side evaluates explicitly as \int_{-\pi}^\pi x^2 \, dx = 2 \int_0^\pi x^2 \, dx = 2 \left[ \frac{x^3}{3} \right]_0^\pi = \frac{2 \pi^3}{3}, so \frac{1}{\pi} \cdot \frac{2 \pi^3}{3} = \frac{2 \pi^2}{3}. Equating both sides gives 4 \sum_{n=1}^\infty 1/n^2 = 2 \pi^2 / 3, hence \sum_{n=1}^\infty 1/n^2 = \pi^2 / 6. This derivation assumes familiarity with the convergence of the Fourier series in L² and the validity of term-by-term integration for the coefficients.[27][28]Generalizations and Recurrence Relations from Parseval
The Parseval identity applied to the Fourier series of higher-degree polynomials on the interval [- \pi, \pi] provides a systematic way to evaluate \zeta(2k) for k \geq 1. For the function f(x) = x^{2k}, which is even, the Fourier cosine series has coefficients a_n that can be computed explicitly via integration by parts, resulting in expressions of the form a_n = (-1)^n \sum_{j=1}^{k} c_j / n^{2j} for suitable constants c_j related to powers of \pi. Applying Parseval's identity then equates the L^2 norm \frac{1}{\pi} \int_{-\pi}^{\pi} x^{4k} \, dx = \frac{2 \pi^{4k}}{4k+1} to \frac{a_0^2}{2} + \sum_{n=1}^{\infty} a_n^2, where the squared coefficients expand to a linear combination of even zeta values \zeta(2), \zeta(4), \dots, \zeta(4k). This yields recurrence relations that allow computation of higher \zeta(2k) once lower values are known. These recurrences are intimately connected to Bernoulli numbers B_{2m}, which arise naturally in the explicit forms of the coefficients through the generating function for powers or the Fourier expansions of Bernoulli polynomials. The general solution to the recurrences gives the closed form \zeta(2k) = (-1)^{k+1} \frac{[B_{2k}](/page/Bernoulli_numbers) (2\pi)^{2k}}{2 (2k)!}, where B_{2k} are the Bernoulli numbers of even index (with B_2 = 1/6, B_4 = -1/30, etc.). For instance, using B_4 = -1/30 yields \zeta(4) = \frac{\pi^4}{90}. Multiple applications of integration and Parseval thus generate a chain of relations linking all even zeta values via the Bernoulli sequence. To illustrate the process starting from the Basel case (k=1), begin with the Fourier sine series for the odd function f(x) = x on [-\pi, \pi]:x = 2 \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \sin(nx).
Parseval's identity applied here gives \zeta(2) = \pi^2 / 6, as detailed in the prior section. To reach k=2, integrate the series term by term from -\pi to x (adjusting constants for periodicity and evenness):
x^2 = \frac{\pi^2}{3} + 4 \sum_{n=1}^{\infty} \frac{(-1)^n}{n^2} \cos(nx).
The coefficients are a_0 = 2\pi^2 / 3 and a_n = 4 (-1)^n / n^2 for n \geq 1, derived without invoking \zeta(2) via direct integration by parts on \int x^2 \cos(nx) \, dx. Now apply Parseval's identity to this series for f(x) = x^2:
\frac{1}{\pi} \int_{-\pi}^{\pi} x^4 \, dx = \frac{a_0^2}{2} + \sum_{n=1}^{\infty} a_n^2.
The left side evaluates to \frac{2}{\pi} \cdot \frac{2 \pi^5}{5} = \frac{2 \pi^4}{5}. The right side is
\frac{(2\pi^2 / 3)^2}{2} + \sum_{n=1}^{\infty} \left( \frac{4 (-1)^n}{n^2} \right)^2 = \frac{2 \pi^4}{9} + 16 \sum_{n=1}^{\infty} \frac{1}{n^4} = \frac{2 \pi^4}{9} + 16 \zeta(4).
Solving \frac{2 \pi^4}{5} = \frac{2 \pi^4}{9} + 16 \zeta(4) yields $16 \zeta(4) = 2 \pi^4 \left( \frac{1}{5} - \frac{1}{9} \right) = \frac{8 \pi^4}{45}, so \zeta(4) = \frac{\pi^4}{90}. This step relies on the integrated series from the k=1 case but computes \zeta(4) directly, without needing intermediate zeta values. For k>2, the coefficients for x^{2k} include lower-order terms (e.g., $1/n^2, 1/n^4, \dots), producing equations that couple multiple zeta values and require solving the resulting system recursively. In a modern perspective, Parseval's identity arises as a consequence of the completeness of the trigonometric system in the Hilbert space L^2([-\pi, \pi]) with inner product \langle f, g \rangle = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(x) \overline{g(x)} \, dx. The normalized basis functions \{1/\sqrt{2\pi}, \sqrt{2/\pi} \cos(nx), \sqrt{2/\pi} \sin(nx)\}_{n=1}^{\infty} satisfy \|f\|^2 = \sum | \langle f, e_j \rangle |^2, which, upon rescaling to the classical form, enables these zeta evaluations as sums of squared Fourier coefficients in the orthonormal expansion. This framework unifies the method with broader operator theory and functional analysis.