REVIEW 3 major objections 4 minor 3 references
An Elementary Proof for the Basel Problem
T0 review · 3 major / 4 minor · reviewed 2026-08-07 · deepseek-v4-flash
Pith's one-line read A double integral, read two ways, proves the Basel sum π²/6.
desk verdict The main double integral is evaluated correctly, but the paper stops at the odd-square sum and never finishes the derivation of π²/6; the applications section also contains a false identity. read the letter →
The pith
A machine-rendered reading of the paper's core claim, the machinery that carries it, and where it could break.
The reading
What carries the argument
The load-bearing object is the improper double integral $I = \iint_{[0,\infty)^2} \frac{x}{(1+x^2)(y^2+x^2)}\,dx\,dy$. It carries the argument because its nonnegative integrand allows the order of integration to be swapped, and the two resulting iterated integrals link $\pi^2/4$ to the odd reciprocal-square sum. The identities that make the link work are the partial-fraction decomposition $\frac{x}{(1+x^2)(y^2+x^2)} = \frac{1}{1-y^2}\left(\frac{x}{y^2+x^2} - \frac{x}{1+x^2}\right)$ and the geometric-series expansion $\frac{1}{1-y^2} = \sum_{k=0}^\infty y^{2k}$ on $(0,1)$, together with the elementary integral $\int_0^\infty \frac{dy}{y^2+x^2} = \frac{\pi}{2x}$.
What would settle it
Numerically integrate $I_R = \int_0^R\int_0^R \frac{x}{(1+x^2)(y^2+x^2)}\,dy\,dx$ and the reversed iterated integral for increasing $R$; agreement with $\pi^2/4$ and with each other as $R\to\infty$ would support the proof, while a mismatch would pinpoint a failure of the interchange of integration, the load-bearing step.
Extended reading notes
Core claim
The central discovery is that the double integral $I = \iint_{[0,\infty)^2} \frac{x}{(1+x^2)(y^2+x^2)}\,dx\,dy$ is tractable in two orders. Integrating first in $y$ and then in $x$ gives $I = \pi^2/4$. Integrating first in $x$ using partial fractions reduces the integrand to $-\frac{\ln y}{1-y^2}$; expanding $1/(1-y^2)$ as a geometric series and integrating term by term gives $I = 2\sum_{k=0}^\infty 1/(2k+1)^2$. Since the two iterated integrals must be equal, $\sum_{k=0}^\infty 1/(2k+1)^2 = \pi^2/8$, and the standard decomposition of the full reciprocal-square sum into odd and even parts gives $\sum_{n=1}^\infty 1/n^2 = \pi^2/6$.
Load-bearing premise
The proof assumes, without stating a justification, that the order of the two improper integrations may be interchanged and that the geometric series may be integrated term by term over $(0,1)$; these steps are valid because the integrand is nonnegative, but the paper does not say so.
Editorial extensions
If this is right
- The Basel sum $\pi^2/6$ is established without Fourier series, complex analysis, or special functions.
- The proof provides the intermediate identity $\sum_{k=0}^\infty 1/(2k+1)^2 = \pi^2/8$ as a by-product.
- The same double-integral technique is used in Section 3 to derive further definite integrals, including $\int_0^1 (\ln y)^2/(1-y^2)\,dy = \pi^3/16$.
- Because the argument is elementary, it can be presented in a first course on calculus or analysis.
Reading between the lines
- Inference: The order swap can be justified rigorously by a standard convergence theorem for nonnegative integrands, such as the monotone convergence theorem; adding this justification would close the only gap in the elementary presentation.
- Inference: The same double-integral construction may generalize to other even zeta values: the Section 3 results suggest a family of identities linking integrals of powers of logarithms against $1-y^2$ to sums over odd powers.
- Inference: An instructor could turn the proof into a guided exercise: compute both iterated integrals and reconcile the two answers, making the Basel problem a self-contained discovery exercise.
Editorial analysis
A structured set of objections, weighed in public.
Referee Report
Summary. The manuscript claims to give an elementary proof of the Basel problem using only a double integral and basic calculus. The core of Section 2 computes I = ∫∫ x/((1+x^2)(y^2+x^2)) dx dy in two orders, obtaining I = π^2/4 by first integrating in y and I = 2∑_{k≥0}1/(2k+1)^2 by integrating in x, then using a geometric series. The displayed conclusion is the odd-square sum π^2/8. Section 3 then attempts applications, including an evaluation of a related integral, with the apparent goal of deriving further identities. The paper never carries out the last step needed to pass from the odd-square sum to the full Basel sum ∑ 1/n^2 = π^2/6.
Significance. If completed, the double-integral computation in Section 2 would give a short, elementary evaluation of the sum over odd squares, a standard result that is a stepping stone to the Basel value. That part of the argument is essentially correct, though it lacks justifications for interchanging limits. The advertised theorem, however, is not proved in the manuscript as written, and Section 3 contains an incorrect integral identity that undermines its applications. The paper therefore has a salvageable core but does not currently establish its central claim. No machine-checked proofs or reproducible code are provided.
major comments (3)
- [Section 2, final displays (and Abstract/Introduction)] The paper proves, at most, that ∑_{k≥0}1/(2k+1)^2 = π^2/8, but the abstract and introduction promise that ∑_{n≥1}1/n^2 = π^2/6. The identity connecting these two sums, namely ∑_{n≥1}1/n^2 = (4/3)∑_{k≥0}1/(2k+1)^2, which follows from separating even and odd terms and solving, is never stated or proved. Without this final step, the central theorem of the paper is not established.
- [Section 3, first integral identity] The identity ∫_0^∞ dx/((1+x^2)(x+z)) = (π/2 − ln z)/(1+z^2) is false. A correct evaluation is (π z/2 − ln z)/(1+z^2), which can be obtained by partial fractions and cancellation of the logarithmic divergences. Since this identity is used immediately afterward to compute the left side of the equality, all subsequent formulas in Section 3, including the equation 'π^3/16 + π^2/8 = ...', are invalid. The section must either be corrected and reworked or removed.
- [Section 2, change of order and partial fractions] Two interchanges of limiting operations are made without justification: the order of the double improper integral over R_+^2 is changed, and the geometric series is integrated term by term. The integrand is nonnegative, so Tonelli's theorem justifies both steps, but the paper never cites it or verifies the hypotheses. In addition, after the partial-fraction decomposition, the two x-integrals ∫x dx/(y^2+x^2) and ∫x dx/(1+x^2) individually diverge; only their difference is well defined. A truncation argument or a direct evaluation of ∫_0^∞ x/[(1+x^2)(y^2+x^2)] dx is needed to make this step rigorous.
minor comments (4)
- [Abstract] There is a typo in 'conv erges'; it should read 'converges'.
- [Section 3, introductory lines] The phrase 'By squaring the following integral already evaluated' is unclear: an integral is a number, and the subsequent expression is not the square of that number. The intended operation appears to be squaring the integrand, but this needs to be stated precisely.
- [Section 3, second equality] The transition from an equality over (0,∞) to an equality over (0,1) for the logarithmic integrals is not justified; the terms from the interval (1,∞) do not automatically vanish and require a substitution or a separate argument.
- [Section 4] The claim that this is 'the most elementary proof of the Basel problem ever discovered' is subjective and outside the mathematical content of the paper; it should be toned down or supported by a clear comparison with known elementary proofs.
Circularity Check
No circularity: the double-integral computation is self-contained, though the written conclusion omits the final reduction from the odd-square sum to the full Basel sum.
full rationale
The proof's load-bearing computation is Section 2's evaluation of I = ∫∫ x/((1+x²)(y²+x²)) dx dy in two orders. The first order gives π²/4 by elementary arctangent evaluations; the second order, after a partial-fraction split and geometric-series expansion, gives 2Σ(2k+1)^{-2}. Equating these yields Σ(2k+1)^{-2}=π²/8. This is an independent derivation of Euler's odd-square sum; it does not assume the Basel value π²/6, and no parameter is fitted and no prior result by the author is cited. The only step linking the derived odd-square sum to the claimed Basel value, namely (1−1/4)Σn^{-2}=Σ(2k+1)^{-2}, is never written, so the manuscript does not literally prove the stated theorem. That is a completeness gap, not a circularity. Sections 3 use only the already-proved odd-square evaluation and standard integration by parts, and the references are classical textbooks, not self-citations. Therefore there are no circular steps.
Assumptions & free parameters
assumptions (4)
- standard math Fubini-Tonelli theorem for nonnegative functions justifies changing the order of integration.
- standard math Geometric series expansion 1/(1-y^2) = Σ y^(2k) for 0<y<1.
- standard math The integral ∫_0^∞ dx/(1+x^2) = π/2.
- standard math Term-by-term integration of the geometric series is valid.
Cite this review
Pith. "Pith review of An Elementary Proof for the Basel Problem." pith.science (2026). https://pith.science/paper/JQVYQDDF
@misc{pith2026250611101,
author = {Pith},
title = {Pith review of: An Elementary Proof for the Basel Problem},
year = {2026},
howpublished = {\url{https://pith.science/paper/JQVYQDDF}},
note = {Machine review of arXiv:2506.11101}
}
read the original abstract
We present an astonishingly simple and elegant proof of the celebrated Basel problem.
Reference graph
Works this paper leans on
-
[1]
Theory and Application of Infinite Series
Knopp, Konrad. Theory and Application of Infinite Series. Dover Publications, 1990
work page 1990
-
[2]
Solutio Problematis Basileensis.(1734)
Euler, Leonhard. Solutio Problematis Basileensis.(1734)
-
[3]
Titchmarsh, E. C. The Theory of Functions.Oxford University Press, 1989
work page 1989
Reviewed August 7, 2026 · model on record in the stance chip above.
Discussion (0). Sign in to comment.