Evaluating the 2026 MIT Integration Bee Limit: A Riemann Sum Hiding Inside an Integral

The problem comes from the 2026 MIT Integration Bee finals, as quoted in a Mathematics Stack Exchange question:

L = lim (n → ∞) ∫₀¹ Σ_{k=1}^{n} 1 / (n x² + k x + n) dx

No single term has an obvious limit, and the sum has no closed form you would want to write out. The structure gives it away, though. Every term carries the parameter n twice, and k appears only next to x. Divide the numerator and denominator by n and the ratio k/n appears, along with a factor 1/n in front. That pattern is a Riemann sum.

The answer is 5π²/72, about 0.6854. There are two clean ways to get there. One converts the sum to an integral first and then evaluates a double integral with a parameter. The other integrates each term exactly and only takes the limit at the end. Both are worth knowing, because they fail in different places.

Turning the summation into an integral

Pull a factor of n out of each denominator:

Σ_{k=1}^{n} 1/(n x² + k x + n) = (1/n) Σ_{k=1}^{n} 1/(x² + 1 + x·(k/n))

For fixed x in [0, 1], the right-hand side is a right-endpoint Riemann sum, with step 1/n, for the function t ↦ 1/(x² + 1 + x t) on [0, 1]. As n grows it converges to ∫₀¹ dt/(x² + 1 + x t). The accepted answer also points out the error to avoid at this step. After substituting t = k/n, the denominator still contains x t, not just t. If you drop the x, the variable you are integrating against disappears from the coupling term and the problem falls apart.

That gives a double integral over the unit square:

L = ∫₀¹ ∫₀¹ dt dx / (x² + 1 + x t)

The inner integral in t is elementary because the denominator is linear in t. The result is (1/x) ln((x² + x + 1)/(x² + 1)). The asker reached the same expression. It is correct, but it is not obviously integrable in closed form, and that is where most attempts stall.

Differentiating under the integral sign

Don't attack the logarithm head-on. Promote the coefficient of x inside it to a parameter a:

F(a) = ∫₀¹ (1/x) ln((x² + a x + 1)/(x² + 1)) dx,   F(0) = 0

The quantity you want is L = F(1). Differentiating with respect to a cancels the awkward 1/x, because d/da of the log is x/(x² + a x + 1):

F'(a) = ∫₀¹ dx / (x² + a x + 1)

This is a standard quadratic-denominator integral. For 0 ≤ a ≤ 1 the discriminant a² − 4 is negative, so the antiderivative is an arctangent rather than a logarithm:

F'(a) = (2/√(4 − a²)) · [ arctan((a + 2)/√(4 − a²)) − arctan(a/√(4 − a²)) ]

This is correct, but it is unpleasant to integrate in a. The square root √(4 − a²) suggests the next move.

The 2 sin θ substitution that collapses everything

Put a = 2 sin θ. Then √(4 − a²) = 2 cos θ, and a running from 0 to 1 means θ runs from 0 to π/6. The second arctangent becomes arctan(tan θ) = θ. The first becomes arctan((1 + sin θ)/cos θ). By the half-angle identity, (1 + sin θ)/cos θ = tan(π/4 + θ/2), so that term is π/4 + θ/2. The difference is π/4 − θ/2:

F'(2 sin θ) = (1/cos θ) · (π/4 − θ/2)

Now integrate F' from 0 to 1 in the new variable. The Jacobian da = 2 cos θ dθ cancels the 1/cos θ exactly:

L = F(1) = ∫₀^{π/6} (π/2 − θ) dθ = π²/12 − π²/72 = 5π²/72

All the difficulty sits in choosing the parameter and the substitution. Once both are in place, the final integral is a polynomial in θ. That ratio of setup to computation is typical of Integration Bee finals.

A second route: integrate each term first

Another answer swaps the order and works at finite n. Since 1 ≤ k ≤ n, the quantity 4n² − k² is positive, so every term has an arctangent antiderivative:

∫ dx/(n x² + k x + n) = (2/√(4n² − k²)) · arctan((2n x + k)/√(4n² − k²)) + C

Evaluate from 0 to 1 and apply the arctangent subtraction formula. The difference of two arctangents collapses to one: (2/√(4n² − k²)) · arctan(√((2n − k)/(2n + k))). Pull out 1/n and write everything in terms of k/n, and the sum over k becomes a Riemann sum for

∫₀¹ (2/√(4 − x²)) · arctan(√((2 − x)/(2 + x))) dx

This integral has an exact antiderivative, −2 arctan²(√((2 − x)/(2 + x))). One way to see it is to write x = 2 cos φ. Then (2 − x)/(2 + x) = tan²(φ/2), and the arctangent is just φ/2. At x = 1 the arctangent equals π/6, and at x = 0 it equals π/4. That gives −2(π²/36 − π²/16) = 5π²/72, matching the first method.

Where this approach does not fit, and what to watch for

Both methods swap a limit with an integral, and in a contest nobody asks you to justify it. On paper you should. The first method passes the limit inside the x-integral. That works here because the integrand 1/(x² + 1 + x t) is smooth and bounded on the closed unit square, so the Riemann sums converge uniformly in x. If the denominator could vanish anywhere on [0, 1] × [0, 1], that argument would break, and the swap would need a domination bound or would fail outright.

The arctangent forms also depend on the sign of the discriminant. F'(a) is an arctangent only while a² < 4. The second method's term-by-term antiderivative needs 4n² − k² > 0, which holds only because k never exceeds n. If you change the summation range, for example to k running up to 2n or beyond, part of the sum switches to logarithmic antiderivatives. The neat collapse to 5π²/72 no longer applies.

Finally, the parameter trick depends on choosing F so that F(0) is trivially zero and F' loses the 1/x. Put the parameter somewhere else, such as on the x² term, and F' gets no simpler. Before you invest in differentiating under the integral sign, check that the derivative actually removes the factor causing the trouble.

This article draws on Integrating: $\lim\limits_{n \to \infty} \int_0^1 \sum_{k=1}^{n} \frac{1}{nx^2 + kx + n} \, dx$ and answers to it (1, 2), on the Stack Exchange network, licensed under CC BY-SA 4.0.