MIT 2025 Integration Bee Quarterfinal #2 Problem 1: $\lim_{A \to \infty} \int_{-\infty}^{\infty} \frac{A}{A^2 (x^3 - 3x)^2 + 1} \, dx$

Question

MIT 2025 Integration Bee Quarterfinal #2 Problem 1 (official slides)

$$\lim_{A \to \infty} \int_{-\infty}^{\infty} \frac{A}{A^2 (x^3 - 3x)^2 + 1} \, dx$$

The answer is given as $\frac{2 \pi}{3}$ in the slides. However, Karthik Vedula (the Grand Integrator of 2025) submitted $\frac{\pi}{3}$ instead, and I am unable to identify errors in his method.

The host remarked that the integral behaves like a Dirac delta function and that it is helpful to find its slopes around each of the roots of $x^3 - 3x$, namely $-\sqrt{3}, 0, \sqrt{3}$.

Answers verifying this solution and providing alternative solutions are all welcome.

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Karthik Vedula's method:

Let $y = Ax$. Then \begin{align*} &\lim_{A \to \infty} \int_{-\infty}^{\infty} \frac{A}{A^2 (x^3 - 3x)^2 + 1} \, dx \\ &= \lim_{A \to \infty} \int_{-\infty}^{\infty} \frac{dy}{\Bigl( \frac{y^3}{A^2} - 3y \Bigr)^{\! 2} + 1} \\ &= \int_{-\infty}^{\infty} \frac{dy}{9y^2 + 1} \\ &= \frac{1}{3} \arctan \frac{y}{3} \biggr|_{-\infty}^{\infty} \\ &= \frac{\pi}{3}. \end{align*} The interchange of the limit and the integral can be justified using the dominated convergence theorem. (The application of DCT is incorrect; see Sangchul Lee's answer for a comprehensive discussion.)

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Curiously, Mathematica gives the answer $- \frac{\pi}{3}$, which is clearly erroneous since the integrand is positive.

Numerical integration using $A = 10\,000\,000$, however, supports the $\frac{\pi}{3}$ result.

Answer 1

The failure of Karthik Vedula's method can be explained in many ways.

1. As other users pointed out, the use of DCT is incorrect as there is no integrable dominating function of the integrand. Indeed, for each $y \in \mathbb{R}$ there exists $A > 0$ such that

$$ \frac{y^3}{A^2} - 3y = 0. $$

Indeed, we can set $A = \frac{|y|}{\sqrt{3}}$ when $y \neq 0$, and we can choose any $A > 0$ when $y = 0$. Consequently.

$$ \sup_{A > 0} \frac{1}{\Bigl( \frac{y^3}{A^2} - 3y \Bigr)^2 + 1} = 1 $$

and this is the "minimal" dominating function for the family $\Bigl\{ \frac{1}{( (y^3/A^2) - 3y )^2 + 1} \Bigr\}_{A>0}$, which is clearly non-integrable. Hence DCT is not applicable here.

2. Of course, the failure of DCT does not necessarily imply that interchanging the order of limit and integral is not valid. So, let's take a closer look on how the integrand changes when $A$ grows. The above observation tells that the integrand has three peaks, one at $y = 0$ and the other two at $y = \pm \sqrt{3}A$. Now note that the last two peaks are pushed away towards infinity as $A \to \infty$:

Consequently, the "mass" concentrated about $\pm\sqrt{3}A$ are lost in the pointwise limit of the integrand as $A \to \infty$. This loss of mass accounts for the difference between the correct answer and the value obtained in Karthik Vedula's approach.

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Addendum. Here is a more general version of the problem:

Theorem. Assume $f : \mathbb{R} \to \mathbb{R}$ is a $C^1$-function such that the following conditions hold:

1. $\int_{|x| > r} \frac{\mathrm{d}x}{|f(x)|^2} < \infty$ for some $r > 0$,
2. $f(x)$ has only finitely many zeros $x_1, \ldots, x_N$, and
3. $f'(x_i) \neq 0$ for each $i \in [1:N]$.

Then

$$ \lim_{A \to \infty} \int_{-\infty}^{\infty} \frac{A}{A^2f(x)^2 + 1} \, \mathrm{d}x = \pi \sum_{i=1}^{N} \frac{1}{|f'(x_i)|}. $$

Intuitively, this is because $\frac{A}{(Ay)^2 + 1}$ "converges to" $\pi \delta(y)$, hence

$$ \lim_{A \to \infty} \int_{-\infty}^{\infty} \frac{A}{A^2f(x)^2 + 1} \, \mathrm{d}x = \pi \int_{-\infty}^{\infty} \delta(f(x)) \, \mathrm{d}x = \pi \sum_{i=1}^{N} \frac{1}{|f'(x_i)|}. $$

To make this claim more rigorous, let $\delta > 0$ be such that

1. the open intervals $B_\delta(x_i) = (x_i-\delta, x_i+\delta)$ are disjoint, and

2. each restriction $f_i = f|_{B_{\delta}(x_i)}$ is a $C^1$-bijection from $B_\delta(x_i)$ to its image $f(B_\delta(x_i))$ with $C^1$-inverse for each $i \in [1:N]$.

Such a $\delta > 0$ exists by the inverse function theorem. For simplicity, write $U = \bigcup_{i=1}^{N} B_\delta(x_i)$. Then

$$ \begin{align*} &\int_{-\infty}^{\infty} \frac{A}{A^2f(x)^2 + 1} \, \mathrm{d}x \\ &= \sum_{i=1}^{N} \int_{B_\delta(x_i)} \frac{A}{A^2f(x)^2 + 1} \, \mathrm{d}x + \int_{B_r(0) \setminus U} \frac{A}{A^2f(x)^2 + 1} \, \mathrm{d}x + \int_{B_r(0)^c \setminus U} \frac{A}{A^2f(x)^2 + 1} \, \mathrm{d}x, \end{align*} $$

where $r > 0$ is as in the statement of the theorem. Now we estimate each term separately:

• For each $i$, substituting $y = A f_i(x)$ yields $$ \begin{align*} \int_{B_\delta(x_i)} \frac{A}{A^2f(x)^2 + 1} \, \mathrm{d}x &= \int_{A\cdot f(B_\delta(x_i))} \frac{|(f_i^{-1})'(y/A)|}{y^2 + 1} \, \mathrm{d}y. \end{align*} $$ Note that $f(B_\delta(x_i))$ is an open interval containing $0$, hence $A\cdot f(B_\delta(x_i))$ expands to $(-\infty, \infty)$ as $A \to \infty$. So by DCT, $$ \begin{align*} \lim_{A\to\infty} \int_{B_\delta(x_i)} \frac{A}{A^2f(x)^2 + 1} \, \mathrm{d}x &= \int_{-\infty}^{\infty} \frac{|(f_i^{-1})'(0)|}{y^2 + 1} \, \mathrm{d}y = \frac{\pi}{|f'(x_i)|}. \end{align*} $$

• For the second term, note that $$ m := \inf_{x \in B_r(0) \setminus U} |f(x)| > 0. $$ (To see why, apply the extremum value theorem to the continuous function $|f|$ on the compact set $\overline{B_r(0)} \setminus U$.) Consequently, $$ \left| \int_{B_r(0) \setminus U} \frac{A}{A^2f(x)^2 + 1} \, \mathrm{d}x \right| \leq \int_{B_r(0) \setminus U} \frac{A}{A^2m^2 + 1} \, \mathrm{d}x \leq \frac{2rA}{A^2m^2 + 1} \to 0. $$

• For the last integral, $$ \int_{B_r(0)^c \setminus U} \frac{A}{A^2f(x)^2 + 1} \, \mathrm{d}x \leq \int_{B_r(0)^c} \frac{A}{A^2f(x)^2} \, \mathrm{d}x = \frac{1}{A} \int_{B_r(0)^c} \frac{\mathrm{d}x}{f(x)^2} \to 0. $$

Combining altogether, the desired claim follows.

Answer 2

To the nice posted solutions we can add one more - the one where the complex integration is used. Although such approach, probably, is not within the frames of the contest, it allows to get the answer quickly, also providing the tool for the integral evaluation with the required accuracy (i.e. smaller corrections evaluation).

Let's take $a>0$ and consider $$I(a,A)=\int_{-\infty}^\infty\frac{A}{A^2(x^3-a^2x)^2+1}dx=\int_{-\infty}^\infty\frac{\frac1A}{x^2(a^2-x^2)^2+\frac1{A^2}}dx$$ Denoting $\,\frac1A=\epsilon$ $$I(a,\epsilon)=\int_{-\infty}^\infty\frac\epsilon{x^2(a^2-x^2)^2+\epsilon^2}dx=\Im\,\int_{-\infty}^\infty\frac{dx}{x(\color{red}x^2-\color{red}a^2)-i\epsilon}dx$$ The polynomial in the denominator has three complex roots $x_1, \,x_2,\,x_3$ $$I(a,\epsilon)=\Im\,\int_{-\infty}^\infty\frac{dx}{(x-x_1)(x-x_2)(x-x_2)}dx$$ We have three equations to find the roots: $$x_1+x_2+x_3=0; \quad x_1x_2+x_2x_3+x_1x_3=-a^2;\quad x_1x_2x_3=i\epsilon$$ Solving the equations by iterations (i.e. $x=x^{(0)}+x^{(1)}+x^{(2)}+\dotsb$) $$x_1=-\frac{i\epsilon}{a^2} +O(\epsilon^3)$$ $$x_2=a+\frac{i\epsilon}{2a^2}+\frac38\frac{\epsilon^2}{a^5} +O(\epsilon^3)$$ $$x_3=-a+\frac{i\epsilon}{2a^2}-\frac38\frac{\epsilon^2}{a^5} +O(\epsilon^3)$$ Closing the contour in the upper half plane we get two simple poles (at $x_2$ and $x_3$; alternatively, closing the contour in the lower half-plane we get one pole at $x_1$ - this is your choice how to integrate). The residues evaluation is straightforward: $$I(a,\epsilon)=4\pi\,\Re\,\frac1{2a+\frac34\frac{\epsilon^2}{a^5}}\,\frac1{a+\frac38\frac{\epsilon^2}{a^5}+\frac{3i\epsilon}{2a^2}}+O(\epsilon^3)$$ $$I=\frac{2\pi}{a^2}\left(1-\frac{3\epsilon^2}{a^6}+O(\epsilon^3)\right)=\frac{2\pi}{a^2}\left(1-\frac3{a^6A^2}+O\Big(\frac1{A^3}\Big)\right)$$

Answer 3

Let $I$ be the integral given by $I=\int_{-\infty}^\infty \frac{A}{A^2(x^3-3x)^2+1}\,dx$. Then, choosing any $\delta\in (0,\sqrt3)$, we can write

$$\begin{align} I&=\int_{-\infty}^\infty \frac{A}{A^2(x^3-3x)^2+1}\,dx\\\\ &=2\int_0^\infty \frac{A}{A^2(x^3-3x)^2+1}\,dx\\\\ &=2\int_0^\delta \frac{A}{A^2(x^3-3x)^2+1}\,dx\\\\ &+2\int_\delta^{\sqrt 3-\delta} \frac{A}{A^2(x^3-3x)^2+1}\,dx\\\\ &+ 2\int_{\sqrt3 -\delta}^{\sqrt3+\delta} \frac{A}{A^2(x^3-3x)^2+1}\,dx\\\\ &+ 2\int_{\sqrt3 +\delta}^\infty \frac{A}{A^2(x^3-3x)^2+1}\,dx\tag1 \end{align}$$

As $A\to \infty$, the second and fourth terms on the right-hand side of $(1)$ vanish.

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To evaluate the first integral on the right-hand side of $(1)$, we note that near $x=0$, $A^2(x^3-3x)^2+1$ is approximately equal to $9A^2x^2+1$. Thus, we write

$$\begin{align} \int_0^\delta \frac{A}{A^2(x^3-3x)^2+1}\,dx&=\int_0^\delta \left(\frac{A}{A^2(x^3-3x)^2+1}-\frac{A}{9A^2x^2+1}\right)\,dx\\\\ &+\int_0^\delta \frac{A}{9A^2x^2+1}\,dx\\\\ &=\int_0^\delta \left(\frac{A}{A^2(x^3-3x)^2+1}-\frac{A}{9A^2x^2+1}\right)\,dx\\\\ &+\frac13 \arctan(3A\delta)\tag2 \end{align}$$

As $A\to \infty$, the integral on the right-hand side of $(2)$ vanishes (since the term in parantheses is $O(1/A)$) while the second term tends to $\pi/6$

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Similarly, we note that near $x=\sqrt3$, $A^2(x^3-3x)^2+1$ is approximately equal to $36A^2(x-\sqrt3)^2+1$. Hence, we find that the limit of the third integral on the right-hand side of $(3)$ is equal to $\pi/6$ also.

Putting it all together, we find that

$$\lim_{A\to\infty}\int_{-\infty}^\infty \frac{A}{A^2(x^3-3x)^2+1}\,dx=\frac{2\pi}3$$

as was to be shown!

Answer 4

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{{\displaystyle #1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\iverson}[1]{\left[\left[\,{#1}\,\right]\right]} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\sr}[2]{\,\,\,\stackrel{{#1}}{{#2}}\,\,\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} & \color{#44f}{\lim_{A\ \to\ \infty}\ \int_{-\infty}^{\infty}{A \over A^{2}\pars{x^{3} - 3x}^{2} + 1}\dd x} \\[5mm] = & \ \pi\lim_{A\ \to\ 0^{+}}\ \int_{-\infty}^{\infty}{A/\pi \over \pars{x^{3} - 3x}^{2} + A^{2}}\dd x \end{align} However, $\ds{{A/\pi \over \pars{x^{3} - 3x}^{2} + A^{2}}}$ is a "representation" of a Delta Dirac as $\ds{A\ \to\ 0^{+}}$ which is related to de $\ds{Poisson\ Kernel}$. Fine details are given in the above mentioned link.

Therefore, \begin{align} & \color{#44f}{\lim_{A\ \to\ \infty}\ \int_{-\infty}^{\infty}{A \over A^{2}\pars{x^{3} - 3x}^{2} + 1}\dd x} = \pi\int_{-\infty}^{\infty}\delta\pars{x^{3} - 3x}\dd x \\[5mm] = & \ \pi\int_{-\infty}^{\infty}\bracks{{\delta\pars{x} \over \verts{3x^{2} - 3}} + {\delta\pars{x + \root{3}} \over \verts{3x^{2} - 3}} + {\delta\pars{x - \root{3}} \over \verts{3x^{2} - 3}}}\dd x \\[5mm] = & \ \pi\pars{{1 \over 3} + {1 \over 6} + {1 \over 6}} = \bbx{\color{#44f}{2\pi \over 3}} \approx 2.0944 \\ & \end{align}

Answer 5

We need to evaluate the limit $$\lim_{A \to \infty} \int_{-\infty}^{\infty} \frac{A}{A^2 (x^3 - 3x)^2 + 1} \, dx$$ Let the integral be $$I(A) = \int_{-\infty}^{\infty} f_A(x) \, dx, \quad \text{where} \quad f_A(x) = \frac{A}{A^2 (x^3 - 3x)^2 + 1}$$

As $A \to \infty$, the term $A^2 (x^3 - 3x)^2$ dominates the denominator unless $x^3 - 3x = 0$. The roots of $x^3 - 3x = 0$ are $\sqrt3,0,-\sqrt3$

Thus, the function $f_A(x)$ is significant only near these roots. We approximate the integral by focusing on small neighborhoods around each root.

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Around $x=0$

Near $x = 0$, $x^3$ is negligible compared to $x$, so $x^3 - 3x \approx -3x$

Thus, the denominator becomes $A^2 (-3x)^2 + 1 = 9A^2 x^2 + 1$

So, the function near $x = 0$ is approximately, $$f_A(x) \approx \frac{A}{9A^2 x^2 + 1}$$ Let $u = 3A x$, so $du = 3A \, dx$, and $dx = \frac{du}{3A}$. The integral near $x = 0$ becomes $$\int_{-\epsilon}^{\epsilon} \frac{A}{9A^2 x^2 + 1} \, dx = \int_{-3A\epsilon}^{3A\epsilon} \frac{A}{u^2 + 1} \cdot \frac{du}{3A} = \frac{1}{3} \int_{-3A\epsilon}^{3A\epsilon} \frac{1}{u^2 + 1} \, du$$ As $A \to \infty$, $3A\epsilon \to \infty$, so the integral becomes $$\frac{1}{3} \int_{-\infty}^{\infty} \frac{1}{u^2 + 1} \, du = \frac{1}{3} \cdot \pi = \frac{\pi}{3}$$

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Around $x=\sqrt3$

Near $x = \sqrt{3}$, let $x = \sqrt{3} + h$, where $h$ is small. Then $x^3 - 3x \approx 6h$

Thus, the denominator becomes $A^2 (6h)^2 + 1 = 36A^2 h^2 + 1$

So, the function near $x = \sqrt{3}$ is approximately $$f_A(x) \approx \frac{A}{36A^2 h^2 + 1}$$ Let $u = 6A h$, so $du = 6A \, dh$, and $dh = \frac{du}{6A}$. The integral near $x = \sqrt{3}$ becomes $$\int_{-\epsilon}^{\epsilon} \frac{A}{36A^2 h^2 + 1} \, dh = \int_{-6A\epsilon}^{6A\epsilon} \frac{A}{u^2 + 1} \cdot \frac{du}{6A} = \frac{1}{6} \int_{-6A\epsilon}^{6A\epsilon} \frac{1}{u^2 + 1} \, du$$ As $A \to \infty$, $6A\epsilon \to \infty$, so the integral becomes $$\frac{1}{6} \int_{-\infty}^{\infty} \frac{1}{u^2 + 1} \, du = \frac{1}{6} \cdot \pi = \frac{\pi}{6}$$

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Around $x = -\sqrt{3}$

The situation near $x = -\sqrt{3}$ is symmetric to $x = \sqrt{3}$. By a similar argument, the contribution to the integral near $x = -\sqrt{3}$ is also $\frac{\pi}{6}$.

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The total integral $I(A)$ is approximately the sum of the contributions from around each root $$I(A) \approx \frac{\pi}{3} + \frac{\pi}{6} + \frac{\pi}{6} = \frac{\pi}{3} + \frac{\pi}{3} = \frac{2\pi}{3}$$

Since this approximation becomes exact as $A \to \infty$, we have $$\lim_{A \to \infty} I(A) = \frac{2\pi}{3} $$

Answer 6

The mistake with Karthik's method is where he interchanged the limit $A→∞$ and the integral. This interchange is not valid in this case because the integrand $$\frac{1}{\left(\frac{y^3}{A^2}-3y\right)^2+1}$$ does not converge uniformly to $\frac{1}{9y^2+1}$ as $A\to\infty$

This non-uniform convergence means you cannot simply take the limit inside the integral. The Dominated Convergence Theorem (DCT) does not apply here because there is no integrable function that dominates the integrand for all $A$.

Answer 7

Split the integral like so:

$$I = \int_{-\infty}^{-\frac{\sqrt{3}}{2}}\frac{Adx}{A^2(x^3-3x)^2+1}+ \int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}}\frac{Adx}{A^2(x^3-3x)^2+1} + \int_{\frac{\sqrt{3}}{2}}^\infty\frac{Adx}{A^2(x^3-3x)^2+1}$$

or with even symmetry we can simplify as

$$I = \int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}}\frac{Adx}{A^2(x^3-3x)^2+1} + 2\int_{\frac{\sqrt{3}}{2}}^\infty\frac{Adx}{A^2(x^3-3x)^2+1}$$

Then use the substitutions $y=Ax$ and $y=A(x-\sqrt{3})$, respectively, for the integrals

$$I = \int_{-\frac{A\sqrt{3}}{2}}^{\frac{A\sqrt{3}}{2}}\frac{dy}{\left(\frac{y^3}{A^2}-3y\right)^2+1} + 2\int_{-\frac{A\sqrt{3}}{2}}^\infty\frac{dy}{\left(\frac{y^3}{A^2}+\frac{3\sqrt{3}y^2}{A}+6y\right)^2+1}$$

and take the limits

$$\longrightarrow \int_{-\infty}^\infty \frac{dy}{9y^2+1}+\int_{-\infty}^\infty\frac{2dy}{36y^2+1} = \boxed{\frac{2\pi}{3}}$$

by dominated convergence.