2026 年麻省理工积分决赛

问题 1:

$\lim_{n\to \infty}\int_0^1\left(\sum_{k=1}^{n}\frac1{nx^2+kx+n}\right)\ud x.$

提示

用定积分的定义,

$\begin{align*} \lim_{n\to\infty}\int_{0}^{1}\left(\sum_{k=1}^{n}\frac{1}{nx^{2}+kx+n}\right)\ud x & =\int_{0}^{1}\lim_{n\to\infty}\frac{1}{n}\left(\sum_{k=1}^{n}\frac{1}{x^{2}+\frac{k}{n}x+1}\right)\ud x\\ & =\int_{0}^{1}\int_{0}^{1}\frac{1}{x^{2}+xy+1}\ud y\ud x\\ & =\int_{0}^{1}\frac{1}{x}\ln\left(\frac{x^{2}+x+1}{x^{2}+1}\right)\ud x\\ & =\int_{0}^{1}\frac{\ln\left(x^{2}+x+1\right)}{x}\ud x-\int_{0}^{1}\frac{\ln\left(x^{2}+1\right)}{x}\ud x. \end{align*}$

令

$J(a)=\int_{0}^{1}\frac{\ln\left(x^{2}+ax+1\right)}{x}\ud x\Longrightarrow J'(a)=\int_{0}^{1}\frac{1}{x^{2}+ax+1}\ud x=\frac{2}{\sqrt{4-a^{2}}}\arctan\frac{\sqrt{4-a^{2}}}{2+a}.$

所以原式等于求

$\begin{align*} J(1)-J(0) & =\int_{0}^{1}J'(a)\ud a=\int_{0}^{1}\frac{2}{\sqrt{4-a^{2}}}\arctan\frac{\sqrt{4-a^{2}}}{2+a}\ud a\\ & \xlongequal{a=2\cos\theta}\int_{\pi/3}^{\pi/2}\theta\ud\theta=\frac{5\pi^{2}}{72}. \end{align*}$

问题 2: 求解

$\int\frac{\ud x}{(x-1)\sqrt[4]{x^3+x}}.$

提示

注意到

$\begin{align*} \int\frac{\ud x}{(x-1)\sqrt[4]{x^{3}+x}} & =8^{1/4}\int\frac{\ud x}{(x-1)\sqrt[4]{\left(x+1\right)^{4}-\left(x-1\right)^{4}}}\\ & =8^{1/4}\int\frac{\ud x}{\left(x-1\right)^{2}\sqrt[4]{\left(\frac{x+1}{x-1}\right)^{4}-1}}\\ & \xlongequal{y=\frac{x+1}{x-1}}-2^{-1/4}\int\frac{\ud y}{\sqrt[4]{y^{4}-1}}\\ & \xlongequal{z=\frac{\sqrt[4]{y^{4}-1}}{y}}2^{-1/4}\int\frac{z^{2}}{z^{4}-1}\ud z\\ & =\frac{1}{4\sqrt[4]{2}}\left[\ln\left|\frac{x+1-\sqrt[4]{8x^{3}+8x}}{x+1+\sqrt[4]{8x^{3}+8x}}\right|+2\arctan\left(\frac{\sqrt[4]{8x^{3}+8x}}{x+1}\right)\right]+C. \end{align*}$

问题 3: 求

$\lim_{n\to\infty}\int_{0}^{2026}\underbrace{\log_{\sqrt{2}}\left(x+\log_{\sqrt{2}}\left(x+\cdots+\log_{\sqrt{2}}\left(x+2026\right)\cdots\right)\right)}_{n\text{个}\log}\ud x.$

提示

记被积函数为 $f_{n}$, 则有递推关系 $f_{n+1}(x)=\log_{\sqrt{2}}\left(x+f_{n}(x)\right)$.
可以证明, 对于任意固定的 $x\in\left[0,2026\right]$, $f_{n}(x)$ 关于 $n$ 单调有界, 所以有极限函数 $f(x)$, 且满足 $2^{f/2}=x+f$. 积分的极限化为

$\lim_{n\to\infty}\int_{0}^{2026}f\ud x=\lim_{n\to\infty}\int_{0}^{2026}f\ud\left(2^{f/2}-f\right)=\lim_{n\to\infty}\int_{4}^{22}\left(2^{f/2}f\cdot\frac{\ln2}{2}-f\right)\ud f=44806-\frac{4088}{\ln2}.$

问题 4: 求

$\int_{0}^{1/4}\left(\left\lfloor \sqrt[4]{\frac{1}{x}-4}\right\rfloor ^{2}+\left\lfloor \sqrt[4]{\frac{1}{x}-4}\right\rfloor \right)\ud x.$

提示

变量替换,

$\begin{align*} \int_{0}^{1/4}\left(\left\lfloor \sqrt[4]{\frac{1}{x}-4}\right\rfloor ^{2}+\left\lfloor \sqrt[4]{\frac{1}{x}-4}\right\rfloor \right)\ud x & \xlongequal{y=\sqrt[4]{\frac{1}{x}-4}}\int_{0}^{\infty}\frac{\lfloor t\rfloor^{2}+\lfloor t\rfloor}{(t^{4}+2)^{2}}\ud t\\ & =\sum_{k=1}^{\infty}k(k+1)\left(\frac{1}{k^{4}+4}-\frac{1}{(k+1)^{4}+4}\right)\\ & =\frac{1}{4}\sum_{k=1}^{\infty}\left(-\frac{k}{k^{2}+1}+\frac{k+1}{(k-1)^{2}+1}-\frac{k+1}{(k+1)^{2}+1}+\frac{k}{(k+2)^{2}+1}\right)\\ & =\frac{3}{4}. \end{align*}$

问题 5: 求

$\int_{-\infty}^{\infty}\left(\left(\frac{1}{x-2}+\frac{3}{x-4}+\frac{5}{x-6}\right)^{-2}+1\right)^{-1}\ud x.$

提示

经计算

$u(x)\coloneqq\left(\frac{1}{x-2}+\frac{3}{x-4}+\frac{5}{x-6}\right)^{-1}=\frac{x}{9}-\frac{44}{81}-\frac{8(19x-64)}{81\left(9x^{2}-64x+100\right)},$

所以被积函数为

$\left(u^{2}(x)+1\right)^{-1}=\frac{1}{1+\left(\frac{x}{9}-\frac{44}{81}-\cdots\right)^{2}}.$

应用 Glasser 主定理,

$\begin{align*} \int_{-\infty}^{\infty}\left(\left(\frac{1}{x-2}+\frac{3}{x-4}+\frac{5}{x-6}\right)^{-2}+1\right)^{-1}\ud x & =9\int_{-\infty}^{\infty}\left(u^{2}(x)+1\right)^{-1}\ud\left(\frac{x}{9}\right).\\ & =\int_{-\infty}^{\infty}\frac{1}{1+u^{2}}\ud u\\ & =9\pi. \end{align*}$