2026 年麻省理工积分初赛

问题 1: 求积分

$\int_{-\pi}^{\pi}\sin^{2025}x\cos^{2026}x\ud x.$

提示

被积函数是奇函数, 答案是 $0$.

问题 2: 求积分

$\int\ue^{2026\ue^{x}+x}\ud x.$

提示

$\begin{align*} \int\ue^{2026\ue^{x}+x}\ud x & =\int\ue^{2026\ue^{x}}\ue^{x}\ud x\\ & =\int\ue^{2026\ue^{x}}\ud\ue^{x}\\ & =\frac{\ue^{2026\ue^{x}}}{2026}+C. \end{align*}$

问题 3: 求积分

$\int_{0}^{2026}\left\{ \frac{\lfloor x\rfloor}{3}\right\} \ud x.$

提示

$\begin{align*} \int_{0}^{2026}\left\{ \frac{\lfloor x\rfloor}{3}\right\} \ud x & =\sum_{k=1}^{2025}\int_{k}^{k+1}\left\{ \frac{k}{3}\right\} \ud x\\ & =\sum_{k=1}^{2025}\left\{ \frac{k}{3}\right\} =675. \end{align*}$

问题 4: 求积分

$\int_{-1}^{1}\underbrace{\left|x+\left|x+\cdots+\left|x+\left|x\right|\right|\right|\right|}_{2026\ \text{个}x}\ud x.$

提示

$\text{原式}=\int_{0}^{1}2026x\ud x=1013.$

问题 5: 求积分

$\int\frac{\ud x}{\sqrt{x+1}-\sqrt{x-1}}.$

提示

$\int\frac{\ud x}{\sqrt{x+1}-\sqrt{x-1}}=\frac{1}{3}\left(\left(x+1\right)^{3/2}+\left(x-1\right)^{3/2}\right)+C.$

问题 6: 求积分

$\int\sqrt{1+\cosh x}\ud x.$

提示

$\int\sqrt{1+\cosh x}\ud x=\int\sqrt{2\cosh^{2}\frac{x}{2}}\ud x=2\sqrt{2}\sinh\frac{x}{2}+C.$

问题 7: 求积分

$\int\frac{2^{\log(x)}}{x^{2}}\ud x.$

提示

$\int\frac{2^{\log(x)}}{x^{2}}\ud x=\int\frac{x^{\ln2}}{x^{2}}\ud x=\frac{x^{\ln2-1}}{\ln2-1}+C.$

问题 8: 求积分

$\int_{0}^{1/2}\left(\sum_{n=2}^{\infty}x^{n}\right)\ud x.$

提示

$\int_{0}^{1/2}\left(\sum_{n=2}^{\infty}x^{n}\right)\ud x=\int_{0}^{1/2}\left(\frac{1}{1-x}-1-x\right)\ud x=\ln2-\frac{5}{8}.$

问题 9: 求积分

$\int x^{2}\sin x\ud x.$

提示

$\int x^{2}\sin x\ud x=-x^{2}\cos x+2x\sin x+2\cos x+C.$

问题 10: 求积分

$\int\frac{\left(x-1\right)^{2}}{2\ue^{x}+x^{2}+1}\ud x.$

提示

$\int\frac{\left(x-1\right)^{2}}{2\ue^{x}+x^{2}+1}\ud x=\int\frac{\ue^{-x}\left(x-1\right)^{2}}{2+\ue^{-x}\left(x^{2}+1\right)}\ud x=-\int\frac{\ud\left(2+\ue^{-x}\left(x^{2}+1\right)\right)}{2+\ue^{-x}\left(x^{2}+1\right)}=-\ln\left(2+\ue^{-x}\left(x^{2}+1\right)\right)+C.$

问题 11: 求积分

$\int_{-1}^{1}\max\left(0,\sqrt{1-x^{2}}-\frac{1}{2}\right)\ud x.$

提示

$\frac{\pi}{3}-\frac{\sqrt{3}}{4}$.

问题 12: 求积分

$\int_{0}^{1}\sqrt{x^{2}+x+\sqrt{x^{2}+x+\sqrt{\cdots}}}\ud x.$

提示

根据单调收敛定理, 记被积函数为 $y$, 则满足 $y=\sqrt{x^{2}+x+y}$, 解得 $y=x+1$. 原积分结果为 $\frac{3}{2}$.

问题 13: 求积分

$\int\left(\cos^{5}x-10\cos^{3}x\sin^{2}x+5\cos x\sin^{4}x\right)\ud x.$

提示

注意到

$\frac{\left(\cos x+\ui\sin x\right)^{5}+\left(\cos x-\ui\sin x\right)^{5}}{2}=\cos^{5}x-10\cos^{3}x\sin^{2}x+5\cos x\sin^{4}x.$

所以

$\text{原式}=\int\frac{\ue^{5\ui x}+\ue^{-5\ui x}}{2}\ud x=\int\cos(5x)\ud x=\frac{\sin(5x)}{5}+C.$

问题 14: 求积分

$\int\arctan\sqrt{x}\ud x.$

提示

$\int\arctan\sqrt{x}\ud x\xlongequal{t=\sqrt{x}}2\int t\arctan t\ud t=(x+1)\arctan\sqrt{x}-\sqrt{x}+C.$

问题 15: 求积分

$\int_{0}^{1000}\left(\left\lfloor \left\lceil x\right\rceil \right\rfloor +\left\lceil \left\lfloor x\right\rfloor \right\rceil +\left\lfloor \left\{ x\right\} \right\rfloor +\left\lceil \left\{ x\right\} \right\rceil +\left\{ \left\lceil x\right\rceil \right\} \right)\ud x.$

提示

$\text{原式}=\sum_{k=1}^{1000}\int_{k-1}^{k}\left(k+(k-1)+0+1+0\right)\ud x=1001000.$

问题 16: 求积分

$\int\sqrt{\frac{\cos x\cot x\csc x}{\sin x\tan x\sec x}}\ud x.$

提示

$\int\sqrt{\frac{\cos x\cot x\csc x}{\sin x\tan x\sec x}}\ud x=\int\cot^{2}x\ud x=-x-\cot x+C.$

问题 17: 求积分

$\int_{-\infty}^{\infty}\frac{\ue^{-x^{2}}}{1+\ue^{2x}}\ud x.$

提示

$\int_{-\infty}^{\infty}\frac{\ue^{-x^{2}}}{1+\ue^{2x}}\ud x\xlongequal{x\to-x}\frac{1}{2}\left(\int_{-\infty}^{\infty}\frac{\ue^{-x^{2}}}{1+\ue^{2x}}\ud x+\int_{-\infty}^{\infty}\frac{\ue^{-x^{2}}}{1+\ue^{-2x}}\ud x\right)=\int_{0}^{\infty}\ue^{-x^{2}}\ud x=\frac{\sqrt{\pi}}{2}.$

问题 18: 求积分

$\int\left(\frac{\sin^{2}x}{x^{2}}-\frac{\sin(2x)}{x}\right)\ud x.$

提示

$\begin{align*} \int\left(\frac{\sin^{2}x}{x^{2}}-\frac{\sin(2x)}{x}\right)\ud x & =\int\frac{\sin^{2}x}{x^{2}}\ud x-\int\frac{\ud\sin^{2}x}{x}\\ & =\int\frac{\sin^{2}x}{x^{2}}\ud x-\frac{\sin^{2}x}{x}-\int\frac{\sin^{2}x}{x^{2}}\ud x\\ & =-\frac{\sin^{2}x}{x}+C. \end{align*}$

问题 19: 求积分

$\int\frac{\ln\left(\ln x\right)\ln\left(\ln\left(\ln x\right)\right)}{x\ln x}\ud x.$

提示

问题 20: 求积分

$\int\frac{\ln\left(\ln x\right)\ln\left(\ln\left(\ln x\right)\right)}{x\ln x}\ud x.$

提示

$\begin{align*} \int\frac{\ln\left(\ln x\right)\ln\left(\ln\left(\ln x\right)\right)}{x\ln x}\ud x & =\int\frac{\ln\left(\ln x\right)\ln\left(\ln\left(\ln x\right)\right)}{\ln x}\ud\ln x\\ & =\int\ln\left(\ln x\right)\ln\left(\ln\left(\ln x\right)\right)\ud\ln\left(\ln x\right)\\ & \xlongequal{t=\ln\left(\ln x\right)}\int t\ln t\ud t\\ & =\frac{\left(\ln\left(\ln x\right)\right)^{2}\left(2\ln\left(\ln\left(\ln x\right)\right)-1\right)}{4}+C. \end{align*}$

问题 21: 求积分

$\int_{0}^{\pi/2}\cos^{2}\left(\frac{\pi}{2}\cos^{2}\left(\frac{\pi}{2}\cos^{2}(x)\right)\right)\ud x.$

提示

$\begin{align*} \int_{0}^{\pi/2}\cos^{2}\left(\frac{\pi}{2}\cos^{2}\left(\frac{\pi}{2}\cos^{2}(x)\right)\right)\ud x & \xlongequal{x\to\frac{\pi}{2}-x}\int_{0}^{\pi/2}\sin^{2}\left(\frac{\pi}{2}\cos^{2}\left(\frac{\pi}{2}\cos^{2}x\right)\right)\ud x\\ & =\frac{1}{2}\int_{0}^{\pi/2}\ud x=\frac{\pi}{4}. \end{align*}$

2026 年麻省理工积分常规赛

问题 1: 求积分

$\int_{0}^{2026}\left(\sum_{k=0}^{\infty}\frac{x^{2026k+2025}}{k!}\right)\ud x.$

提示

$\int_{0}^{2026}\left(\sum_{k=0}^{\infty}\frac{x^{2026k+2025}}{k!}\right)\ud x=\int_{0}^{2026}x^{2025}\ue^{x^{2026}}\ud x=\frac{\ue^{2026^{2026}}-1}{2026}.$

问题 2: 求积分

$\int_{0}^{\ln34}\left(2+\frac{1}{2\ue^{x}-1}+\frac{1}{3\ue^{x}-1}\right)\ud x.$

提示

$\begin{align*} \int_{0}^{\ln34}\left(2+\frac{1}{2\ue^{x}-1}+\frac{1}{3\ue^{x}-1}\right)\ud x & =\int_{0}^{\ln34}\left(\frac{2\ue^{x}}{2\ue^{x}-1}+\frac{3\ue^{x}}{3\ue^{x}-1}\right)\ud x\\ & =\left[\ln\left|2\ue^{x}-1\right|+\ln\left|3\ue^{x}-1\right|\right]_{0}^{\ln34}=\ln\frac{6767}{2}. \end{align*}$

问题 3: 求积分

$\int_{0}^{1/2}\left(\cos(\pi x)-\pi\left(\frac{1}{4}-x^{2}\right)\left(\frac{5}{4}-x^{2}\right)\right)\ud x.$

提示

$\frac{1}{\pi}-\frac{\pi}{10}$.

问题 4: 求积分

$\int_{\int_{\int_{\int_{0}^{1}x\ud x}^{1}x\ud x}^{1}x\ud x}^{\int_{0}^{\int_{0}^{\int_{0}^{1}x\ud x}x\ud x}x\ud x}x\ud x.$

提示

$\int_{\int_{\int_{\int_{0}^{1}x\ud x}^{1}x\ud x}^{1}x\ud x}^{\int_{0}^{\int_{0}^{\int_{0}^{1}x\ud x}x\ud x}x\ud x}x\ud x=\int_{\int_{\int_{1/2}^{1}x\ud x}^{1}x\ud x}^{\int_{0}^{\int_{0}^{1/2}x\ud x}x\ud x}x\ud x=\int_{\int_{3/8}^{1}x\ud x}^{\int_{0}^{1/8}x\ud x}x\ud x=\int_{\frac{55}{128}}^{\frac{1}{128}}x\ud x=-\frac{189}{2048}.$

问题 5: 求积分

$\int\frac{4x^{2}-1}{x^{2}(x^{2}-1)}\ud x.$

提示

$-\frac{1}{x}+\frac{3}{2}\ln\left(\frac{x-1}{x+1}\right)+C$.

问题 6: 求积分

$\int\frac{\ud x}{27x-x^{-1/3}}.$

提示

$\int\frac{\ud x}{27x-x^{-1/3}}=\int\frac{x^{1/3}\ud x}{27x^{4/3}-1}=\frac{1}{36}\int\frac{\ud27x^{4/3}-1}{27x^{4/3}-1}=\frac{1}{36}\ln\left(27x^{4/3}-1\right)+C.$

问题 7: 求积分

$\int\frac{2\sqrt{x}+1}{\sqrt{x^{2}+x\sqrt{x}}}\ud x.$

提示

$\int\frac{2\sqrt{x}+1}{\sqrt{x^{2}+x\sqrt{x}}}\ud x\xlongequal{t=\sqrt{x}}2\int\frac{2t+1}{\sqrt{t^{2}+t}}\ud t=4\sqrt{x+\sqrt{x}}+C.$

问题 8: 求积分

$\int\frac{\ue^{x}}{\ue^{2x}-\ue^{-2x}}\ud x.$

提示

$\int\frac{\ue^{x}}{\ue^{2x}-\ue^{-2x}}\ud x\xlongequal{t=\ue^{x}}\int\frac{\ud t}{t^{2}-t^{-2}}=\frac{1}{2}\left(\arctan\ue^{x}-\mathrm{arctanh}\ue^{x}\right)+C.$

问题 9: 求积分

$\int_{0}^{\pi}\frac{\left|\sin x+\sin2x+\sin3x+\sin4x\right|}{\sin x+\sin2x+\sin3x+\sin4x}\ud x.$

提示

$\begin{align*} \int_{0}^{\pi}\frac{\left|\sin x+\sin2x+\sin3x+\sin4x\right|}{\sin x+\sin2x+\sin3x+\sin4x}\ud x & =\int_{0}^{\pi}\mathrm{sgn}\left(\sin x+\sin2x+\sin3x+\sin4x\right)\ud x\\ & =\int_{0}^{\pi}\mathrm{sgn}\left(\sin\frac{5x}{2}\cos x\cos\frac{x}{2}\right)\ud x\\ & =\left(\int_{0}^{2\pi/5}-\int_{2\pi/5}^{\pi/2}+\int_{\pi/2}^{4\pi/5}-\int_{4\pi/5}^{\pi}\right)\ud x=\frac{2\pi}{5}. \end{align*}$

问题 10: 求积分

$\int_{0}^{1}\frac{1-x}{\sqrt{\ue^{2x}-x^{2}}}\ud x.$

提示

$\begin{align*} \int_{0}^{1}\frac{1-x}{\sqrt{\ue^{2x}-x^{2}}}\ud x & =\int_{0}^{1}\frac{\ue^{-x}\left(1-x\right)}{\sqrt{1-(x\ue^{-x})^{2}}}\ud x\\ & =\int_{0}^{1}\frac{\ud\left(x\ue^{-x}\right)}{\sqrt{1-(x\ue^{-x})^{2}}}=\arcsin(x\ue^{-x})|_{0}^{1}=\arcsin\frac{1}{\ue}. \end{align*}$

问题 11: 求积分

$\int_{0}^{2026}\left\{ x+\frac{1}{2}\left\lfloor \frac{x}{2}\right\rfloor +\frac{1}{3}\left\lfloor \frac{x}{3}\right\rfloor +\cdots\right\} \ud x.$

提示

设 $f(x)=x+\frac{1}{2}\left\lfloor \frac{x}{2}\right\rfloor +\frac{1}{3}\left\lfloor \frac{x}{3}\right\rfloor +\cdots$, 设 $n\in\NN$, 则对于任意的 $x\in[n,n+1)$, 有 $\left\lfloor \frac{x}{k}\right\rfloor =\left\lfloor \frac{n}{k}\right\rfloor $. 所以, 在区间 $[n,n+1)$ 上, $f(x)=x+C_{n}$, 有

$\text{原积分}=\sum_{n=0}^{2025}\int_{n}^{n+1}\left\{ x+C_{n}\right\} \ud x=\sum_{n=0}^{2025}\frac{1}{2}=1013.$

问题 12: 求积分

$\int_{1/2}^{2}\frac{x^{8}}{x^{8}-x^{6}+x^{4}-x^{2}+1}\ud x.$

提示

$\begin{align*} \int_{1/2}^{2}\frac{x^{8}}{x^{8}-x^{6}+x^{4}-x^{2}+1}\ud x & =\int_{1/2}^{2}\frac{x^{8}(x^{2}+1)}{(x^{10}+1)}\ud x\\ & \xlongequal{t=x^{-1}}\int_{1/2}^{2}\frac{1+t^{-2}}{1+t^{10}}\ud t\\ & =\frac{1}{2}\int_{1/2}^{2}\frac{1+t^{-2}}{1+t^{10}}\left(1+t^{10}\right)\ud t\\ & =\frac{3}{2}. \end{align*}$

问题 13: 求积分

$\int\left(\mathrm{arcsinh} x\right)^{2}\ud x.$

提示

$\begin{align*} \int\left(\mathrm{arcsinh} x\right)^{2}\ud x & =x\left(\mathrm{arcsinh} x\right)^{2}-2\int\frac{x\mathrm{arcsinh} x}{\sqrt{1+x^{2}}}\ud x\\ & =x\mathrm{arcsinh}^{2}x-2\sqrt{1+x^{2}}\mathrm{arcsinh} x+2\int1\ud x\\ & =x\mathrm{arcsinh}^{2}x-2\sqrt{1+x^{2}}\mathrm{arcsinh} x+2x+C. \end{align*}$

问题 14: 求积分

$\int\cos^{2}(2026x)\cos(1013x)\ud x.$

提示

$\frac{\sin(1013x)}{2026}+\frac{\sin(3039x)}{12156}+\frac{\sin(5065x)}{20260}$.

问题 15: 求积分

$\int\ue^{x}\arcsin\left(\tanh x\right)\ud x.$

提示

$\begin{align*} \int\ue^{x}\arcsin\left(\tanh x\right)\ud x & \xlongequal{t=\ue^{x}}\int\arcsin\frac{t^{2}-1}{t^{2}+1}\ud t\\ & =t\arcsin\frac{t^{2}-1}{t^{2}+1}-\int\frac{2t}{t^{2}+1}\ud t\\ & =\ue^{x}\arcsin\left(\tanh x\right)-\ln(\ue^{2x}+1)+C. \end{align*}$

问题 16: 求积分

$\int\frac{x\left(\left(1-x^{2}\right)\cos x+2x\sin x\right)}{\left(1+x^{2}\right)^{2}}\ud x.$

提示

用凑微分和分部积分

$\begin{align*} \text{原积分} & =\frac{1}{2}\int\frac{\left(\left(1-x^{2}\right)\cos x+2x\sin x\right)}{\left(1+x^{2}\right)^{2}}\ud(1+x^{2})\\ & =-\frac{1}{2}\int\left(\left(1-x^{2}\right)\cos x+2x\sin x\right)\ud(1+x^{2})^{-1}\\ & =-\frac{1}{2}\left(\frac{\left(1-x^{2}\right)\cos x+2x\sin x}{1+x^{2}}-\int\sin x\ud x\right)\\ & =-\frac{x\sin x+\cos x}{1+x^{2}}+C. \end{align*}$

问题 17: 求积分

$\int_{0}^{1}\sin\left(\left(x-1\right)\left(5x-1\right)\right)\sin\left(x\left(3x-2\right)\right)\ud x.$

提示

用积化和差公式

$\begin{align*} \text{原积分} & =\frac{1}{2}\int_{0}^{1}\cos\left(2(x-1)^{2}-1\right)\ud x-\frac{1}{2}\int_{0}^{1}\cos\left(2(2x-1)^{2}-1\right)\ud x\\ & =\frac{1}{2}\int_{0}^{1}\cos\left(2x^{2}-1\right)\ud x-\frac{1}{4}\int_{-1}^{1}\cos\left(2x^{2}-1\right)\ud x\\ & =0. \end{align*}$

问题 18: 求积分

$\int_{1/4}^{\sqrt{2}}\frac{\ud x}{x\left(x^{x^{x^{\cdots}}}\ln x-1\right)}.$

提示

迭指数在区间 $[\ue^{-\ue},\ue^{1/\ue}]$ 上收敛¹, 设收敛极限为 $y$, 则有 $y=x^{y}$, 于是有 $y’=\frac{y^{2}}{x\left(1-y\ln x\right)}$,

$\int_{1/4}^{\sqrt{2}}\frac{\ud x}{x(y\ln x-1)}=-\int_{1/4}^{\sqrt{2}}\frac{y'(x)\ud x}{y^{2}(x)}=-\int_{1/2}^{2}\frac{\ud y}{y^{2}}=-\frac{3}{2}.$

问题 19: 求积分

$\int_{0}^{1}\frac{1+\left((2026+2x-x^{2})^{2}-2\right)\left(2026+4x-4x^{2}\right)^{2}}{\left(2025+2x-x^{2}\right)\left(2027+2x-x^{2}\right)\left(2025+4x-4x^{2}\right)\left(2027+4x-4x^{2}\right)}\ud x.$

提示

记 $A=(2025+2x-x^{2})(2027+2x-x^{2})$, $B=(2025+4x-4x^{2})(2027+4x-4x^{2})$, 于是原积分等于

$\int_{0}^{1}\left(\frac{1+(A-1)(B+1)}{AB}\right)\ud x=\int_{0}^{1}\left(1+\frac{1}{A}-\frac{1}{B}\right)\ud x.$

再记 $f(x)=\frac{1}{\left(2026-t^{2}\right)\left(2028-t^{2}\right)}$, 则有 $\frac{1}{A}=f(1-x)$, $\frac{1}{B}=f(2x-1)$. 由于 $f$ 是偶函数, 所以

$\int_{0}^{1}f(2x-1)\ud x=\frac{1}{2}\int_{-1}^{1}f(x)\ud x=\int_{0}^{1}f(x)\ud x=\int_{0}^{1}f(1-x)\ud x.$

因此上述积分结果为 $1$.

问题 20: 求积分

$\int_{1/2}^{1}4^{x-1}\left(1+4^{4^{x-1}-1}\left(1+4^{4^{4^{x-1}-1}-1}\left(1+\cdots\right)\right)\right)\ud x.$

提示

积分重现法

$\begin{align*} \text{原积分}=I & \xlongequal{t=4^{x-1}}\int_{1/2}^{1}t\left(1+4^{t-1}\left(1+4^{4^{t}-1}\left(1+\cdots\right)\right)\right)\frac{\ud t}{\ln4\cdot t}\\ & =\frac{1}{\ln4}\int_{1/2}^{1}1+4^{t-1}\left(1+4^{4^{t}-1}\left(1+\cdots\right)\right)\ud t\\ & =\frac{1}{\ln4}\left(\frac{1}{2}+I\right), \end{align*}$

解得 $I=\frac{1}{2\left(\ln4-1\right)}$.

2026 年麻省理工积分四分之一决赛

问题: 求积分

$\int\frac{\sin x+\cos x}{\sqrt{25\sin^{2}x+16\cos^{2}x}}\ud x.$

提示

$\begin{align*} \int\frac{\sin x+\cos x}{\sqrt{25\sin^{2}x+16\cos^{2}x}}\ud x & =\int\frac{\sin x}{\sqrt{25-9\cos^{2}x}}\ud x+\int\frac{\cos x}{\sqrt{16+9\sin^{2}x}}\ud x\\ & =-\frac{1}{3}\arcsin\left(\frac{3}{5}\cos x\right)+\frac{1}{3}\mathrm{arcsinh}\left(\frac{3}{4}\sin x\right)+C. \end{align*}$

问题: 求积分

$\int_{0}^{\infty}\frac{x^{3}}{\ue^{x}+1}\ud x.$

提示

$\begin{align*} \int_{0}^{\infty}\frac{x^{3}}{\ue^{x}+1}\ud x & =\int_{0}^{\infty}x^{3}\sum_{n=0}^{\infty}\left(-1\right)^{n}\ue^{-(n+1)x}\ud x\\ & =\sum_{n=0}^{\infty}(-1)^{n}\int_{0}^{\infty}x^{3}\ue^{-(n+1)x}\ud x\\ & =\sum_{n=0}^{\infty}(-1)^{n}\frac{6}{(n+1)^{4}}\\ & =\frac{21}{4}\zeta(4)=\frac{7\pi^{4}}{120}. \end{align*}$

问题: 求积分

$\int_{0}^{1}\frac{1-x^{5/2}}{1-x}\ud x.$

提示

$\begin{align*} \int_{0}^{1}\frac{1-x^{5/2}}{1-x}\ud x & \xlongequal{t=\sqrt{x}}2\int_{0}^{1}t\cdot\frac{1-t^{5}}{1-t^{2}}\ud t\\ & =2\int_{0}^{1}t\cdot\frac{1+t+t^{2}+t^{3}+t^{4}}{1+t}\ud t\\ & =2\int_{0}^{1}t\cdot\left(t^{3}+t+\frac{1}{1+t}\right)\ud t\\ & =\frac{46}{15}-2\ln2. \end{align*}$

问题: 求积分

$\int_{0}^{\infty}\frac{\ud x}{1+x^{6}}.$

提示

一般地,

$\begin{align*} \int_{0}^{\infty}\frac{x^{p-1}}{1+x^{q}}\ud x & \xlongequal{t=x^{q}}\frac{1}{q}\int_{0}^{\infty}\frac{t^{p/q-1}}{1+t}\ud t\\ & =\frac{1}{q}B\left(\frac{p}{q},1-\frac{p}{q}\right)\\ & =\frac{\pi}{q\sin\frac{p\pi}{q}}. \end{align*}$

问题: 求积分

$\int_{0}^{1/2}\frac{\ln(1-x)}{x}\ud x.$

提示

$\begin{align*} A=\int_{0}^{1/2}\frac{\ln(1-x)}{x}\ud x & =\int_{\frac{1}{2}}^{1}\frac{\ln t}{1-t}\ud t\\ & =\int_{0}^{1/2}\ln(1-x)\ud\ln x=\ln^{2}2+\int_{0}^{1/2}\frac{\ln x}{1-x}\ud x\\ & =\ln^{2}2+\int_{0}^{1}\frac{\ln x}{1-x}\ud x-\int_{1/2}^{1}\frac{\ln t}{1-t}\ud t\\ & =\ln^{2}2+\int_{0}^{1}\frac{\ln(1-x)}{x}\ud x-A, \end{align*}$

所以

$A=\frac{1}{2}\left(\ln^{2}2+\int_{0}^{1}\frac{\ln(1-x)}{x}\ud x\right)=\frac{1}{2}\left(\ln^{2}2-\frac{\pi^{2}}{6}\right).$

问题: 求积分

$\int\frac{\sqrt{x}}{1+x^{2}}\ud x.$

提示

$\begin{align*} \int\frac{\sqrt{x}}{1+x^{2}}\ud x & \xlongequal{u=\sqrt{x}}2\int\frac{u^{2}}{1+u^{4}}\ud u=\int\frac{\left(u^{2}+1\right)+\left(u^{2}-1\right)}{u^{4}+1}\ud u\\ & =\int\frac{1+u^{-2}}{u^{2}+u^{-2}}\ud u+\int\frac{1-u^{-2}}{u^{2}+u^{-2}}\ud u=\int\frac{\ud(u-u^{-1})}{(u-u^{-1})^{2}+2}+\int\frac{\ud(u+u^{-1})}{(u+u^{-1})^{2}-2}\\ & =\frac{1}{\sqrt{2}}\arctan\left(\frac{u-u^{-1}}{\sqrt{2}}\right)+\frac{1}{2\sqrt{2}}\ln\left|\frac{u+u^{-1}-\sqrt{2}}{u+u^{-1}+\sqrt{2}}\right|+C\\ & =\frac{1}{\sqrt{2}}\arctan\left(\frac{x-1}{\sqrt{2x}}\right)+\frac{1}{2\sqrt{2}}\ln\left|\frac{x+1-\sqrt{2x}}{x+1+\sqrt{2x}}\right|+C. \end{align*}$

问题: 求积分

$\int_{1}^{\infty}\frac{\left(-1\right)^{\left\lfloor x\right\rfloor +\left\lfloor \frac{x}{2}\right\rfloor +\left\lfloor \frac{x}{3}\right\rfloor +\cdots}}{x^{2}}\ud x.$

提示

设 $[x]=n$, 则有

$\left\lfloor x\right\rfloor +\left\lfloor \frac{x}{2}\right\rfloor +\left\lfloor \frac{x}{3}\right\rfloor +\cdots=[n]+\left[\frac{n}{2}\right]+\left[\frac{n}{3}\right]+\cdots=\sum_{k=1}^{n}\left[\frac{n}{k}\right]=\sum_{m=1}^{n}d(m),$

其中 $d(m)$ 表示 $m$ 的因子个数. 上式最后一步可以解释为

$I=\left\{ \left(k,i\right):k=1,2,\cdots n,\ i\le\frac{n}{k}\right\} =\left\{ \left(k,i\right):ki\le n\right\} =\left\{ \left(k,i\right):m=ki,\ m\le n\right\} ,$

因此

$\sum_{k=1}^{n}\left[\frac{n}{k}\right]=\sum_{k=1}^{n}\sum_{i:\ ik\le n}1=\sum_{(k,i)\in I}1=\sum_{m=1}^{n}\sum_{(k,i):\ ki=m}1=\sum_{m=1}^{n}d(m).$

由于当 $n$ 为平方数时, $d(n)$ 才为偶数, 其他情况均为奇数, 因此被积函数中 $(-1)$ 的指数只在 $x$ 是平方数时改变一次符号.

$\begin{align*} \text{原积分} & =\int_{1}^{\infty}\frac{(-1)^{\sum_{m=1}^{n}d(m)}}{x^{2}}\ud x\\ & =\sum_{n=1}^{\infty}\int_{n}^{n+1}\frac{(-1)^{\sum_{m=1}^{n}d(m)}}{x^{2}}\ud x\\ & =\sum_{n=1}^{\infty}\left(-1\right)^{\sum_{m=1}^{n}d(m)}\frac{1}{n(n+1)}\\ & =-1+2\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}+\frac{1}{4^{2}}-\cdots\right)=1-\frac{\pi^{2}}{6}. \end{align*}$

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,

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