2023 MIT Integration Bee Finals 해설

Problem 1

${\displaystyle {\int }_0^{\frac{\pi }{2}}\frac{\sqrt[3]{\tan x}}{(\sin x+\cos x{)}^2}dx}$

풀이 1

$u=\tan x$로 치환하자. $du={\sec }^{2}xdx$이고,

$$(\sin x+\cos x{)}^2=1+\sin (2x)=1+\frac{2\tan x}{1+{\tan }^{2}x}$$

이다. 오일러의 반사 공식에 의해 $z\notin \mathbb{Z}$에 대하여 $\mathrm{\Gamma }(z)\mathrm{\Gamma }(1-z)=\pi \csc (\pi z)$가 성립한다. 따라서

$$\begin{array}{rl}{\int }_0^{\frac{\pi }{2}}\frac{\sqrt[3]{\tan x}}{(\sin x+\cos x{)}^2}dx& ={\int }_0^{\mathrm{\infty }}\frac{u^{1/3}}{1+{\displaystyle \frac{2u}{1+u^2}}}\cdot \frac{1}{1+u^2}du={\int }_0^{\mathrm{\infty }}\frac{u^{1/3}}{(1+u{)}^2}du\\ & =\mathrm{B}(4/3,2/3)=\frac{\mathrm{\Gamma }(4/3)\mathrm{\Gamma }(2/3)}{\mathrm{\Gamma }(2)}=\frac{1}{3}\cdot \frac{\mathrm{\Gamma }(1/3)\mathrm{\Gamma }(2/3)}{1}\\ & =\boxed{{\displaystyle \frac{2\pi }{3\sqrt{3}}}}\end{array}$$

이다.

풀이 2

$u^3=\tan x$로 치환하면 $3u^{2}du={\sec }^{2}xdx$이므로

$$\begin{array}{rl}{\int }_0^{\frac{\pi }{2}}\frac{\sqrt[3]{\tan x}}{(\sin x+\cos x{)}^2}dx& ={\int }_0^{\mathrm{\infty }}\frac{u}{1+{\displaystyle \frac{2u^3}{1+u^6}}}\cdot \frac{3u^2}{1+u^6}du={\int }_0^{\mathrm{\infty }}\frac{3u^2}{(1+u^3{)}^2}\cdot udu\\ & ={[-\frac{1}{1+u^3}\cdot u]}_0^{\mathrm{\infty }}-{\int }_0^{\mathrm{\infty }}(-\frac{1}{1+u^3})du={\int }_0^{\mathrm{\infty }}\frac{1}{1+u^3}du\end{array}$$

이다. $t=1/u$로 치환하면

$${\int }_0^{\mathrm{\infty }}\frac{1}{1+u^3}du={\int }_{\mathrm{\infty }}^0\frac{1}{1+(1/t{)}^3}(-\frac{dt}{t^2})={\int }_0^{\mathrm{\infty }}\frac{t}{1+t^3}dt$$

이므로

$$\begin{array}{rl}{\int }_0^{\frac{\pi }{2}}\frac{\sqrt[3]{\tan x}}{(\sin x+\cos x{)}^2}dx& =\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{1+u}{1+u^3}du=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{1}{u^2-u+1}du\\ & =\frac{1}{2}{[\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{2x-1}{\sqrt{3}}\right)]}_0^{\mathrm{\infty }}\\ & =\boxed{{\displaystyle \frac{2\pi }{3\sqrt{3}}}}\end{array}$$

이다.

Problem 2

${\displaystyle {\int }_0^{\pi }{\left(\frac{\sin (2x)\sin (3x)\sin (5x)\sin (30x)}{\sin (x)\sin (6x)\sin (10x)\sin (15x)}\right)}^{2}dx}$

$\sin (2x)=2\sin (x)\cos (x)$이고, $\cos (2x)=2{\cos }^2(x)-1$로부터

$$\frac{\cos (3x)}{\cos (x)}=4{\cos }^2(x)-3=2\cos (2x)-1$$

이다. 오일러 공식에 의해

$$\begin{array}{rl}\frac{\sin (2x)\sin (3x)\sin (5x)\sin (30x)}{\sin (x)\sin (6x)\sin (10x)\sin (15x)}& =\frac{\cos (x)}{\cos (3x)}\cdot \frac{\cos (15x)}{\cos (5x)}=\frac{2\cos (10x)-1}{2\cos (2x)-1}\\ & =\frac{e^{i\cdot 10x}+e^{-i\cdot 10x}-1}{e^{i\cdot 2x}+e^{-i\cdot 2x}-1}=\frac{t^5+t^{-5}-1}{t+t^{-1}-1}\\ & =\frac{1}{t^4}\cdot \frac{t^{10}-t^5+1}{t^2-t+1}\\ & =\frac{1}{t^4}(t^8+t^7-t^5-t^4-t^3+t+1)\\ & =t^4+t^{-4}+t^3+t^{-3}-(t+t^{-1})-1\\ & =2\cos (8x)+2\cos (6x)-2\cos (2x)-1\end{array}$$

이다. 따라서

$$\begin{array}{rl}& {\int }_0^{\pi }{\left(\frac{\sin (2x)\sin (3x)\sin (5x)\sin (30x)}{\sin (x)\sin (6x)\sin (10x)\sin (15x)}\right)}^{2}dx\\ & ={\int }_0^{\pi }(2\cos (8x)+2\cos (6x)-2\cos (2x)-1{)}^{2}dx\\ & ={\int }_0^{\pi }(4{\cos }^2(8x)+4{\cos }^2(6x)+4{\cos }^2(2x)+1+\cdots )dx\\ & ={\int }_0^{\pi }[2(1+\cos (16x))+2(1+\cos (12x))+2(1+\cos (4x))+1+\cdots ]dx\\ & =\boxed{{\displaystyle 7\pi }}\end{array}$$

이다. 생략한 항의 적분은 0이다.

Problem 3

${\displaystyle {\int }_{-1/2}^{1/2}\sqrt{x^2+1+\sqrt{x^4+x^2+1}}dx}$

${\displaystyle \sqrt{x^2+1+\sqrt{x^4+x^2+1}}=\sqrt{f(x)}+\sqrt{g(x)}}$으로 두자.

$$\begin{array}{c}{x}^2+1+\sqrt{x^4+x^2+1}=f(x)+g(x)+2\sqrt{f(x)g(x)},\\ (f(x)-g(x){)}^2=(f(x)+g(x){)}^2-4f(x)g(x)={(x^2+1)}^2-(x^4+x^2+1)=x^2\end{array}$$

이므로 $f(x)-g(x)=x$를 고르자. 따라서

$$\begin{array}{rl}f(x)& =\frac{(f(x)+g(x))+(f(x)-g(x))}{2}=\frac{x^2+x+1}{2},\\ g(x)& =\frac{(f(x)+g(x))-(f(x)-g(x))}{2}=\frac{x^2-x+1}{2}\end{array}$$

이므로

$$\begin{array}{rl}{\int }_{-1/2}^{1/2}\sqrt{x^2+1+\sqrt{x^4+x^2+1}}dx& =2{\int }_0^{1/2}(\sqrt{\frac{x^2+x+1}{2}}+\sqrt{\frac{x^2-x+1}{2}})dx\\ & =\sqrt{2}{\int }_0^{1/2}(\sqrt{{(x+\frac{1}{2})}^2+\frac{3}{4}}+\sqrt{{(x-\frac{1}{2})}^2+\frac{3}{4}})dx\\ & =\sqrt{2}{\int }_{1/2}^1\sqrt{x^2+\frac{3}{4}}dx+\sqrt{2}{\int }_{-1/2}^0\sqrt{x^2+\frac{3}{4}}dx\\ & =\sqrt{2}{\int }_0^1\sqrt{x^2+\frac{3}{4}}dx\end{array}$$

이다. $x={\displaystyle \frac{\sqrt{3}}{2}}\sinh u$로 치환하자. ${\displaystyle {\sinh }^{-1}x=\log (x+\sqrt{x^2+1})}$임을 이용하면

$$\begin{array}{rl}{\int }_0^1\sqrt{x^2+\frac{3}{4}}dx& ={\int }_0^{\log \left(\frac{2+\sqrt{7}}{\sqrt{3}}\right)}\frac{3}{4}{\cosh }^{2}udu\\ & =\frac{3}{4}{\int }_0^{\log \left(\frac{2+\sqrt{7}}{\sqrt{3}}\right)}\frac{1+\cosh (2u)}{2}du\\ & =\frac{3}{8}[\log \left(\frac{2+\sqrt{7}}{\sqrt{3}}\right)+\frac{1}{2}\sinh (2{\sinh }^{-1}\frac{2}{\sqrt{3}})]\end{array}$$

이므로

$$\begin{array}{rl}{\int }_{-1/2}^{1/2}\sqrt{x^2+1+\sqrt{x^4+x^2+1}}dx& =\frac{3\sqrt{2}}{8}\log \left(\frac{2+\sqrt{7}}{\sqrt{3}}\right)+\frac{3\sqrt{2}}{8}\cdot \frac{2}{\sqrt{3}}\cdot \sqrt{\frac{7}{3}}\\ & =\boxed{{\displaystyle \frac{3\sqrt{2}}{8}\log \left(\frac{2+\sqrt{7}}{\sqrt{3}}\right)+\frac{\sqrt{14}}{4}}}\end{array}$$

이다.

Problem 4

${\displaystyle \lfloor {10}^{20}{\int }_2^{\mathrm{\infty }}\frac{x^9}{x^{20}-48x^{10}+575}dx\rfloor }$

적분을 풀면

$$\begin{array}{rl}{10}^{20}{\int }_2^{\mathrm{\infty }}\frac{x^9}{x^{20}-48x^{10}+575}dx& ={10}^{19}{\int }_2^{\mathrm{\infty }}\frac{d\left(x^{10}\right)}{(x^{10}-24{)}^2-1}={10}^{19}{\int }_{2^{10}}^{\mathrm{\infty }}\frac{du}{(u-24{)}^2-1}\\ & ={10}^{19}{\int }_{{10}^3}^{\mathrm{\infty }}\frac{du}{u^2-1}=\frac{{10}^{19}}{2}\log \frac{{10}^3+1}{{10}^3-1}\\ & =\frac{{10}^{19}}{2}\log \frac{1+{10}^{-3}}{1-{10}^{-3}}\end{array}$$

이다. $\left|x\right|<1$에 대하여 테일러 급수

$$\log (1\pm x)=\pm x-\frac{x^2}{2}\pm \frac{x^3}{3}-\cdots$$

이 수렴하므로

$$\begin{array}{rl}\frac{{10}^{19}}{2}\log \frac{1+{10}^{-3}}{1-{10}^{-3}}& =\frac{{10}^{19}}{2}\cdot 2({10}^{-3}+\frac{{10}^{-9}}{3}+\frac{{10}^{-15}}{5}+\frac{{10}^{-21}}{7}+\cdots )\\ & ={10}^{16}+\frac{{10}^{10}}{3}+\frac{{10}^4}{5}+\frac{{10}^{-2}}{7}+\cdots \end{array}$$

이다. 따라서

$$\begin{array}{rl}\lfloor {10}^{20}{\int }_2^{\mathrm{\infty }}\frac{x^9}{x^{20}-48x^{10}+575}dx\rfloor & =\lfloor \frac{{10}^{19}}{2}\log \frac{1+{10}^{-3}}{1-{10}^{-3}}\rfloor \\ & =\lfloor {10}^{16}+\frac{{10}^{10}}{3}+\frac{{10}^4}{5}+\frac{{10}^{-2}}{7}+\cdots \rfloor \\ & ={10}^{16}+\frac{{10}^{10}-1}{3}+\frac{{10}^4}{5}\\ & =\boxed{{\displaystyle 10000003333335333}}\end{array}$$

이다.

Problem 5

${\displaystyle {\int }_0^1{\left(\sum _{n=1}^{\mathrm{\infty }}\frac{\lfloor 2^{n}x\rfloor }{3^n}\right)}^{2}dx}$

$x=2u$로 치환하자. $0<u<1/2$에 대하여 $\lfloor 2^{1}u\rfloor =0$이므로

$$\begin{array}{rl}{\int }_0^1{\left(\sum _{n=1}^{\mathrm{\infty }}\frac{\lfloor 2^{n}x\rfloor }{3^n}\right)}^{2}dx& =2{\int }_0^{1/2}{\left(3\sum _{n=1}^{\mathrm{\infty }}\frac{\lfloor 2^{n+1}u\rfloor }{3^{n+1}}\right)}^{2}du=18{\int }_0^{1/2}{\left(\sum _{n=2}^{\mathrm{\infty }}\frac{\lfloor 2^{n}u\rfloor }{3^n}\right)}^{2}du\\ & =18{\int }_0^{1/2}{\left(\sum _{n=1}^{\mathrm{\infty }}\frac{\lfloor 2^{n}u\rfloor }{3^n}\right)}^{2}du\end{array}$$

이다. 이번에는 $x=t+1/2$로 치환하면

$$\begin{array}{rl}{\int }_{1/2}^1{\left(\sum _{n=1}^{\mathrm{\infty }}\frac{\lfloor 2^{n}x\rfloor }{3^n}\right)}^{2}dx& ={\int }_0^{1/2}{\left(\sum _{n=1}^{\mathrm{\infty }}\frac{\lfloor 2^{n}t\rfloor +2^{n-1}}{3^n}\right)}^{2}dt\\ & ={\int }_0^{1/2}{(\sum _{n=1}^{\mathrm{\infty }}\frac{\lfloor 2^{n}t\rfloor }{3^n}+\frac{1}{2}\cdot \frac{2/3}{1-2/3})}^{2}dt\\ & ={\int }_0^{1/2}[{\left(\sum _{n=1}^{\mathrm{\infty }}\frac{\lfloor 2^{n}t\rfloor }{3^n}\right)}^2+2\sum _{n=1}^{\mathrm{\infty }}\frac{\lfloor 2^{n}t\rfloor }{3^n}+1]dt\end{array}$$

이다. 따라서

$$\begin{array}{rl}{\int }_0^1{\left(\sum _{n=1}^{\mathrm{\infty }}\frac{\lfloor 2^{n}x\rfloor }{3^n}\right)}^{2}dx& ={\int }_0^{1/2}{\left(\sum _{n=1}^{\mathrm{\infty }}\frac{\lfloor 2^{n}x\rfloor }{3^n}\right)}^{2}dx+{\int }_{1/2}^1{\left(\sum _{n=1}^{\mathrm{\infty }}\frac{\lfloor 2^{n}x\rfloor }{3^n}\right)}^{2}dx\\ & =2{\int }_0^{1/2}{\left(\sum _{n=1}^{\mathrm{\infty }}\frac{\lfloor 2^{n}x\rfloor }{3^n}\right)}^{2}dx+2{\int }_0^{1/2}\sum _{n=1}^{\mathrm{\infty }}\frac{\lfloor 2^{n}x\rfloor }{3^n}dx+\frac{1}{2}\\ & =\frac{1}{9}{\int }_0^1{\left(\sum _{n=1}^{\mathrm{\infty }}\frac{\lfloor 2^{n}x\rfloor }{3^n}\right)}^{2}dx+2\sum _{n=1}^{\mathrm{\infty }}\frac{1}{3^n}{\int }_0^{1/2}\lfloor 2^{n}x\rfloor dx+\frac{1}{2}\end{array}$$

이다. 이때, 단조 수렴 정리에 따라 적분과 급수의 순서를 바꿀 수 있다. 마지막으로 $2^{n}x=y$로 치환하면

$$\begin{array}{rl}{\int }_0^1{\left(\sum _{n=1}^{\mathrm{\infty }}\frac{\lfloor 2^{n}x\rfloor }{3^n}\right)}^{2}dx& =\frac{9}{8}(2\sum _{n=1}^{\mathrm{\infty }}\frac{1}{6^n}{\int }_0^{2^{n-1}}\lfloor y\rfloor dy+\frac{1}{2})\\ & =\frac{9}{4}\sum _{n=1}^{\mathrm{\infty }}\frac{1}{6^n}\sum _{k=1}^{2^{n-1}-1}k+\frac{9}{16}\\ & =\frac{9}{4}\sum _{n=1}^{\mathrm{\infty }}\frac{1}{6^n}\frac{2^{n-1}(2^{n-1}-1)}{2}+\frac{9}{16}\\ & =\frac{9}{8}\sum _{n=1}^{\mathrm{\infty }}[\frac{1}{4}{\left(\frac{2}{3}\right)}^n-\frac{1}{2}{\left(\frac{1}{3}\right)}^n]+\frac{9}{16}\\ & =\frac{9}{8}(\frac{1}{4}\cdot \frac{2/3}{1-2/3}-\frac{1}{2}\cdot \frac{1/3}{1-1/3})+\frac{9}{16}\\ & =\boxed{{\displaystyle \frac{27}{32}}}\end{array}$$

이다.