数分每日一题
8.1
Fejer积分
记
$$\displaystyle I_n = \int_{0} ^ \frac{\pi}{2}(\frac{\sin nx}{\sin x})^2\mathrm{d}x$$
证明
$$\displaystyle I_n = \frac{n \pi}{2}$$
prove:
$$\displaystyle I_n - I_{n - 1} = \int_{0} ^ {\frac{\pi}{2}}\frac{\sin ^ 2 nx - \sin^2{(n-1)x}}{\sin ^2 x}\mathrm dx$$
$$\displaystyle =\frac{1}{2} \int_{0} ^ \frac{\pi}{2} \frac{\cos(2n - 2)x - \cos 2nx}{\sin ^ 2x}\mathrm dx$$
$$\displaystyle = \int_{0} ^ \frac{\pi}{2} \frac{\sin (2n - 1)x}{\sin x}\mathrm dx = J_{n}$$
$$\displaystyle J_n - J_{n-1} = \int_0^\frac{\pi}{2} \frac{\sin(2n - 1)x - \sin(2n - 3)x}{\sin x}\mathrm dx$$
$$\displaystyle = \int_0^\frac{\pi}{2} \frac{2\sin\frac{2n - 1 -(2n - 3)}{2} x \cdot \cos \frac{(2n - 1 ) + (2n - 3)}{2}x}{\sin x}\mathrm dx$$
$$\displaystyle = 2\int_0^\frac{\pi}{2} \cos(2n - 2)x \mathrm dx = 0$$
即得出
$$J_n = J_{n-1} = \cdots = J_{1}$$
又
$$\displaystyle J_1 = \int_{0}^\frac{\pi}{2} \mathrm dx = \frac{\pi}{2} , I_1 = \frac{\pi}{2}$$
因此由等差数列求和公式得出
$$\displaystyle I_n = I_1 + (n - 1)\cdot \frac{\pi}{2} = \frac{n\pi}{2}$$
8.2
IMC 2022 Day1 problem1
Problem 1 : Let f :[0,1]->(0,∞) be an integrable function such that f(x) · f(1-x) = 1 for all x ∈ [0,1]. Prove that
$$\displaystyle \int_0^{1} f(x)\mathrm{d}x \geq 1$$
译：设**f[0,1] -> (0,+∞)**是一个可积函数，满足f(x)f(1-x) = 1，对于所有x∈[0,1],证明：
$$\displaystyle \int_0^{1} f(x)\mathrm{d}x \geq 1$$
Prove1:
由 AM-GM 不等式得
$$\displaystyle f(x) + f(1-x) \geq 2\sqrt{f(x)f(1-x)}=2$$
同时我们考虑
$$\displaystyle \int_0^{1}f(x)\mathrm{d}x = \int_0^\frac{1}{2}f(x)\mathrm{d}x+\int_0^\frac{1}{2}f(1-x)\mathrm{d}x$$
$$\displaystyle =\int_0^\frac{1}{2}(f(x)+f(1-x))\mathrm{d}x\geq\int_0^\frac{1}{2}2\mathrm{d}x=1$$
Prove2:
注意到:
$$\displaystyle \int_0^1 f(x)\mathrm{d}x=\int_0^1 f(1-x)\mathrm{d}x =\int_0^1\frac{1}{f(x)}\mathrm{d}x$$
所以有
$$\displaystyle (\int_0^1 f(x)\mathrm{d}x)^2 = \int_0^1 f(x)\mathrm{d}x\cdot\int_0^1\frac{1}{f(x)}\mathrm{d}x \geq (\int_0^1 1\mathrm{d}x)^2\geq1$$
8.3
3