设f(x)=∑x^n/n^2,证明f(x)+f(1-x)+lnxln(1-x)=∑1/n^2

题目

$$\text{设}f\left(x\right)=\sum _{n=1}^{\infty }\frac{x^n}{n^2}\text{，证明：}f\left(x\right)+f\left(1-x\right)+\ln x\ln \left(1-x\right)=\sum _{n=1}^{\infty }\frac{1}{n^2}$$

解答

$$f^{\prime }\left(x\right)=\sum _{n=1}^{\infty }\frac{x^{n-1}}{n}=\frac{1}{x}\sum _{n=1}^{\infty }\frac{x^n}{n}=\frac{1}{x}{\int }_0^x\sum _{n=1}^{\infty }{t}^{n-1}dt=\frac{1}{x}{\int }_0^x\sum _{n=0}^{\infty }{t}^{n}dt=\frac{1}{x}{\int }_0^x\frac{1}{1-t}dt=-\frac{\ln \left(1-x\right)}{x}$$
$$f^{\prime }\left(x\right)-f^{\prime }\left(1-x\right)=-\frac{\ln \left(1-x\right)}{x}+\frac{\ln x}{1-x}$$
$$\because {\left[\ln x\ln \left(1-x\right)\right]}^{\prime }=\frac{\ln \left(1-x\right)}{x}-\frac{\ln x}{1-x}$$
$$\therefore f^{\prime }\left(x\right)-f^{\prime }\left(1-x\right)+{\left[\ln x\ln \left(1-x\right)\right]}^{\prime }=0$$
$$\text{令}g\left(x\right)=f\left(x\right)+f\left(1-x\right)+\ln x\ln \left(1-x\right)x\in \left(0,1\right)$$
$$\therefore g^{\prime }\left(x\right)=0$$
$$\because {\int }_0^x{g}^{\prime }\left(u\right)du=g\left(x\right)-\underset{t\to 0^{+}}{\lim }g\left(t\right)=0$$
$$\therefore g\left(x\right)=\underset{t\to 0^{+}}{\lim }g\left(t\right)$$
$$\because \underset{t\to 0^{+}}{\lim }g\left(t\right)=\underset{t\to 0^{+}}{\lim }f\left(t\right)+\underset{t\to 0^{+}}{\lim }f\left(1-t\right)+\underset{t\to 0^{+}}{\lim }\ln t\cdot \ln \left(1-t\right)$$
$$\text{其中}\underset{t\to 0^{+}}{\lim }f\left(t\right)=0,\underset{t\to 0^{+}}{\lim }f\left(1-t\right)=\sum _{n=1}^{\infty }\frac{1}{n^2},\underset{t\to 0^{+}}{\lim }\ln t\cdot \ln \left(1-t\right)=\underset{t\to 0^{+}}{\lim }t\ln t=\underset{t\to 0^{+}}{\lim }\frac{\ln t}{\frac{1}{t}}=\underset{t\to 0^{+}}{\lim }\frac{\frac{1}{t}}{-\frac{1}{t^2}}=0$$
$$\therefore \underset{t\to 0^{+}}{\lim }g\left(t\right)=\sum _{n=1}^{\infty }\frac{1}{n^2}$$
$$\therefore g\left(x\right)=\sum _{n=1}^{\infty }\frac{1}{n^2}$$
$$\therefore f\left(x\right)+f\left(1-x\right)+\ln x\ln \left(1-x\right)=\sum _{n=1}^{\infty }\frac{1}{n^2}$$