中册 6.5 傅里叶级数 第14题

\section*{解题过程：} （1）由收玫定理知它可以展成傅里叶级数．函数 $f(x)=|x|+\sin ^{2} \pi x$ 在 $[-1,1]$ 上为偶函数，故 $$ \begin{aligned} b_{n} & =0, n=1,2, \cdots \\ a_{0} & =\frac{2}{1} \int_{0}^{1}\left(x+\sin ^{2} \pi x\right) \mathrm{d} x=2 \\ a_{n} & =\frac{2}{1} \int_{0}^{1}\left(x+\sin ^{2} \pi x\right) \cos n \pi x \mathrm{~d} x=2\left[\int_{0}^{1} x \cos n \pi x \mathrm{~d} x+\frac{1}{2} \int_{0}^{1}(1-\cos 2 \pi x) \cos n \pi x \mathrm{~d} x\right] \\ & =\left.2\left(\frac{x}{n \pi} \sin n \pi x+\frac{1}{n^{2} \pi^{2}} \cos n \pi x\right)\right|_{0} ^{1}+\left.\frac{1}{n \pi} \sin n \pi x\right|_{0} ^{1}-\int_{0}^{1} \cos 2 \pi x \cos n \pi x \mathrm{~d} x=\frac{2}{n^{2} \pi^{2}}\left[(-1)^{n}-1\right] \end{aligned} $$ 由于 $f(x)=|x|+\sin ^{2} \pi x$ 在 $[-1,1]$ 上连续，所以 $\displaystyle f(x)=1+\sum_{n=1}^{\infty} \frac{2}{n^{2} \pi^{2}}\left[(-1)^{n}-1\right] \cos n \pi x, x \in[-1,1]$ ． （2）不妨设 $a \neq 0$ ，函数的傅氏系数： $$ \begin{aligned} a_{0} & =\frac{1}{\pi} \int_{0}^{2 \pi} f(x) \mathrm{d} x=\frac{1}{\pi} \int_{0}^{2 \pi} \mathrm{e}^{a x} \mathrm{~d} x=\frac{\mathrm{e}^{2 a \pi}-1}{a \pi} \\ a_{n} & =\frac{1}{\pi} \int_{0}^{2 \pi} f(x) \cos n x \mathrm{~d} x=\frac{1}{\pi} \int_{0}^{2 \pi} \mathrm{e}^{a x} \cos n x \mathrm{~d} x \\ & =\frac{1}{\pi} \frac{1}{a^{2}+n^{2}}\left[(n \sin n x+a \cos n x) \mathrm{e}^{a x}\right]_{0}^{2 \pi}=\frac{a\left(\mathrm{e}^{2 a \pi}-1\right)}{\left(n^{2}+a^{2}\right) \pi},(n=1,2, \cdots) \\ b_{n} & =\frac{1}{\pi} \int_{0}^{2 \pi} f(x) \sin n x \mathrm{~d} x=\frac{1}{\pi} \int_{0}^{2 \pi} \mathrm{e}^{a x} \sin n x \mathrm{~d} x \\ & =\frac{1}{\pi} \frac{1}{a^{2}+n^{2}}\left[(a \sin n x-n \cos n x) \mathrm{e}^{a x}\right]_{0}^{2 \pi}=-\frac{n\left(\mathrm{e}^{2 a \pi}-1\right)}{\left(a^{2}+n^{2}\right) \pi},(n=1,2, \cdots) \end{aligned} $$ 所以 $f(x)$ 的傅里叶级数展开式为 $\displaystyle \frac{\mathrm{e}^{2 a \pi}-1}{2 a \pi}+\frac{\mathrm{e}^{2 a \pi}-1}{\pi} \sum_{n=1}^{\infty} \frac{1}{\left(n^{2}+a^{2}\right)}(a \cos n x-n \sin n x)$ ． 当 $0