(1)

フーリエ係数\(C_{k}\)は
\(k\ne0\)のとき、
\begin{align*} C_{k} & =\frac{1}{2\pi}\int_{-\pi}^{\pi}\sgn\left(x\right)e^{-ikx}dx\\ & =\frac{1}{2\pi}\left(\int_{-\pi}^{0}\left(-1\right)e^{-ikx}dx+\int_{0}^{\pi}e^{-ikx}dx\right)\\ & =\frac{1}{2\pi}\left(\frac{1}{ik}\left[e^{-ikx}\right]_{-\pi}^{0}+\frac{-1}{ik}\left[e^{-ikx}\right]_{0}^{\pi}\right)\\ & =\frac{1}{2\pi}\left\{ \frac{1}{ik}\left(1-e^{ik\pi}\right)-\frac{1}{ik}\left(e^{-ik\pi}-1\right)\right\} \\ & =\frac{1}{2\pi ik}\left(1-e^{ik\pi}-e^{-ik\pi}+1\right)\\ & =\frac{1}{2\pi ik}\left(2-\left(e^{ik\pi}+e^{-ik\pi}\right)\right)\\ & =\frac{1}{2\pi ik}\left(2-2\cos\left(k\pi\right)\right)\\ & =\frac{1}{\pi ik}\left(1-\left(-1\right)^{k}\right)\\ & =\begin{cases} \frac{2}{\pi ik} & k\in2\mathbb{Z}-1\\ 0 & k\in2\mathbb{Z}\setminus\left\{ 0\right\} \end{cases} \end{align*} \(k=0\)のとき、
\begin{align*} C_{0} & =\frac{1}{2\pi}\int_{-\pi}^{\pi}\sgn\left(x\right)e^{-i0x}dx\\ & =\frac{1}{2\pi}\int_{-\pi}^{\pi}\sgn\left(x\right)dx\\ & =0 \end{align*} となるので
\[ C_{k}=\begin{cases} \frac{2}{\pi ik} & k\in2\mathbb{Z}-1\\ 0 & k\in2\mathbb{Z} \end{cases} \] となる。
これより、フーリエ級数展開\(F\left(x\right)\)は
\begin{align*} F\left(x\right) & =\sum_{k=-\infty}^{\infty}C_{k}e^{ikx}\\ & =\sum_{k=-\infty}^{\infty}C_{2k-1}e^{i\left(2k-1\right)x}\\ & =\sum_{k=-\infty}^{\infty}\frac{2}{\pi i\left(2k-1\right)}e^{i\left(2k-1\right)x}\\ & =\sum_{k=1}^{\infty}\frac{2}{\pi i\left(2k-1\right)}e^{i\left(2k-1\right)x}+\sum_{k=1}^{\infty}\frac{-2}{\pi i\left(2k-1\right)}e^{-i\left(2k-1\right)x}\\ & =\sum_{k=1}^{\infty}\frac{2}{\pi i\left(2k-1\right)}\left(e^{i\left(2k-1\right)x}-e^{-i\left(2k-1\right)x}\right)\\ & =\sum_{k=1}^{\infty}\frac{4}{\pi\left(2k-1\right)}\sin\left(\left(2k-1\right)x\right) \end{align*} となる。

(2)

周期\(2\pi\)の区間\(\left[\pi,\pi\right]\)で\(f\left(x\right)=\left|\sin\left(x\right)\right|\)となる関数のフーリエ係数\(C_{k}\)は
\(k\ne\pm1,\)のとき
\begin{align*} C_{k} & =\frac{1}{2\pi}\int_{-\pi}^{\pi}\left|\sin\left(x\right)\right|e^{-ikx}dx\\ & =\frac{1}{2\pi}\left(\int_{-\pi}^{0}-\sin\left(x\right)e^{-ikx}dx+\int_{0}^{\pi}\sin\left(x\right)e^{-ikx}dx\right)\\ & =\frac{1}{2\pi}\left(\int_{\pi}^{0}\sin\left(-x\right)e^{ikx}dx+\int_{0}^{\pi}\sin\left(x\right)e^{-ikx}dx\right)\\ & =\frac{1}{2\pi}\left(\int_{0}^{\pi}\sin\left(x\right)e^{ikx}dx+\int_{0}^{\pi}\sin\left(x\right)e^{-ikx}dx\right)\\ & =\frac{1}{2\pi}\int_{0}^{\pi}\sin\left(x\right)\left(e^{ikx}+e^{-ikx}\right)dx\\ & =\frac{1}{\pi}\int_{0}^{\pi}\sin\left(x\right)\cos\left(kx\right)dx\\ & =\frac{1}{\pi}\left\{ \left[-\cos\left(x\right)\cos\left(kx\right)-k\sin\left(x\right)\sin\left(kx\right)\right]_{0}^{\pi}+k^{2}\int_{0}^{\pi}\sin\left(x\right)\cos\left(kx\right)dx\right\} \\ & =\frac{1}{1-k^{2}}\cdot\frac{1}{\pi}\left[-\cos\left(x\right)\cos\left(kx\right)-k\sin\left(x\right)\sin\left(kx\right)\right]_{0}^{\pi}\\ & =\frac{1}{1-k^{2}}\cdot\frac{-1}{\pi}\left(-\cos\left(k\pi\right)-1\right)\\ & =\frac{1}{1-k^{2}}\cdot\frac{1}{\pi}\left(\left(-1\right)^{k}+1\right)\\ & =-\frac{\left(-1\right)^{k}+1}{\left(k^{2}-1\right)\pi}\\ & =\begin{cases} -\frac{2}{\left(k^{2}-1\right)\pi} & k\in2\mathbb{Z}\\ 0 & k\in2\mathbb{Z}-1 \end{cases} \end{align*} \(k=\pm1\)のとき、
\begin{align*} C_{\pm1} & =\frac{1}{\pi}\int_{0}^{\pi}\sin\left(x\right)\cos\left(\pm x\right)dx\\ & =\frac{1}{\pi}\int_{0}^{\pi}\sin\left(x\right)\cos\left(x\right)dx\\ & =\frac{1}{2\pi}\int_{0}^{\pi}\sin\left(2x\right)dx\\ & =\frac{1}{2\pi}\left[-\frac{1}{2}\cos\left(2x\right)\right]_{0}^{2\pi}\\ & =0 \end{align*} これより、
\[ C_{k}=\begin{cases} -\frac{2}{\left(k^{2}-1\right)\pi} & k\in2\mathbb{Z}\\ 0 & k\in2\mathbb{Z}-1 \end{cases} \] となるのでフーリエ級数展開\(F\left(x\right)\)は
\begin{align*} F\left(x\right) & =\sum_{k=-\infty}^{\infty}C_{k}e^{ikx}\\ & =\sum_{k=-\infty}^{\infty}C_{2k}e^{i2kx}\\ & =-\sum_{k=-\infty}^{\infty}\frac{2}{\left(\left(2k\right)^{2}-1\right)\pi}e^{i2kx}\\ & =\frac{2}{\pi}-\sum_{k\ne0}\frac{2}{\left(4k^{2}-1\right)\pi}e^{i2kx}\\ & =\frac{2}{\pi}-\sum_{k=1}^{\infty}\frac{2}{\left(4k^{2}-1\right)\pi}\left(e^{2ikx}+e^{-2ikx}\right)\\ & =\frac{2}{\pi}-4\sum_{k=1}^{\infty}\frac{1}{\left(4k^{2}-1\right)\pi}\cos\left(2kx\right) \end{align*} となる。

(3)

周期\(2\pi\)の区間\(\left[\pi,\pi\right]\)で\(f\left(x\right)=\sin\left(x\right)H\left(x\right)\)となる関数のフーリエ係数\(C_{k}\)は
\(k\ne\pm1,\)のとき
\begin{align*} C_{k} & =\frac{1}{2\pi}\int_{-\pi}^{\pi}\sin\left(x\right)H\left(x\right)e^{-ikx}dx\\ & =\frac{1}{2\pi}\int_{0}^{\pi}\sin\left(x\right)e^{-ikx}dx\\ & =\frac{1}{4\pi i}\int_{0}^{\pi}\left(e^{ix}-e^{-ix}\right)e^{-ikx}dx\\ & =\frac{1}{4\pi i}\int_{0}^{\pi}\left(e^{i\left(1-k\right)x}-e^{-i\left(1+k\right)x}\right)dx\\ & =\frac{1}{4\pi i}\left[\frac{1}{i\left(1-k\right)}e^{i\left(1-k\right)x}+\frac{1}{i\left(1+k\right)}e^{-i\left(1+k\right)x}\right]_{0}^{\pi}\\ & =\frac{-1}{4\pi}\left[\frac{1}{1-k}e^{i\left(1-k\right)x}+\frac{1}{1+k}e^{-i\left(1+k\right)x}\right]_{0}^{\pi}\\ & =\frac{-1}{4\pi}\left\{ \frac{1}{1-k}\left(e^{i\left(1-k\right)\pi}-1\right)+\frac{1}{1+k}\left(e^{-i\left(1+k\right)\pi}-1\right)\right\} \\ & =\frac{-1}{4\pi\left(1-k^{2}\right)}\left\{ \left(1+k\right)\left(-e^{-ik\pi}-1\right)+\left(1-k\right)\left(-e^{-ik\pi}-1\right)\right\} \\ & =\frac{e^{-ik\pi}+1}{4\pi\left(1-k^{2}\right)}\left(1+k+1-k\right)\\ & =\frac{e^{-ik\pi}+1}{2\pi\left(1-k^{2}\right)}\\ & =\frac{\left(-1\right)^{k}+1}{2\pi\left(1-k^{2}\right)}\\ & =\begin{cases} \frac{1}{\pi\left(1-k^{2}\right)} & k\in2\mathbb{Z}\\ 0 & k\in2\mathbb{Z}-1 \end{cases} \end{align*} \(k=\pm1\)のとき
\begin{align*} C_{\pm1} & =\frac{1}{4\pi i}\int_{0}^{\pi}\left(e^{i\left(1\mp1\right)x}-e^{-i\left(1\pm1\right)x}\right)dx\\ & =\frac{1}{4\pi i}\int_{0}^{\pi}\left(e^{\mp2ix}\pm1\right)dx\\ & =\frac{1}{4\pi i}\left[\frac{1}{\mp2i}e^{\mp2ix}\pm x\right]_{0}^{\pi}\\ & =\frac{1}{4\pi i}\left(\frac{1}{\mp2i}e^{\mp2i\pi}\pm\pi-\frac{1}{\mp2i}\right)\\ & =\mp\frac{i}{4} \end{align*} これより、
\[ C_{k}=\begin{cases} \frac{e^{-ik\pi}+1}{2\pi\left(1-k^{2}\right)} & k\ne\pm1\\ \mp\frac{i}{4} & k=\pm1 \end{cases} \] となるのでフーリエ級数展開\(F\left(x\right)\)は
\begin{align*} F\left(x\right) & =\sum_{k=-\infty}^{\infty}C_{k}e^{ikx}\\ & =\sum_{k=-1}^{1}C_{k}e^{ikx}+\sum_{\left|k\right|\geq2}C_{k}e^{ikx}\\ & =\frac{i}{4}e^{-ix}+\frac{1}{\pi}-\frac{i}{4}e^{ix}+\sum_{k=2}^{\infty}\left(C_{k}e^{ikx}+C_{-k}e^{-ikx}\right)\\ & =\frac{1}{\pi}-\frac{i}{4}\left(e^{ix}-e^{-ix}\right)+\sum_{k=1}^{\infty}\left(C_{2k}e^{2ikx}+C_{-2k}e^{-2ikx}\right)\\ & =\frac{1}{\pi}+\frac{1}{2}\sin\left(x\right)+\sum_{k=1}^{\infty}\left(C_{2k}\left(e^{2ikx}+e^{-2ikx}\right)\right)\\ & =\frac{1}{\pi}+\frac{1}{2}\sin\left(x\right)+2\sum_{k=1}^{\infty}\left(C_{2k}\cos\left(2kx\right)\right)\\ & =\frac{1}{\pi}+\frac{1}{2}\sin\left(x\right)+2\sum_{k=1}^{\infty}\left(\frac{1}{\pi\left(1-\left(2k\right)^{2}\right)}\cos\left(2kx\right)\right)\\ & =\frac{1}{\pi}+\frac{1}{2}\sin\left(x\right)-2\sum_{k=1}^{\infty}\left(\frac{1}{\left(4k^{2}-1\right)\pi}\cos\left(2kx\right)\right) \end{align*} となる。

(4)

周期\(2\pi\)の区間\(\left[\pi,\pi\right]\)で\(f\left(x\right)=x^{n}\)となる関数のフーリエ係数\(C_{k}\)は
\(k\ne0\)のとき、
\begin{align*} C_{k} & =\frac{1}{2\pi}\int_{-\pi}^{\pi}x^{n}e^{-ikx}dx\\ & =\frac{1}{2\pi}\sum_{j=1}^{n+1}\left[\left(-1\right)^{j-1}\left(\frac{i}{k}\right)^{j}P\left(n,j-1\right)x^{n-\left(j-1\right)}e^{-ikx}\right]_{-\pi}^{\pi}\\ & =\frac{1}{2\pi}\sum_{j=1}^{n+1}\left\{ \left(-1\right)^{j-1}\left(\frac{i}{k}\right)^{j}P\left(n,j-1\right)\left(\pi^{n-\left(j-1\right)}e^{-ik\pi}-\left(-\pi\right)^{n-\left(j-1\right)}e^{ik\pi}\right)\right\} \\ & =\frac{1}{2\pi}\sum_{j=1}^{n+1}\left\{ \left(-1\right)^{j-1}\left(\frac{i}{k}\right)^{j}P\left(n,j-1\right)\left(\pi^{n-\left(j-1\right)}\left(-1\right)^{k}-\left(-1\right)^{n-\left(j-1\right)}\pi^{n-\left(j-1\right)}\left(-1\right)^{k}\right)\right\} \\ & =\frac{1}{2\pi}\sum_{j=1}^{n+1}\left\{ \left(-1\right)^{j-1+k}\left(\frac{i}{k}\right)^{j}P\left(n,j-1\right)\pi^{n-\left(j-1\right)}\left(1-\left(-1\right)^{n-\left(j-1\right)}\right)\right\} \\ & =\frac{1}{2\pi}\sum_{j=0}^{n}\left\{ \left(-1\right)^{n+1-j-1+k}\left(\frac{i}{k}\right)^{n+1-j}P\left(n,n+1-j-1\right)\pi^{j}\left(1-\left(-1\right)^{j}\right)\right\} \cmt{n+1-j\rightarrow j}\\ & =\frac{1}{2\pi}\sum_{j=0}^{n}\left\{ \left(-1\right)^{n-j+k}\left(\frac{i}{k}\right)^{n+1-j}P\left(n,n-j\right)\pi^{j}\left(1-\left(-1\right)^{j}\right)\right\} \\ & =\frac{1}{\pi}\sum_{j=0}^{\left\lfloor \frac{n-1}{2}\right\rfloor }\left\{ \left(-1\right)^{n-\left(2j+1\right)+k}\left(\frac{i}{k}\right)^{n+1-\left(2j+1\right)}P\left(n,n-\left(2j+1\right)\right)\pi^{2j+1}\right\} \\ & =\frac{1}{\pi}\sum_{j=0}^{\left\lfloor \frac{n-1}{2}\right\rfloor }\left\{ \left(-1\right)^{n-1+k}\left(\frac{i}{k}\right)^{n-2j}P\left(n,n-2j-1\right)\pi^{2j+1}\right\} \end{align*} \(k=0\)のとき
\begin{align*} C_{0} & =\frac{1}{2\pi}\int_{-\pi}^{\pi}x^{n}e^{-i0x}dx\\ & =\frac{1}{2\pi}\int_{-\pi}^{\pi}x^{n}dx\\ & =\frac{1}{2\pi}\cdot\frac{1}{n+1}\left[x^{n+1}\right]_{-\pi}^{\pi}\\ & =\frac{1}{2\pi\left(n+1\right)}\left(\pi^{n+1}-\left(-\pi\right)^{n+1}\right)\\ & =\frac{\pi^{n}}{2\left(n+1\right)}\left(1-\left(-1\right)^{n+1}\right) \end{align*} となるのでフーリエ級数展開\(F_{n}\left(x\right)\)は
\begin{align*} F_{n}\left(x\right) & =\sum_{k=-\infty}^{\infty}C_{k}e^{ikx}\\ & =\frac{\pi^{n}}{2\left(n+1\right)}\left(1-\left(-1\right)^{n+1}\right)+\sum_{k\ne0}\frac{1}{\pi}\sum_{j=0}^{\left\lfloor \frac{n-1}{2}\right\rfloor }\left\{ \left(-1\right)^{n-1+k}\left(\frac{i}{k}\right)^{n-2j}P\left(n,n-2j-1\right)\pi^{2j+1}\right\} e^{ikx}\\ & =\begin{cases} \frac{\pi^{2m}}{\left(2m+1\right)}+\sum_{k\ne0}\frac{1}{\pi}\sum_{j=0}^{m-1}\left\{ \left(-1\right)^{-1+k}\left(\frac{-1}{k^{2}}\right)^{m-j}P\left(2m,2m-2j-1\right)\pi^{2j+1}\right\} e^{ikx} & n=2m,m\in\mathbb{N}_{0}\\ \sum_{k\ne0}\frac{1}{\pi}\sum_{j=0}^{m}\left\{ \left(-1\right)^{k}\left(\frac{i}{k}\right)^{2m+1-2j}P\left(2m+1,2m-2j\right)\pi^{2j+1}\right\} e^{ikx} & n=2m+1,m\in\mathbb{N}_{0} \end{cases}\\ & =\begin{cases} \frac{\pi^{2m}}{\left(2m+1\right)}+\sum_{k=1}^{\infty}\frac{1}{\pi}\sum_{j=0}^{m-1}\left(-1\right)^{-1+k}\left(\frac{-1}{k^{2}}\right)^{m-j}P\left(2m,2m-2j-1\right)\pi^{2j+1}\left(e^{ikx}+e^{-ikx}\right) & n=2m,m\in\mathbb{N}_{0}\\ \sum_{k=1}^{\infty}\frac{i}{\pi}\sum_{j=0}^{m}\left\{ \left(-1\right)^{k}\left(\frac{i}{k}\right)^{2\left(m-j\right)}P\left(2m+1,2m-2j\right)\pi^{2j+1}\right\} \left(\frac{1}{k}e^{ikx}-\frac{1}{k}e^{-ikx}\right) & n=2m+1,m\in\mathbb{N}_{0} \end{cases}\\ & =\begin{cases} \frac{\pi^{2m}}{\left(2m+1\right)}+\frac{2}{\pi}\sum_{k=1}^{\infty}\sum_{j=0}^{m-1}\left(-1\right)^{-1+k}\left(\frac{-1}{k^{2}}\right)^{m-j}P\left(2m,2m-2j-1\right)\pi^{2j+1}\cos\left(kx\right) & n=2m,m\in\mathbb{N}_{0}\\ \frac{2}{\pi}\sum_{k=1}^{\infty}\sum_{j=0}^{m}\left\{ \left(-1\right)^{k+1}\left(\frac{-1}{k^{2}}\right)^{m-j}P\left(2m+1,2m-2j\right)\pi^{2j+1}\right\} \frac{\sin\left(kx\right)}{k} & n=2m+1,m\in\mathbb{N}_{0} \end{cases} \end{align*} となる。
\(n=0\)のときは、
\begin{align*} F_{0}\left(x\right) & =1+0\\ & =1 \end{align*} \(n=1\)のときは、
\begin{align*} F_{1}\left(x\right) & =\frac{2}{\pi}\sum_{k=1}^{\infty}\sum_{j=0}^{0}\left\{ \left(-1\right)^{k+1}\left(\frac{-1}{k^{2}}\right)^{\left(0-j\right)}P\left(1,-2j\right)\pi^{2j+1}\right\} \frac{\sin\left(kx\right)}{k}\\ & =\frac{2}{\pi}\sum_{k=1}^{\infty}\left\{ \left(-1\right)^{k+1}P\left(1,0\right)\pi\right\} \frac{\sin\left(kx\right)}{k}\\ & =2\sum_{k=1}^{\infty}\left(-1\right)^{k+1}\frac{\sin\left(kx\right)}{k} \end{align*} \(n=2\)のときは、
\begin{align*} F_{2}\left(x\right) & =\frac{\pi^{2}}{3}+\frac{2}{\pi}\sum_{k=1}^{\infty}\sum_{j=0}^{0}\left(-1\right)^{-1+k}\left(\frac{-1}{k^{2}}\right)^{1-j}P\left(2,2-2j-1\right)\pi^{2j+1}\cos\left(kx\right)\\ & =\frac{\pi^{2}}{3}+\frac{2}{\pi}\sum_{k=1}^{\infty}\left(-1\right)^{-1+k}\left(\frac{-1}{k^{2}}\right)P\left(2,1\right)\pi\cos\left(kx\right)\\ & =\frac{\pi^{2}}{3}+4\sum_{k=1}^{\infty}\frac{\left(-1\right)^{k}}{k^{2}}\cos\left(kx\right) \end{align*} \(n=3\)のときは
\begin{align*} F_{3}\left(x\right) & =\frac{2}{\pi}\sum_{k=1}^{\infty}\sum_{j=0}^{1}\left\{ \left(-1\right)^{k+1}\left(\frac{-1}{k^{2}}\right)^{1-j}P\left(3,2-2j\right)\pi^{2j+1}\right\} \frac{\sin\left(kx\right)}{k}\\ & =\frac{2}{\pi}\sum_{k=1}^{\infty}\left\{ \frac{\left(-1\right)^{k}}{k^{2}}P\left(3,2\right)\pi+\left(-1\right)^{k+1}P\left(3,0\right)\pi^{3}\right\} \frac{\sin\left(kx\right)}{k}\\ & =2\sum_{k=1}^{\infty}\left\{ \frac{\left(-1\right)^{k}6}{k^{2}}+\left(-1\right)^{k+1}\pi^{2}\right\} \frac{\sin\left(kx\right)}{k}\\ & =2\sum_{k=1}^{\infty}\left(-1\right)^{k}\left(\frac{6}{k^{2}}-\pi^{2}\right)\frac{\sin\left(kx\right)}{k} \end{align*} \(n=4\)のときは
\begin{align*} F_{4}\left(x\right) & =\frac{\pi^{4}}{5}+\frac{2}{\pi}\sum_{k=1}^{\infty}\sum_{j=0}^{1}\left(-1\right)^{-1+k}\left(\frac{-1}{k^{2}}\right)^{2-j}P\left(4,4-2j-1\right)\pi^{2j+1}\cos\left(kx\right)\\ & =\frac{\pi^{4}}{5}+\frac{2}{\pi}\sum_{k=1}^{\infty}\left(\left(-1\right)^{-1+k}\left(\frac{-1}{k^{2}}\right)^{2}P\left(4,3\right)\pi+\left(-1\right)^{-1+k}\left(\frac{-1}{k^{2}}\right)P\left(4,1\right)\pi^{3}\right)\cos\left(kx\right)\\ & =\frac{\pi^{4}}{5}+\frac{2}{\pi}\sum_{k=1}^{\infty}\left(-1\right)^{-1+k}\left(\frac{24}{k^{4}}\pi-\frac{4}{k^{2}}\pi^{3}\right)\cos\left(kx\right)\\ & =\frac{\pi^{4}}{5}+2\sum_{k=1}^{\infty}\left(-1\right)^{k}\left(4\pi^{2}-\frac{24}{k^{2}}\right)\frac{\cos\left(kx\right)}{k^{2}} \end{align*}