Define $f(x)=x\ $ for $\ \pi\leq x\leq \pi$. Find the complex form of the Fourier series $Sf(x)$.
My attempt so far.
$$
Sf(x) = \sum_{k = -\infty}^{\infty} C_k[f] e^{-ikx}
$$
where
$$
C_k[f] = \frac{1}{2\pi}\int_{-\pi}^{\pi} f(x) e^{-ikx} dx\
$$
Now,
$$C_k[f]=\frac{1}{2\pi}\int_{-\pi}^{\pi} x e^{-ikx} dx\
$$
Integrating by parts,
$$
C_k[f]= \frac{1}{2\pi} \bigg( \left[ -\frac{xe^{-ikx}}{ik} \right]_{-\pi}^{\pi} - \left[ \frac{1}{i^2 k^2} e^{-inx} \right]_{-\pi}^{\pi}\bigg)
$$
$$
C_k[f]= \frac{1}{2\pi} \bigg( \frac{-\pi}{ik} \left( e^{ikx}+e^{-ikx} \right) + \frac{1}{i^2 k^2} \left(e^{ikx}-e^{-ikx} \right)\bigg)
$$
$$
C_k[f]= \frac{1}{\pi} \bigg( \frac{-\pi}{ik} \left( \frac{e^{ikx}+e^{-ikx}}{2} \right) + \frac{1}{i k^2} \left(\frac{e^{ikx}-e^{-ikx}}{2i} \right)\bigg)
$$
$$
C_k[f]= \frac{1}{\pi} \bigg( \frac{-\pi}{ik} \cos(k\pi) + \frac{1}{i k^2} \sin(kx))\bigg)
$$
Now $\sin(kx)=0.$
Hence,
$$
C_k[f]= \frac{-1}{ik} \cos(k\pi)
$$
$$
C_k[f]= \frac{-1}{ik} (-1)^k
\\ \\$$
So substituting this expression for $C_k[f]$ into $Sf(x)$, yields, $$ Sf(x) = \frac{-1}{i}\sum_{k \neq 0} \frac{(-1)^k}{k} e^{ikx} $$ But the answer is $$ Sf(x) = i\sum_{k \neq 0} \frac{(-1)^k}{k} e^{ikx} $$ Where have I made a mistake. I don't understand how this is the answer.