Part 5(e):We use an even continuation for this function. Integration was done using the angle-sum identity as in
Problem 3.
\begin{equation*}
b_n = 0.
\end{equation*}
\begin{equation*}
a_0 = \frac{2}{\pi} \int_{0}^{\pi} \sin{((m-\frac{1}{2})x)}dx \\
= \frac{-2}{\pi (m - \frac{1}{2}} \cos{(m-\frac{1}{2})x} \big|_{0}^{\pi}\\
=\frac{-2}{\pi (m- \frac{1}{2})}.
\end{equation*}
\begin{equation*}
a_n = \frac{2}{\pi} \int_{0}^{\pi} \sin{((m-\frac{1}{2})x)}\cos{nx} dx\\
= -\frac{1}{\pi}\left[\frac{\cos{m + n - \frac{1}{2}} x }{m + n - \frac{1}{2}} + \frac{\cos{m - n - \frac{1}{2}} x }{m - n - \frac{1}{2}} \right]_{0}^{\pi}\\
= \frac{1}{\pi}\left(\frac{1}{m + n - \frac{1}{2}} + \frac{1}{m - n - \frac{1}{2}} \right).
\end{equation*}
\begin{equation*}
\sin{((m-\frac{1}{2})x)} = \frac{1}{\pi (m- \frac{1}{2})} + \frac{1}{\pi}\sum_{n=1}^{\infty} \left(\frac{1}{m + n - \frac{1}{2}} + \frac{1}{m - n - \frac{1}{2}} \right) \cos{n x}.
\end{equation*}
