Fourier sine series expansion 
The function $f(x)$ is defined as $$f(x)=1\qquad0<x<\pi$$
Sketch the odd extension and show that the Fourier sine series expansion is
$$f(x)~\frac4\pi\sum_{n=1}^\infty\frac{\sin((2n-1)x)}{2n-1}$$

In this question, $f(x)=1$, $a_0$ is obviously $2$ and $a_m$ and $b_m$ are zero when plugged into Fourier series equation. Could you please explain why the Fourier series can be expanded in such this form?
 A: I don't know how you got $a_0 = 2$ and $A_n, B_n = 0$ so I'll post the solution.
You have
$$\begin{align}
f(x) &= 1 \\
&= \sum_{n=1}^{\infty} B_n \sin\bigg(\frac{n\pi x}{L}\bigg) \ \ \text{(because we're doing the odd extension)} \\
&= \sum_{n=1}^{\infty} B_n \sin(nx) \ \ \text{(with $L = \pi$)} \\
\end{align}$$
Integrating over our domain and using orthogonality, we find
$$\begin{align}
\int_{0}^{\pi} f(x)\sin(mx)dx &= \int_{0}^{\pi} \sin(mx)dx \ \ (1)\\
&= \sum_{n = 1}^{\infty} B_n \int_{o}^{\pi} \sin(nx)\sin(mx) dx \\
&= \sum_{n = 1}^{\infty} \frac{B_n}{2} \int_{-\pi}^{\pi} \sin(nx)\sin(mx) dx \ \ \text{(as an odd function $\times$ odd function is an even function)} \\
&= \frac{B_m \pi}{2},\ \ n = m \ \ (2)
\end{align} $$
Equating $(1)$ and $(2)$, we find
$$\begin{align}
\frac{B_m \pi}{2} &= \int_{0}^{\pi} \sin(mx)dx \\
\implies B_m &= \frac{2}{\pi} \int_{0}^{\pi} \sin(mx)dx \\
&= \frac{2}{\pi} \bigg[\frac{-\cos(mx)}{m} \bigg]_{0}^{\pi} \\
&= \frac{-2}{m \pi}\bigg[(-1)^{m} - 1 \bigg] \\
&= \begin{cases} 
      0 & m = \text{even} \\
      \frac{4}{m \pi} & m = \text{odd} \\ 
\end{cases}
\end{align}$$
Hence we should set $m = 2j - 1$ for $j \ge 1$ to keep only those cases that are non-zero.
Hence, our solution is given by
$$\begin{align}
f(x) &= \sum_{n=1}^{\infty} B_n \sin(nx) \\
&= \sum_{m=1}^{\infty} B_m \sin(mx) \ \ \text{(using $(2)$)} \\
&= \sum_{m=1}^{\infty} \frac{4}{m \pi} \sin(mx) \\
&= \frac{4}{\pi} \sum_{m=1}^{\infty} \frac{\sin(mx)}{m} \\
&= \frac{4}{\pi} \sum_{j=1}^{\infty} \frac{\sin((2j - 1)x)}{2j - 1} \\
\end{align}$$
EDIT
$$\begin{align}
f(x) &= 1 \\
&= \sum_{n=1}^{\infty} B_n \sin(nx) \\
\end{align}$$
Multiply both sides by $\sin(mx)$ for orthogonality
$$\begin{align}
\implies f(x)\sin(mx) &= \sin(mx) \sum_{n=1}^{\infty} B_n \sin(nx) \\
&= \sum_{n=1}^{\infty} B_n \sin(nx)\sin(mx) \\
\end{align}$$
where we can take the $\sin(mx)$ term inside the series because we aren't summing over $m$ so it can be thought of almost like a constant.
Integrate both sides over $[0, \pi]$
$$\begin{align}
\implies \int_{0}^{\pi} f(x)\sin(mx) dx &= \int_{0}^{\pi} \bigg( \sum_{n=1}^{\infty} B_n \sin(nx)\sin(mx) \bigg) dx \\
&= \sum_{n=1}^{\infty} B_n \int_{0}^{\pi} \sin(nx)\sin(mx) dx \\
\end{align}$$
Replace $f(x)$ with $1$
$$\begin{align}
\implies \int_{0}^{\pi} f(x)\sin(mx) dx &= \int_{0}^{\pi} \sin(mx) dx \\ &= \int_{0}^{\pi} \bigg( \sum_{n=1}^{\infty} B_n \sin(nx)\sin(mx) \bigg) dx \\
&= \sum_{n=1}^{\infty} B_n \int_{0}^{\pi} \sin(nx)\sin(mx) dx \\
\end{align}$$
