I need to solve the Fourier coefficients of $\mathrm{f}\left(x\right) = \sin\left(\pi x\right)$. I know, since $\mathrm{f}$ is an $odd$ function, that the coefficients are given by:
\begin{align} b_{n} & = \frac{2}{L}\int_{0}^{L}\mathrm{f}\left(t\right) \sin\left(\frac{2n\pi t}{L}\right)\,\mathrm{d}t = \int_{0}^{2}\sin\left(\pi t\right)\sin\left(n\pi t\right)\,\mathrm{d}t = \\[5mm] & = -\frac{1}{2}\int_{0}^{2}\sin\left(\left(n + 1\right)\pi t\right)\,\mathrm{d}t + \frac{1}{2}\int_{0}^{2}\sin\left(\left(n - 1\right)\pi t\right)\,\mathrm{d}t = 0 \end{align}
I however, found that it should be equal to $b_n=-\,\frac{2}{\left(4n^{2} - 1\right)\pi}$.
My question is, what am I doing wrong ?.
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\mrm{f}\pars{x} \equiv \sin\pars{\pi x} = \sum_{n = 1}^{\infty}b_{n}\sin\pars{\pi n x} \\[5mm] \implies &\ \int_{-1}^{1}\mrm{f}\pars{x}\sin\pars{n\pi x}\,\dd x \\[5mm] = &\ \sum_{m = 1}^{\infty} b_{m}\int_{-1}^{1}\sin\pars{\pi m x}\sin\pars{\pi n x}\,\dd x \\[5mm] \implies &\ \int_{-1}^{1}\sin\pars{\pi x}\sin\pars{n\pi x}\,\dd x \\[5mm] = &\ \sum_{m = 1}^{\infty} b_{m}\int_{-1}^{1}\sin\pars{\pi m x}\sin\pars{\pi n x}\,\dd x \\[5mm] \implies &\ 2\,\delta_{n1}\int_{0}^{1}{1 - \cos\pars{2\pi x} \over 2}\,\dd x \\[5mm] = &\ \sum_{m = 1}^{\infty} b_{m}\bracks{% 2\,\delta_{mn}\int_{0}^{1}{1 - \cos\pars{2\pi nx} \over 2}\,\dd x} \\[5mm] \implies &\ 2\delta_{n1} = 2b_{n} \implies \bbox[10px,border:1px groove navy]{b_{n} = \delta_{n1}} \\ & \end{align}