Prove that $$\hat{f}(n)={\frac{2}{\pi(1-4n^2)}},\ given\ thatf(x)=|sin(\pi x)||,\int_{0}^{1}sin\pi(x)dx={\frac{2}\pi}\\where\ \hat{f}(x)=\int_{0}^{1}f(x)e(-nx)dx. \ Use\ the\ fact\ that\\sin\pi(x)={\frac{e^{x/2}-e^{-x/2}}{2i}}$$
Here's my attempt $$ as\sin(\pi x)={\frac{e^{x/2}-e^{-x/2}}{2i}},\ \ f(x)= |{\frac{e^{x/2}-e^{-x/2}}{2i}}|=\ {\frac{e^{x/2}-e^{-x/2}}{2}}\mathbf{(I\ think\ this \ might\ be\ wrong, perhaps\ I\ should\ find \ the\ norm\ because\ {\frac{e^{x/2}-e^{-x/2}}{2}}\ is\ a \ complex\ number } )\\\hat{f}(x)={\frac{1}{2}}\int_{0}^{1}{\frac{e^{x/2}-e^{-x/2}}{2}}={\frac{1}{2}}\int_{0}^{1}f(x)e(-nx)dx={\frac{e^{x/2}-e^{-x/2}}{2}}e^{-nx}={\frac{1}{2}}\int_{0}^{1} e^{(x(1-2n))/2}- e^{(-x(1+2n))/2}={\frac{1}{2}}(({\frac{2}{1-2n}})(e^{((1-2n)/2)}-1)+({\frac{2}{1 +2n}})(e^{(-(1+2n))/2)}-1)) $$ If I expanded the expression, The nonexponential term comes out to be $\frac{2}{1-4n^{2}}$ but I'm still left with with exponential terms that I can't seem to manipulate $({\frac{2}{1-2n}})(e^{((1-2n)/2)})$ and $({\frac{2}{1 +2n}})(e^{(-(1+2n))/2)})$ into anything simpler.
I would get
\begin{eqnarray*} \sin (\pi x) &=&\frac{e^{i\pi x}-e^{-i\pi x}}{2i} \\ \hat{f}(n) &=&\int_{0}^{1}dx\exp [-nx]\frac{e^{i\pi x}-e^{-i\pi x}}{2i} \\ &=&\frac{1}{2i}\int_{0}^{1}dx\{\exp [-(n-i\pi )x]-\exp [-(n+i\pi )x]\} \end{eqnarray*} where the integral is trivial