I got the Fourier series as $(e-e^{-1})(\frac 1 2+\sum \limits _{n=1} ^\infty \frac {(-1)^n(\cos(n\pi x)-n\pi \sin(n\pi x)} {1+n^2\pi^2})$.
Although I've seen the answer online as being $\sum \limits _{n=0} ^\infty \frac {(-1)^n(e-e^-1)(\cos(n\pi x)-n\pi \sin(n\pi x))} {1+n^2\pi^2}$ but I don't see why there's no $\frac 1 2$ coefficient on $a_0$ in the answer I found on the internet.
Am I right or have I made a mistake?
The correct coefficients are $a_0 = \frac {\mathbb e - \mathbb e ^{-1}} 2, \space a_n = (\mathbb e - \mathbb e ^{-1}) \frac {(-1)^n} {1+n^2 \pi^2}, \space b_n = -(\mathbb e - \mathbb e ^{-1})n\pi \frac {(-1)^n} {1+n^2 \pi^2}$.