I could use some help with the basic problem of finding the fourier series of the $2\pi$-periodic function $f(t) = t+1$, for $-\pi < t < \pi$.
Edit: I removed my calculations, since they were based on the fact that it was an odd function. $f(t)$ is neither an odd or an even function.
The function $1$ has a simple Fourier series -- it is $1$. That allows you to reduce to an odd function $f(t)$ because the Fourier series of $1+t$ is the Fourier series of $1$ plus the Fourier series of $t$. For $t$, $$ t \sim \sum_{n=1}^{\infty}\left[\frac{2}{\pi}\int_{0}^{\pi}u\sin(nu)\right]\sin(nu)du $$ And, \begin{align} \int u\sin(nu)du&=-u\frac{\cos(nu)}{n}+\int\frac{\cos(nu)}{n}du \\ &=-u\frac{\cos(nu)}{n}+\frac{\sin(nu)}{n^2}+C \end{align} That gives coefficients \begin{align} \frac{2}{\pi}\int_{0}^{\pi}u\sin(nu)du&=\left.\frac{2}{\pi}\left[-u\frac{\cos(nu)}{n}+\frac{\sin(nu)}{n^2}\right]\right|_{0}^{\pi} \\ &=\frac{2}{\pi}\left[-\pi\frac{(-1)^{n}}{n}\right]=\frac{2(-1)^{n+1}}{n} \end{align} So your Fourier series for $f(t)=1+t$ is $$ f(t) \sim 1 + \sum_{n=1}^{\infty}\frac{2(-1)^{n+1}}{n}\sin(nx). $$