$$F(f′)(ξ) =\int^\infty_{-\infty} e^{−2\pi i\xi} tf′(t)dt \\ =\left.e^{−2\pi i\xi t}f(t)\right\vert^\infty_{t=−\infty}−\int^\infty_{-\infty} −2\pi i\xi e^{−2\pi i\xi t}f(t)dt =2\pi i\xi\cdot F(f)(\xi) $$
(The first term must vanish, as we assume $f$ is absolutely integrable on $\mathbb R$.) This is for fourier transform and we have a similar relation for fourier series. But why is the first term equal to zero in fourier series it will have $1/T$ multiplied to it and the limits will change. The first term will be zero for every function with time period equal to $2\pi$. But in standard books the relation is valid for all functions, how?
For continuous functions it should be obvious since $f(0) = f(1)$ and $e^{-i2n\pi 1} = e^{-i2n\pi 0}=1$ which means that the first term will cancel out.
The case of a non-continuous function complicates things a bit, but we can do the integration over any interval of length $1$ and for this to me meaningful $f$ must be derivable and therefore continuous almost everywhere so we can certainly find an interval so that $f$ is continuous at the endpoint (and the first term will therefore cancel anyway).
To put it in formula language we have:
$$c_n = \int_0^1 f'(t) e^{-2in\pi t} dt = \int_0^\tau f'(t) e^{-2in\pi t} dt + \int_\tau^1f'(t) e^{-2in\pi t} dt \\ = \int_1^{\tau+1}f'(t) e^{-2in\pi t} dt + \int_\tau^1f'(t) e^{-2in\pi t}dt \\ = \int_\tau^{\tau}f'(t) e^{-2in\pi t} dt \\ = \left.f(t)e^{-2in\pi t}\right\vert_{t=\tau}^{1+\tau}+{2i\pi}\int_\tau^{\tau}f'(t) e^{-2in\pi t} dt $$
For any $\tau$ where $f$ is continouos at $\tau$. Therefore the first term becomes because $f(1+\tau) = f(\tau)$:
$$f(\tau+1)e^{-2in\pi(\tau+1)} - f(\tau)e^{-2in\pi\tau} = f(\tau+1)e^{-2in\pi}e^{-2in\pi\tau} - f(\tau)e^{-2in\pi\tau} \\ \left(f(\tau+1)-f(\tau)\right) e^{-2in\pi\tau} = 0$$