I want to show that the Fourier coefficients $\int_{-\pi}^\pi e^{ij \lambda} f'(\lambda) d \lambda$ of the derivative of a continuously differentiable function $f: [ - \pi, \pi] \rightarrow \mathbb{R}$ with $f(-\pi) = f( \pi)$, are equal to $-ij f_j$, for $f_j$ the Fourier coefficients of $f$.
My attempt of proving this: By partial integration $$\int_{-\pi}^\pi e^{ij \lambda} f'(\lambda) d \lambda = \left[e^{ij \lambda} f(\lambda) \right]_{-\pi}^\pi - \int_{-\pi}^\pi ije^{ij \lambda} f(\lambda)d\lambda.$$ We can then write out the first part but as we have the "$j$" it does not cancel out, I think. Could anyone help me with the details? Thanks alot!
As @Daniel Fischer pointed out it does cancel in contrary to what I thought then the Fourier coefficients of $f$ are given by $f_j = \int_{-\pi}^\pi e^{ij\lambda} f(\lambda) d \lambda$. To this end by integration by parts \begin{align*} \int_{-\pi}^\pi e^{ij\lambda} f'(\lambda) d \lambda &= \left[ e^{ij\lambda} f(\lambda) \right]_{-\pi}^\pi - \int_{-\pi}^\pi ije^{ij\lambda} f(\lambda) d\lambda \\ &= e^{ij\pi} f(\pi) - e^{-ij\pi} f(-\pi) - ij\int_{-\pi}^\pi e^{ij\lambda} f(\lambda) d\lambda \\ &= e^{ij\pi} f(\pi) - e^{ij\pi} f(\pi) - ij\int_{-\pi}^\pi e^{ij\lambda} f(\lambda) d\lambda\\ &=- ij\int_{-\pi}^\pi e^{ij\lambda} f(\lambda) d\lambda = -ij f_j. \end{align*} Which verifies my claim.