I am trying to solve the following problem.
I have a known Fourier coefficient, given by the formula $\alpha_k=\frac{1}{2a}\int_{-a}^af(t)\cos\omega kt\text{d}t$
and I would like to show how to isolate $f(t)$ when I know $\alpha_k$.
My attempt, which I am not sure of,is with $\alpha_k=2\frac{\sin\omega ak}{\omega k}$:
$$2\frac{\sin\omega ak}{\omega k}=\frac{1}{2a}\int_{-a}^af(t)\cos\omega kt\text{d}t$$
$$\frac{d}{dt}2\frac{\sin\omega ak}{\omega k}=\frac{1}{2a}f(t)\cos\omega kt\text{d}t$$
However the left hand side is zero. How can I prove that f(t) can be "recovered" from the integral, when knowing $\alpha_k$?
This is for a proof, and that is why it is a little bit strange question.
The alternative is to use indefinite integral, then the operation of differentiating both sides will work out,but I am not sure since an indefinite integral is not the correct Fourier coefficient formula.
Any ideas appreciated
Thanks
Assuming $f(t)=m=const.$, and solving the integral, yields:
$$a_k=\frac{m}{a}\cdot\frac{\sin(wak)}{wk}$$
From here you can equate this expression to your given value of $a_k$ and solve for $m$, which is the same as $f(t)$.
As @GreginGre had mentioned in their comment, recovering $f(t)$ in the general case (where it isn't necessarily equal to a constant), is probably impossible without more information.