From linear algebra we know that if a set of vectors form a basis for a space, their is a unique linear combination of the basis to form any vector in that space.
I'm assuming this extends to scalar functions too - if there is a set of functions that form a basis for a set of functions, then their is a unique linear combination of the basis to form any function in that set.
In the case of Fourier series, the basis are the complex exponentials with the same base period and the set of functions they span are periodic functions.
Now consider some periodic function $f$ with fundamental period $T_0$. Now say we use the following formula to find the Fourier coefficients: $$a_k = \int_0^{T_0}f \dot {}e^{\frac{-itk}{T_0}}dt$$
The basis in this case is $e^{\frac{-itk}{T_0}}$ for integer $k$. Call this basis 1.
Now consider finding the coefficients in another way - by integrating over some period that isn't the fundamental period - for instance $2T_0$ (my prof said this was valid).
$$b_k = \int_0^{2T_0}f \dot {}e^{\frac{-itk}{2T_0}}dt$$
In this case, the basis is $e^{\frac{-itk}{2T_0}}$ for integer $k$. Call this basis 2. Note that basis 1 is a subset of basis 2.
Now consider two different ways of constructing $f$ from basis 2.
1) Use the $b_k$ coefficients. This obviously constructs $f$ since it's just a Fourier series.
2) If a function in basis 2 is also in basis 1, set it's coefficient to $a_k$. Otherwise set it to $0$. In other words, you are constructing $f$ from the Fourier coefficients of basis 1, and "not using" any other functions that are in basis 2.
It seems like I have constructed the same function from one basis in two different ways - how is this possible? Doesn't this violate the rule that their is a unique combination of the basis to form a function in it's span?
The only way I can think of for this rule to not be violated is if the set of coefficients $b_k$ is the same as the set of coefficients $a_k$ for functions that exist in both basis and zero for functions unique to basis 2. (In other words both constructions use the same coefficients). But my prof said this wasn't the case - the coefficients are different.
I think the problem is that those two was are actually not different.
I try to give an example of what I mean: If $f$ is a $T_0$-periodic function and you integrate
$$b_k = \int_0^{2T_0} f \exp\left(\frac{2\pi i t}{2T_0}\right)dt = \int_0^{T_0} f \exp\left(\frac{2\pi i t}{2T_0}\right)dt+\int_{T_0}^{2T_0} f \exp\left(\frac{2\pi i t}{2T_0}\right)dt$$
$$= \int_0^{T_0} f \exp\left(\frac{2\pi i t}{2T_0}\right)dt + \int_{0}^{T_0} f \exp\left(\frac{2\pi i (t+T_0)}{2T_0}\right)dt$$
$$= \int_0^{T_0} f \exp\left(\frac{2\pi i t}{2T_0}\right)dt - \int_{0}^{T_0} f \exp\left(\frac{2\pi i t}{2T_0}\right)dt = 0$$
Because: $$\exp\left(\frac{2\pi i (t+T_0)}{2T_0}\right) = \exp\left(\frac{2\pi i t}{2T_0}\right)\cdot \underbrace{\exp\left(\frac{2\pi i T_0}{2T_0}\right)}_{-1}$$
So I think every 2nd coefficient $b_k$ will be zero, that means you still have the same linear combination. Does this make sense?
PS: I think you need the factor $2\pi$ in the exponent in order to get the fourier coefficients? (Well apart from the leading constants.) I now just assumed you forgot those.