I have a question from homework, which is really not understandable..
the question:
$$\int _0^{2\pi }f\left(ax\right)\cdot f\left(x+b\right)dx$$
$f$ is integrable and $2\pi$ periodic.
express the expression above as in Fourier coefficient of $f$.
I tried:
$g(x):= f(ax)$
$$\hat{g}{\left(n\right)}=\frac{1}{2\pi }\int _0^{\frac{2\pi }{a}}f\left(t\right)e^{-in\cdot \frac{a}{t}}dt\cdot \frac{1}{a}=\hat{f}{\left(n\right)\cdot \frac{1}{a}}$$
Is it true to say something like that? because the boundary of the integral is not $[0,2\pi]$ as it should be.
If it is true, then I can do the same with $f(x+b)$ and its easy.
If not, I am in a problem and did not understand :\
If my question is not understandable, I will give an example:
express fourier coefficient of $f'(x)$ as fourier coefficient of $f(x)$, how will I do that?:
$i\cdot n\cdot \hat{f}{\left(n\right)}$
Hope it was understandable.
EDIT: Tried what I have been told: received this: $$2\pi \cdot \sum _{m,n=-\infty }^{\infty }\hat{f}\left(m\right)\cdot \hat{f}\left(n\right)\cdot e^{inb}, ma+n =0$$ else, it is 0 ( it is Kroneker delta's )
The $2 \pi$ periodic Fourier series for $f(x)$ is
$$f(x)=\sum\limits_{n=-\infty}^{\infty} c_n\,e^{i n x}\tag{1}$$
where
$$c_n=\frac{1}{2 \pi}\int\limits_0^{2 \pi} f(t)\,e^{-i n t}\,dt\tag{2}$$
(see Exponential_form) so
$$f(a x) \cdot f(x+b)=\left(\sum\limits_{n=-\infty}^{\infty}c_n\,e^{i n a x}\right) \cdot \left(\sum\limits_{m=-\infty}^{\infty} c_m\,e^{i m (x+b)}\right)=\sum\limits_{n=-\infty}^{\infty}c_n \sum\limits_{m=-\infty}^{\infty} c_m\,\,e^{i n a x}\,e^{i m (x+b)}\tag{3}$$
and since
$$\int\limits_0^{2\pi} e^{i n a x}\, e^{i m (x+b)}\,dx=\frac{i e^{i b m} \left(1-e^{2 i \pi (a n+m)}\right)}{a n+m}\tag{4}$$
term-wise integration leads to:
$$\int\limits_0^{2\pi} f(a x) \cdot f(x+b)\,dx=\sum\limits_{n=-\infty}^{\infty}c_n\ \sum\limits_{m=-\infty}^{\infty} c_m\, \frac{i e^{i b m} \left(1-e^{2 i \pi (a n+m)}\right)}{a n+m}\tag{5}$$