Why the integral in $[-\pi-\frac{\pi}{k}, \pi-\frac{\pi}{k}]$ equals the integral in $[-\pi, \pi]$?

Question

I'm showing that the coefficients of the Fourier series of a $2\pi$-periodic function $f$ can be written as $$a_k = \frac{1}{2\pi} \int\limits_{-\pi}^{\pi} \left(f(x) - f\left(x - \frac{\pi}{k}\right)\right)\cos(kx) dx$$ and $$b_k = \frac{1}{2\pi} \int\limits_{-\pi}^{\pi} \left(f(x) - f\left(x - \frac{\pi}{k}\right)\right)\sin(kx) dx.$$ Firstly, I consider the change $x = u - \frac{\pi}{k}$. Then $$a_k = \frac{1}{\pi} \int\limits_{-\pi}^{\pi} f(x)\cos(kx) dx = -\frac{1}{\pi} \int\limits_{-\pi-\frac{\pi}{k}}^{\pi-\frac{\pi}{k}} f\left(x - \frac{\pi}{k}\right)\cos(kx) dx.$$ So, considering the mean $a_k = \frac{a_k + a_k}{2}$, $$a_k = \frac{1}{2\pi}\left(\int\limits_{-\pi}^\pi f\left(x\right)\cos(kx)dx - \int\limits_{-\pi-\frac{\pi}{k}}^{\pi-\frac{\pi}{k}}f\left(x - \frac{\pi}{k}\right)\cos(kx)dx\right).$$ Now I have a problem justifying that $\int\limits_{-\pi-\frac{\pi}{k}}^{\pi-\frac{\pi}{k}}f\left(x - \frac{\pi}{k}\right)\cos(kx)dx = \int\limits_{-\pi}^\pi f\left(x-\frac{\pi}{k}\right)\cos(kx)dx$. Is it because $f$ is $2\pi$-periodic? Is it the same for $b_k$?

Answer 1

$\int\limits_{-\pi-\frac{\pi}{k}}^{\pi-\frac{\pi}{k}}f\left(x - \frac{\pi}{k}\right)\cos(kx)dx = \int\limits_{-\pi-\frac{\pi}{k}}^{-\pi} f\left(x-\frac{\pi}{k}\right)\cos(kx)dx\space\space+\int\limits_{-\pi}^{\pi} f\left(x-\frac{\pi}{k}\right)\cos(kx)dx\space\space-\int\limits_{\pi-\frac{\pi}{k}}^{{\pi}} f\left(x-\frac{\pi}{k}\right)\cos(kx)dx\space\space$

Now by doing a change of variable from $x$ to $x+2\pi$, the above equation becomes

$\int\limits_{\pi-\frac{\pi}{k}}^{\pi-\frac{\pi}{k}}f\left(x - \frac{\pi}{k}\right)\cos(kx)dx = \int\limits_{\pi-\frac{\pi}{k}}^{\pi} f\left(x+2\pi-\frac{\pi}{k}\right)\cos(kx)dx\space\space+\int\limits_{-\pi}^{\pi} f\left(x-\frac{\pi}{k}\right)\cos(kx)dx\space\space-\int\limits_{\pi-\frac{\pi}{k}}^{{\pi}} f\left(x-\frac{\pi}{k}\right)\cos(kx)dx\space\space$

and as you rightly guessed , since $f(x)$ is $2\pi$ periodic , hence $f(x)=f(x+2\pi)$ and therefore first and third integral cancel out resulting in

$\int\limits_{\pi-\frac{\pi}{k}}^{\pi-\frac{\pi}{k}}f\left(x - \frac{\pi}{k}\right)\cos(kx)dx = \int\limits_{-\pi}^{\pi} f\left(x-\frac{\pi}{k}\right)\cos(kx)dx\space\space$

which now completes your proof.

Similarly the result can be proved for $b_k$

Answer 2

The Fourier series for a $P$-periodic function is

$$f(x)=\frac{\text{a0}}{2}+\lim_{N\to\infty}\left(\sum\limits_{n=1}^N \left(a(n) \cos\left(\frac{2 \pi n x}{P}\right)+b(n) \sin\left(\frac{2 \pi n x}{P}\right)\right)\right)\tag{1}$$

where

$$a_n=\frac{2}{P} \int\limits_P f(x)\, \cos(\frac{2 \pi}{P } n x) dx\,,\quad n\ge 0\tag{2}$$

and

$$b_n=\frac{2}{P} \int\limits_P f(x)\, \sin(\frac{2 \pi}{P } n x) dx\,,\quad n\ge 1\tag{3}$$

so for the integrals in (2) and (3) above you only need to select the lower and upper evaluation limits such that their difference is the period $P$ (see Common forms of the Fourier series).