Can I change limits of integration from $[-P/2,P/2]$ to $[0,P]$?

Question

For a function $f(x+2\pi)=f(x)$, I have the usual FS coefficient $a_n=\frac{1}{\pi}\int_{-\pi}^{\pi}f(t)\cos(nt)\, dt$ and equivalent for $b_n$.

Say I now have a function with period $P$, $g(x+P)=g(x)$ and use the change of variable $t=\frac{2\pi}{P}x$, $$ a_n=\frac{2}{P}\int_{-P/2}^{P/2}g(x)\cos\Big(\frac{2\pi n }{P}x\Big)\, dx $$

Is it correct to change the limits from $[-P/2,P/2]$ to $[0,P]$, i.e. \begin{align} a_n&=\frac{2}{P}\int_{-P/2}^{P/2}g(x)\cos\Big(\frac{2\pi n }{P}x\Big)\, dx\\ &=\frac{2}{P}\int_{0}^{P}g(x)\cos\Big(\frac{2\pi n }{P}x\Big)\, dx \qquad ? \end{align}

Answer 1

Yes. Since the function is periodic with period P, the graph of the function over the interval (-P/2,0) is identical to the graph over the interval (P/2,P).

Answer 2

Here's a traditional proof. Observe that

$$a_n = \frac{2}{P}\int_{-P/2}^{0}g(x)\cos\Big(\frac{2\pi n }{P}x\Big)\, dx + \frac{2}{P}\int_{0}^{P/2}g(x)\cos\Big(\frac{2\pi n }{P}x\Big)\, dx$$

Now, consider

\begin{align} \alpha_n&=\frac{2}{P}\int_{0}^{P}g(x)\cos\Big(\frac{2\pi n }{P}x\Big)\, dx\\ &=\frac{2}{P}\int_{0}^{P/2}g(x)\cos\Big(\frac{2\pi n }{P}x\Big)\, dx + \frac{2}{P}\int_{P/2}^{P}g(x)\cos\Big(\frac{2\pi n }{P}x\Big)\, dx \end{align}

We have hence that $\alpha_n=a_n$ if and only if

$$ \int_{P/2}^{P}g(x)\cos\Big(\frac{2\pi n }{P}x\Big)\, dx = \int_{-P/2}^{0}g(x)\cos\Big(\frac{2\pi n }{P}x\Big)\, dx\tag{$*$} $$

Now, consider the change of variables $u=x+P$. Then $du=dx$, $u=P/2$ when $x=-P/2$ and $u=P$ when $x=0$. Applying the substitution to the RHS on $(*)$ yields:

\begin{align} \int_{-P/2}^{0}g(x)\cos\Big(\frac{2\pi n }{P}x\Big)\, dx &= \int_{P/2}^{P}g(u-P)\cos\Big(\frac{2\pi n }{P}\,(u-P)\Big)\, dx\\ &= \int_{P/2}^{P}g(u)\cos\Big(\frac{2\pi n }{P}\,u - 2\pi n\Big)\, dx\\ &= \int_{P/2}^{P}g(u)\cos\Big(\frac{2\pi n }{P}\,u\Big)\, dx \end{align}

which is precisely $(*)$.