2.3

Given that $f$ is $2\pi$-periodic, i.e. $f(x) = f(x + 2n\pi)$ for any $n \in \mathbb{Z}$, just as the hint prescribes, observe that, for some $k \in \mathbb{Z}$, the interval: $$I = (a,b] \subset (k\pi, (k+4)\pi]$$ Define $c = (k+2)\pi$. Certainly, if $c \neq a$, then $c \in (a,b]$. Now, $$\int_{(a,b]} f(x)dx = \int_{(a,c]} f(x)dx + \int_{(c,b]} f(x)dx$$ Since $f(x)|_{(c,b]} = f(x - 2\pi)|_{(k\pi,a]}$: $$\int_{(c,b]} f(x)dx = \int_{(k\pi, a]} f(x)dx$$ So we can certainly re-write: $$\int_{(a,b]} f(x)dx = \int_{(k\pi, c]} f(x)dx = \int_{(k\pi,(k+2)\pi]}f(x)dx$$ Finally, since we can certainly break up the integral: $$\int_{(k\pi,(k+2)\pi]}f(x)dx = \int_{(k\pi,(k+1)\pi]}f(x)dx + \int_{((k+1)\pi,(k+2)\pi]}f(x)dx$$ ... by $2\pi$-periodicity, again, it's apparent that: $$\int_{(a,b]} f(x)dx = \int_{(k\pi,(k+2)\pi]}f(x)dx = \int_{((k+1)\pi,(k+3)\pi]}f(x)dx$$ Thus, we can say this equality holds for any integer $k$, specifically, $k= -1$. Therefore: $$\int_{(a,b]} f(x)dx = \int_{(-\pi, \pi]}f(x)dx$$