Basic properties of Fourier series .1

Question

Suppose $f$ is $2\pi$-peridic and integrable on any finite interval.

Prove that if $a,b \in \mathbb{R}$, then $$ \int_{a}^{b} f(x)\ dx=\int_{a+2\pi}^{b+2\pi} f(x)\ dx =\int_{a-2\pi}^{b-2\pi}f(x)\ dx . $$ Also prove that $$\int_{-\pi}^{\pi} f(x+a)\ dx=\int_{-\pi}^{\pi} f(x)\ dx =\int_{-\pi +a}^{\pi+a}f(x)\ dx . $$

Answer 1

For the first line we use the substitution with $\varphi(x) = x \pm 2\pi$ to obtain $$\int_{a \pm 2\pi}^{b \pm 2\pi} f(x) \mathrm{d}x = \int_{\varphi(a)}^{\varphi(b)} f(x) \mathrm{d}x = \int_{a}^{b} f(x \pm 2 \pi) \mathrm{d}x = \int_{a}^{b} f(x) \mathrm{d}x.$$ For the second line we have using substitution with $\varphi(x) = x + a$ $$\int_{-\pi}^\pi f(x+a) \mathrm{d}x = \int_{-\pi + a}^{\pi + a} f(x) \mathrm{d}x.$$ Splitting up the integral yields $$\int_{-\pi + a}^{\pi + a} f(x) \mathrm{d}x = \int_{-\pi + a}^\pi f(x) \mathrm{d}x + \int_\pi^{\pi + a} f(x) \mathrm{d}x.$$ By the first assertion we have $$\int_{-\pi + a}^\pi f(x) \mathrm{d}x + \int_\pi^{\pi + a} f(x) \mathrm{d}x = \int_{-\pi + a}^\pi f(x) \mathrm{d}x + \int_{-\pi}^{-\pi + a} f(x) \mathrm{d}x = \int_{-\pi}^\pi f(x) \mathrm{d}x.$$ This shows the second line.

Answer 2

The prove of the first line:$\\$ Since, $d(x+ 2\pi) = dx$ and since $f(x) = f(x+2\pi)$, then if $x = a$ then $x + 2\pi = a + 2\pi$.$\\$ Also if $x = b$ then then $x + 2\pi = b + 2\pi$.Then $$\int_{a}^{b} f(x)dx = \int_{a + 2\pi }^{b +2\pi} f(x + 2\pi)dx = \int_{a + 2\pi }^{b +2\pi} f(x )dx $$ as $f$ is periodic.