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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 $$\int_{-\pi}^{\pi} f(x+a)\ dx=\int_{-\pi}^{\pi} f(x)\ dx =\int_{-\pi +a}^{\pi+a}f(x)\ dx . $$

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  • $\begingroup$ What did you try? Did you try integration by substitution? $\endgroup$
    – user301452
    Feb 26, 2017 at 10:01
  • $\begingroup$ yeah, I used the definition of a periodic function which is f (x + \2pi) = f(x) , and I adjust the boundary to be x + 2pi and also x-2pi and I proved the first line, thank u. but what about the second line ? $\endgroup$
    – Emptymind
    Feb 26, 2017 at 19:35
  • $\begingroup$ Assume $a \in [-\pi, \pi]$ (why?) and try to split up the integral into two integrals over smaller intervals, i.e. $$\int_{-\pi}^\pi f(x+a) \mathrm{d}x = \int_{-\pi}^a f(x+a) \mathrm{d}x + \int_{a}^\pi f(x+a) \mathrm{d}x.$$ Now use periodicity to glue it together at another point again. $\endgroup$
    – user301452
    Feb 26, 2017 at 19:56
  • $\begingroup$ Because the assumption of the problem tells us that the function is integrable on a finite interval say [\-pi , \pi ] so we have to take a in this interval. but for me I could n`t see clearly what do you mean by using periodicity to glue it together at another point again. $\endgroup$
    – Emptymind
    Feb 27, 2017 at 5:04
  • $\begingroup$ My last line was a bit misleading, I'm sorry. Wrote down a proof as an answer now. $\endgroup$
    – user301452
    Feb 27, 2017 at 8:19

2 Answers 2

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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.

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  • $\begingroup$ why did you use the first line in your answer as a given, Although it is required to be proved? $\endgroup$
    – Emptymind
    Feb 27, 2017 at 20:44
  • $\begingroup$ I thought you proved it already? It is just a substitution and a use of preiodicity. $\endgroup$
    – user301452
    Feb 27, 2017 at 21:15
  • $\begingroup$ Could u clarify this step please? thanks. $\endgroup$
    – Emptymind
    Feb 27, 2017 at 21:58
  • $\begingroup$ In the solution of the second line why $f(x + a) = f(x)$? Is not the function $2\pi- periodic$ i.e. the function repeats it`s value after $2\pi$ not after a. $\endgroup$
    – Emptymind
    Feb 28, 2017 at 9:14
  • $\begingroup$ Can you specify your problem? I do a substitution there. $\endgroup$
    – user301452
    Feb 28, 2017 at 9:31
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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.

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