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Suppose $f: \mathbb R \to \mathbb R$ is Riemann integrable on every finite interval and periodic with period $T>0$. Then for every interval $[a,b]$: $$ \int_a^b f = \int_c^d f,$$ where $c = a+T$ and $d = b+T$.

I don't understand why, if so, this is true.

How van I explain it? I could do it if I knew that the antiderivative is periodic, but do we even know that there is an antiderivative?

Caleb Stanford
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CnR
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  • Very related but slightly different: http://math.stackexchange.com/questions/98409/integral-of-periodic-function-question?rq=1 – Caleb Stanford Jul 17 '16 at 16:17

2 Answers2

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By the change variable $x=u-T$ so $du=dx$ we have $$\int_a^b f(x)dx=\int_{a+T}^{b+T}f(u-T)du=\int_c^df(u)du$$

  • is every periodic function continuous? I understand it but if f is not continuous so we can't change variable and I don't know what to do – CnR Jul 22 '13 at 09:49
  • No of course, but we needn't the continuity hypothesis. –  Jul 22 '13 at 09:51
  • to change variables I think f must be continuous : I checked in Wikipedia to reassure it http://en.wikipedia.org/wiki/Integration_by_substitution – CnR Jul 22 '13 at 09:57
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    The assumption $f$ continuous in Wikipedia is just to ensure that the integral has meaning but you can weaken this hypothesis and just assume that $f$ is Riemann integrable and to convince your self you should return to the definition of "Riemann integrable" and see how the integral was built. –  Jul 22 '13 at 10:04
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Okay, let's think the graphical interpretation of integration .It's very logical and I don't know a formal proof for this . But I can explain it to you with the help of an example.
for eg: Consider - $$f(x)= sin x$$
If I were to integrate sin x from 0 to 2pi . Without even working out we can say that it is 0 as the area above x-axis is the same as the area below x axis. Now lets take this integral one step further.
Integrate this from 0+100pi to 2pi+100pi.
Quite obviously again area above x axis will be equal to the area below x axis.(Number of cycles does not make a difference) . And so,we can conclude:

$$\int_{0}^{2pi}sin(x)=\int_{0+nT}^{2pi+nT}sin(x)dx=0 $$where n is any arbit natural number.

Simar
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