Suppose $f(x)$ is periodic with period $l$ and integrable. Prove that, for any a,
$\int_{0}^{l}$$f(x+a)$dx = $\int_{0}^{l}$$f(x)$dx
I tried to change the values and define y=a+x so that dy=dx and the limits of the integrals are as we want.
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Avoid "no clue" questions. – Martin R Sep 24 '19 at 09:29
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See https://math.stackexchange.com/questions/94233/an-integrable-and-periodic-function-fx-satisfies-int-0tfxdx-int-a?rq=1 – Benigno Sep 24 '19 at 09:40
2 Answers
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For 1) Use the fact that $$\int_{a}^{a+l}f(x)dx = \int_{a}^{l}f(x)dx + \int_{l}^{l+a}f(x)dx$$ Now use the periodicity with $l$ to get $$\int_{l}^{l+a}f(x)dx=\int_{0}^{0+a}f(x+l)dx=\int_{0}^{a}f(x)dx$$
Altogether $$\int_{a}^{a+l}f(x)dx =\int_{a}^{l}f(x)dx + \int_{l}^{l+a}f(x)dx = \int_{a}^{l}f(x)dx+\int_{0}^{a}f(x)dx=\int_{0}^{l}f(x)dx$$
For 2) Use the substitution $y=x+a$ to get $$\int_{0}^{l}f(x+a)dx = \int_{a}^{a+l}f(y)dy $$ Now you have a form that behaves exactly as the start of 1). Repeating the complete procedure again will lead to the same result, thus $$\int_{0}^{l}f(x+a)dx = \int_{0}^{l}f(y)dy $$
Nurator
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Please mark the answer as accepted if you are happy with it to show other people that this question is answered. – Nurator Sep 24 '19 at 10:01
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For the second I found:
$\int_{0}^{l}$$f(x+a)$dx = $\int_{0}^{l}$$f(x)$dx
$\int_{0}^{l}$$f(x+a)$dx $-$ $\int_{0}^{l}$$f(x)$dx = $\int_{0}^{l}$$f(x+a)-f(x)$dx = $\int_{0}^{l}$$0$dx = $0$
Mathlover
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Unfortunately that is not correct. $f(x+a)\neq f(x)$. You only have periodicity with $l$, not with $a$! – Nurator Sep 24 '19 at 10:00
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Use the substitution $y=a+x$ as you described as your own idea and rewrite part 2) to a version that will look exactly like part 1). Then the proof is as before. – Nurator Sep 24 '19 at 10:37
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It seems stupid but I still can't get the right answer after using different substitutions. – Mathlover Sep 25 '19 at 10:00
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I edited my solution to show part 2). Glad to see you took the effort to try yourself! – Nurator Sep 25 '19 at 12:28
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Thankyou very much, I forgot to use the substitution in the upper and lower bound – Mathlover Sep 25 '19 at 12:51