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An integrable and periodic function $f(x)$ satisfies $\int_{0}^{T}f(x)dx=\int_{a}^{a+T}f(x)dx$.

I have to show if $f\in \mathcal L[-\pi,\pi]$ and if $$f(u+2\pi)=f(u), (-\infty<u<\infty),$$ then, for any real $x$, $$\int_{-\pi+x}^{\pi+x}f(u) du=\int_{-\pi}^{\pi}f(u) du.$$ I am approaching the answer of this question in the following manner, $$ \int_{-\pi}^{\pi}f(u) du=\int_{-\pi}^{-\pi+x}f(u)du+\int_{-\pi+x}^{\pi}f(u)du= I_{1}+I_{2}.\tag{1}$$ Now, $$I_{1}=\int_{-\pi}^{-\pi+x}f(u)du$$ Taking $u=t-2\pi$, we get, $$I_{1}=\int_{\pi}^{\pi+x}f(t-2\pi) dt=\int_{\pi}^{\pi+x}f(t)dt. (\because f\ \text{is periodic})$$ Thus, $$I_{1}=\int_{\pi}^{\pi+x}f(t) dt.$$ From $(1)$, we get, $$ \int_{-\pi}^{\pi}f(u)du=\int_{\pi}^{\pi+x}f(u)du+\int_{-\pi+x}^{\pi}f(u)du=\int_{-\pi+x}^{\pi+x}f(u)du.$$ Am i approach right?

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marked as duplicate by robjohn, t.b., Rudy the Reindeer, no identity, Asaf Karagila Jun 3 '12 at 15:59

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I don't know that this question already asked! Sorry for that! But you can tell me am i justify the answer of this question? –  Kns Jun 3 '12 at 11:57

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It looks correct and pretty nice, indeed.

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