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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 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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    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?2012-06-03

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It looks correct and pretty nice, indeed.