If $f$ is continuous and $2\pi$-periodic, then $\int_{-\pi}^{\pi}f(y)\,dy=\int_{-\pi}^{\pi}f(x-y)\,dy$ for any $x\in\mathbb{R}$. In order to verify the identity, first of all, let $y=-y$, we get the following identity $\int_{-\pi}^{\pi}f(y)\,dy=\int_{-\pi}^{\pi}f(-y)\,dy$, and then we let $y=y-x$,we get the next identity $\int_{-\pi}^{\pi}f(y)\,dy=\int_{-\pi-x}^{\pi-x}f(x-y)\,dy$. But there is some difference to the right anwer. Can you explain where the problem is?
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Since your last integral is over an interval of the length of the period ($\pi-x-(-\pi-x)=2\pi$), it is equal to $\int_{-\pi}^\pi f(x-y) \, dy$ and you're done.
See here: Integral of periodic function over the length of the period is the same everywhere (add $\pi+x$ to both limits of integration to see it clearly.)
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