Why integrate over $2\pi$ in inverse DTFT?
There are two reasons:


As you say, the sequence is periodic so all of the information is contained in the $-\pi &lt; w &lt;\pi$ region.  Thus, integrating over that region tells you everything that there is to be learned from the inverse transform.
Integrating from $-\infty &lt; w&lt; \infty$ would only give answers of $0$, $\infty$, or $-\infty$.  The reason for that is that integrating over $-\pi &lt; w &lt;\pi$ will generally give a finite answer which can be something negative (e.g. -4), 0, or something positive (e.g. 2.5).  If you integrate over $-\infty &lt; w&lt; \infty$ each $2\pi$ region will give the exact same answer, so the negative results will accumulate to $-\infty$, the zeros will sum to $0$, and the positive results will sum to $\infty$.  (Note that the results are often complex, but that doesn't change the primary point.  It just increases the dimensionality of the infinities.)