Why integrate over $2\pi$ in inverse DTFT? In DTFT of a signal, the spectrum of a sequence is periodic with period $2\pi$ and all the information needed for derivation of the original signal from its spectrum is contained in $\pi &lt;\omega &lt;\pi$.

But , why do they integrate over $2\pi$ and not from $-\infty$ to $+\infty$ , as in continuous Fourier transform in inverse DTFT? After all, when you derive a transform of a signal, the inverse of it is not arbitrary and to recover the original signal, you can't integrate over an other  interval because all the information is contained in there.
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.)