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A vanishing moment is a cancellation condition: positive and negative parts of an analyzing function balance when weighted by a monomial. Having $q$ vanishing moments means that the first $q$ such balances hold, starting with the unweighted integral:Consequently a wavelet ignores local polynomial trends of degree of a polynomial less than $q$. A smooth feature is close to its Taylor polynomial, so only the small Taylor remainder contributes to its detail coefficient. A jump discontinuity cannot be approximated this way across its edge; its jump response of a localized analyzing function can remain large relative to details from a smooth region.^[ref]

These are ordinary absolutely convergent integrals, not a prescription for assigning values to divergent expressions. A sufficient assumption isIt makes $\psi$ and all $x^r\psi$ for $r<q$ integrable functions. Being square-integrable alone does not ensure these weighted integrals exist. For a square-integrable function with compact support, the Cauchy-Schwarz inequality guarantees every such weighted integral is finite. Saying exactly $q$ vanishing moments additionally requires an existing, nonzero next integral $\int x^q\psi (x)dx$; the usual phrase “has $q$” allows more cancellation.

Use the Fourier transform convention $\hat{\psi }(\omega )=\int \psi (x)e^{-i\omega x}dx$. If $\psi$ and $x^r\psi$ for $r<q$ are integrable functions, moment differentiation of the Fourier transform yields a $C^{q-1}$ Fourier transform andThe dominated convergence theorem justifies these derivatives. Under these assumptions, $q$ vanishing moments are equivalent to vanishing derivatives through order $q-1$ at zero. For $q>1$ this gives a Peano zero $\hat{\psi }(\omega )=o(|\omega {|}^{q-1})$; for $q=1$ it gives $\hat{\psi }(\omega )=o(1)$. It does not by itself justify a $q$th derivative or an $O(|\omega {|}^q)$ estimate.

If also $\int |x{|}^q|\psi (x)|dx<\infty$, subtract the degree-$(q-1)$ Taylor polynomial of $e^{-i\omega x}$ inside the integral. Its Taylor remainder gives the stronger boundFor an integrable function with compact support, every weighted integral exists and its Fourier transform extends to an entire function. The language “zero of order at least $q$” is then literal analytic function terminology. If the next integral is nonzero, the order is exactly $q$. The smooth-mask vanishing-moment criterion relates these cancellations to a refinement mask under its additional hypotheses.^[ref]