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 Incoming links: Radial sine-power kernel integrability

Assign any value, for example zero, at

(0, 0)

, where the displayed quotient is undefined. A single point is a Lebesgue null set, so this choice does not affect Lebesgue integrability. Elsewhere the function is continuous, hence it is Lebesgue measurable on the plane after this extension.

x^{2} + y^{2} = r^{2}

and the area element is

r d r d θ

\int_{R^{2}} ∣ k (x, y) ∣ d x d y = 2 π \int_{0}^{\infty} ∣ sin (r^{2}) ∣ r^{1 - 2 α} d r .

For

0 < r \leq 1

∣ sin (r^{2}) ∣ \leq r^{2}

, so the radial integrand is bounded by

r^{3 - 2 α}

. This is integrable at zero when

α < 2

. For

r \geq 1

∣ sin (r^{2}) ∣ \leq 1

, giving the integrable tail bound

r^{1 - 2 α}

when

α > 1

. In particular,

∥ k ∥_{1} \leq 2 π (\int_{0}^{1} r^{3 - 2 α} d r + \int_{1}^{\infty} r^{1 - 2 α} d r) = π (\frac{1}{2 - α} + \frac{1}{α - 1}) < \infty.

Therefore

k \in L^{1} (R^{2}) (1 < α < 2) .

Equivalently, the substitution rule

u = r^{2}

reduces its exact L1 norm to

∥ k ∥_{1} = π \int_{0}^{\infty} \frac{∣ s i n u ∣}{u ^{α}} d u .