Skip to main content
MathOverflow

Return to Question

Added some context and restated the question.
Source Link
zab
zab
  • 222
  • 1
  • 6

Is it possible to haveLet $h: \mathbb{R} \mapsto \mathbb{C}$ be a non-trivialpositive definite function, continuous at the origin. $h: \mathbb{R} \mapsto \mathbb{C}$ so that(In fact, $h$ is the Fourier transform of a finite measure). Define the oscillatory integral $$Q(t) := \int_0^\infty \frac{\text{Im} \{ h(u)^t\}}{u} du, \quad t > 0$$$$Q(t) := \int_0^\infty \frac{\text{Im} \{ h(u)^t\}}{u} du, \quad t > 0.$$ isAre there non-trivial examples of $h$ that makes $Q$ independent of $t$?

Is there a known class of functions $\{ h \}$ that have this property?

Is it possible to have a non-trivial function $h: \mathbb{R} \mapsto \mathbb{C}$ so that the oscillatory integral $$Q(t) := \int_0^\infty \frac{\text{Im} \{ h(u)^t\}}{u} du, \quad t > 0$$ is independent of $t$?

Is there a known class of functions that have this property?

Let $h: \mathbb{R} \mapsto \mathbb{C}$ be a positive definite function, continuous at the origin. (In fact, $h$ is the Fourier transform of a finite measure). Define the oscillatory integral $$Q(t) := \int_0^\infty \frac{\text{Im} \{ h(u)^t\}}{u} du, \quad t > 0.$$ Are there non-trivial examples of $h$ that makes $Q$ independent of $t$?

Is there a known class of functions $\{ h \}$ that have this property?

Source Link
zab
zab
  • 222
  • 1
  • 6

Oscillatory integral independent of a parameter

Is it possible to have a non-trivial function $h: \mathbb{R} \mapsto \mathbb{C}$ so that the oscillatory integral $$Q(t) := \int_0^\infty \frac{\text{Im} \{ h(u)^t\}}{u} du, \quad t > 0$$ is independent of $t$?

Is there a known class of functions that have this property?