English
If CircleIntegrable f, then the integrand function used in circle integral, namely deriv(circleMap(c,R)) θ · f(circleMap(c,R,θ)), is interval-integrable on [0,2π].
Русский
Если CircleIntegrable f, то интегрируемая функция на интервале [0,2π], используемая в circle integral, равна deriv(circleMap(c,R))(θ) · f(circleMap(c,R,θ)).
LaTeX
$$$\text{out}(hf) = \text{IntervalIntegrable}(\theta \mapsto \operatorname{deriv}(\circleMap(c,R))(\theta) \cdot f(\circleMap(c,R,\theta)),\; \text{volume}, 0, 2\pi).$$$
Lean4
nonrec theorem neg (hf : CircleIntegrable f c R) : CircleIntegrable (-f) c R :=
hf.neg