Log-trigonometric integrals on [0, π/4] #
This file collects the lemmas behind A006, the values of ∫₀^{π/4} log cos x dx and
∫₀^{π/4} log sin x dx.
Main results #
LeanPolyLog.LogTrig.integral_pow_mul_neg_log:∫₀¹ t^m (-log t) dt = 1/(m+1)².LeanPolyLog.LogTrig.integral_neg_log_div_one_add_sq:∫₀¹ -log t / (1+t²) dt = G.LeanPolyLog.LogTrig.integral_log_cos_sub_log_sin:∫₀^{π/4} (log cos x - log sin x) dx = G.LeanPolyLog.LogTrig.integral_log_sin_add_integral_log_cos:∫₀^{π/4} log sin x dx + ∫₀^{π/4} log cos x dx = -(π/2) log 2.LeanPolyLog.LogTrig.integral_log_cos_sub_integral_log_sin:∫₀^{π/4} log cos x dx - ∫₀^{π/4} log sin x dx = G.
Proof sketch #
- The sum:
cos x = sin (π/2 - x)turns∫₀^{π/4} log cosinto∫_{π/4}^{π/2} log sin, and Mathlib knows∫₀^{π/2} log sin = -(π/2) log 2. - The difference: the substitution
x = arctan tturns it into∫₀¹ -log t / (1+t²) dt. Expanding1/(1+t²) = ∑ (-1)ⁿ t²ⁿand integrating term by term gives∑ (-1)ⁿ/(2n+1)² = G. The interchange of sum and integral is justified by∫₀¹ t²ⁿ (-log t) dt = 1/(2n+1)², which is summable.
The moments ∫₀¹ t^m (-log t) dt #
An antiderivative of t ^ m * -log t, namely t^(m+1)/(m+1)² - t^(m+1) log t/(m+1).
It is continuous on all of ℝ and vanishes at 0.
Equations
Instances For
theorem
LeanPolyLog.LogTrig.hasDerivAt_powMulNegLogPrimitive
(m : ℕ)
{t : ℝ}
(ht : t ≠ 0)
:
HasDerivAt (powMulNegLogPrimitive m) (t ^ m * -Real.log t) t
Away from 0, the derivative of powMulNegLogPrimitive m is t ^ m * -log t.
The antiderivative powMulNegLogPrimitive m is continuous.
theorem
LeanPolyLog.LogTrig.integrableOn_pow_mul_neg_log
(m : ℕ)
:
MeasureTheory.IntegrableOn (fun (t : ℝ) => t ^ m * -Real.log t) (Set.Ioc 0 1) MeasureTheory.volume
t ^ m * -log t is integrable on (0, 1]: it is the nonnegative derivative of a
continuous function.