Integral representation of the arcsin(a) function
Today we show the proof of this integral posted by @integralsbot ∫10ln(1+ax1−ax)dxx√1−x2=πarcsin(a)|a|≤1 In the proof we use the residue theorem to evaluate it.
Proof:
Recall the logarithmic definition of arcanh(x) arcanh(x)=12ln(1+x1−x)|x|<1 Hence I(a)=∫10ln(1+ax1−ax)dxx√1−x2=2∫10arcanh(ax)x√1−x2dx Hence I′(a)=2∫101√1−x2(1−a2x2)dx=2∫π2011−a2sin2(w)dw(x↦sin(w))=4∫π201a2cos(2w)−a2+2dw=2∫π01a2cosθ−a2+2dθ(θ↦2w)=∫π−π1a2cosθ−a2+2dθ(The integrand is even) If we make the following change of variable cosθ=z+z−12 dθ=dzzi I′(a)=∫π−π1a2cosθ−a2+2dθ=2i∮|z|=1dza2z2+(4−2a2)z+a2 The integrand has two poles: z1=a2−2+2√1−a2a2 z2=a2−2−2√1−a2a2 Only z1 is inside the unit circle. Suppose |a|≤1: |a2−2+2√1−a2a2|<1⟹−a2<a2−2+2√1−a2<a2⟹−2a2<−2+2√1−a2<0⟹1−a2<√1−a2<1⟹a≠0∧a≠1 Then, if we add the assumption |a|<1∧a≠0⟹|z1|<1
By the residue theorem: I′(a)=∫π−π1a2cosθ−a2+2dθ=2i∮|z|=1dza2z2+(4−2a2)z+a2=4πa2limz→a2−2+2√1−a2a21(z−a2−2−2√1−a2a2)=4π(a2−2+2√1−a2−(a2−2−2√1−a2))=π√1−a2 Therefore I(a)=∫10ln(1+ax1−ax)dxx√1−x2=πarcsin(a)+C If we let a→0 then C→0.
The case when a=1 can be obtained as a limitng case easily.
Therefore ∫10ln(1+ax1−ax)dxx√1−x2=πarcsin(a)|a|≤1
No comments:
Post a Comment