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Thursday, March 10, 2022

Integral of the day XXIX

Integral involving the arctanh function

Integral representation of the arcsin(a) function


Today we show the proof of this integral posted by @integralsbot 10ln(1+ax1ax)dxx1x2=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+x1x)|x|<1 Hence I(a)=10ln(1+ax1ax)dxx1x2=210arcanh(ax)x1x2dx Hence I(a)=21011x2(1a2x2)dx=2π2011a2sin2(w)dw(xsin(w))=4π201a2cos(2w)a2+2dw=2π01a2cosθa2+2dθ(θ2w)=ππ1a2cosθa2+2dθ(The integrand is even) If we make the following change of variable cosθ=z+z12 dθ=dzzi I(a)=ππ1a2cosθa2+2dθ=2i|z|=1dza2z2+(42a2)z+a2 The integrand has two poles: z1=a22+21a2a2 z2=a2221a2a2 Only z1 is inside the unit circle. Suppose |a|1: |a22+21a2a2|<1a2<a22+21a2<a22a2<2+21a2<01a2<1a2<1a0a1 Then, if we add the assumption |a|<1a0|z1|<1

By the residue theorem: I(a)=ππ1a2cosθa2+2dθ=2i|z|=1dza2z2+(42a2)z+a2=4πa2limza22+21a2a21(za2221a2a2)=4π(a22+21a2(a2221a2))=π1a2 Therefore I(a)=10ln(1+ax1ax)dxx1x2=πarcsin(a)+C If we let a0 then C0.

The case when a=1 can be obtained as a limitng case easily.

Therefore 10ln(1+ax1ax)dxx1x2=πarcsin(a)|a|1

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