Inverse Hyperbolic Tangent -- From Wolfram MathWorld

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The inverse hyperbolic tangent tanh^(-1)z (Zwillinger 1995, p. 481; Beyer 1987, p. 181), sometimes called the area hyperbolic tangent (Harris and Stocker 1998, p. 267), is the multivalued function that is the inverse function of the hyperbolic tangent.

The function is sometimes denoted arctanhz (Jeffrey 2000, p. 124) or Arthz (Gradshteyn and Ryzhik 2000, p. xxx). The variants Arctanhz or Artanhz (Harris and Stocker 1998, p. 263) are sometimes used to refer to explicit principal values of the inverse hyperbolic tangent, although this distinction is not always made. Worse yet, the notation arctanhz is sometimes used for the principal value, with Arctanhz being used for the multivalued function (Abramowitz and Stegun 1972, p. 87). Note that in the notation tanh^(-1)z, tanhz is the hyperbolic tangent and the superscript -1 denotes an inverse function, not the multiplicative inverse.

The principal value of tanh^(-1)z is implemented in the Wolfram Language as ArcTanh[z] and in the GNU C library as atanh(double x).

InverseHyperbolicTangentBranchCut

The inverse hyperbolic tangent is a multivalued function and hence requires a branch cut in the complex plane, which the Wolfram Language's convention places at the line segments (-infty,-1] and [1,infty). This follows from the definition of tanh^(-1)z as

tanh^(-1)z=1/2[ln(1+z)-ln(1-z)]. (1)

The inverse hyperbolic tangent is given in terms of the inverse tangent by

tanh^(-1)z=1/itan^(-1)(iz) (2)

(Gradshteyn and Ryzhik 2000, p. xxx). For real x<1, this simplifies to

tanh^(-1)x=1/2ln((1+x)/(1-x)). (3)

The derivative of the inverse hyperbolic tangent is

d/(dz)tanh^(-1)z=1/(1-z^2), (4)

and the indefinite integral is

inttanh^(-1)zdz=ztanh^(-1)z+1/2ln(z^2-1)+C. (5)

It has special values

tanh^(-1)0=0 (6)
tanh^(-1)1=infty (7)
tanh^(-1)infty=-1/2pii (8)
tanh^(-1)i=1/4pii. (9)

It has Maclaurin series

tanh^(-1)z=sum_(n=1)^(infty)(z^(2n-1))/(2n-1) (10)
=z+1/3z^3+1/5z^5+1/7z^7+1/9z^9+... (11)
tanh^(-1)z=-1/2pii+sum_(n=1)^(infty)(z^(-2n+1))/(2n-1) (12)
=-1/2pii+z+1/3z^3+1/5z^5+1/7z^7+... (13)

(OEIS A005408).

An indefinite integral involving tanh^(-1)z is given by

int(dx)/(xsqrt(a+bx))=ln[(sqrt(a+bx)-sqrt(a))/(sqrt(a+bx)+sqrt(a))] (14)
=ln[((sqrt(a+bx)-sqrt(a))^2)/((a+bx)-a)] (15)
=ln[((2a+bx)-2sqrt(a(a+bx)))/(bx)] (16)
=2tanh^(-1)(-sqrt(a/(a+bx))) (17)

when a>0.

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