Get Math Help

GET TUTORING NEAR ME!

(800) 434-2582

By submitting the following form, you agree to Club Z!'s Terms of Use and Privacy Policy

    Home / Get Math Help

    Inverse Hyperbolic Tangent

    Plot

    Alternate form

    1/2 log(x + 1) - 1/2 log(1 - x)

    Root

    x = 0

    Properties as a real function

    {x element R : -1

    R (all real numbers)

    bijective from its domain to R

    odd

    Series expansion at x = -1

    1/2 (log(x + 1) - log(2)) + (x + 1)/4 + 1/16 (x + 1)^2 + 1/48 (x + 1)^3 + 1/128 (x + 1)^4 + O((x + 1)^5) (generalized Puiseux series)

    Series expansion at x = 0

    x + x^3/3 + x^5/5 + O(x^6) (Taylor series)

    Series expansion at x = 1

    (-1/2 i (-i log(x - 1) + i log(2) + π) + (x - 1)/4 - 1/16 (x - 1)^2 + 1/48 (x - 1)^3 - 1/128 (x - 1)^4 + 1/320 (x - 1)^5 + O((x - 1)^6)) + i π ceiling(arg(x - 1)/(2 π))

    Series expansion at x = ∞

    -(i π)/2 + 1/x + 1/(3 x^3) + 1/(5 x^5) + O((1/x)^6) (Laurent series)

    Derivative

    d/dx(tanh^(-1)(x)) = 1/(1 - x^2)

    Indefinite integral

    integral tanh^(-1)(x) dx = 1/2 log(1 - x^2) + x tanh^(-1)(x) + constant (assuming a complex-valued logarithm)

    Limit

    lim_(x->-∞) tanh^(-1)(x) = (i π)/2≈1.5708 i

    lim_(x->∞) tanh^(-1)(x) = -(i π)/2≈-1.5708 i

    Alternative representation

    tanh^(-1)(x) = sn^(-1)(x|1)

    tanh^(-1)(x) = coth^(-1)(1/x)

    tanh^(-1)(x) = 1/2 (-log(1 - x) + log(1 + x))

    Definite integral

    integral_0^1 tanh^(-1)(x) dx≈0.693147...

    integral_(-1)^0 tanh^(-1)(x) dx≈-0.69314718056...

    Series representation

    tanh^(-1)(x) = sum_(k=0)^∞ x^(1 + 2 k)/(1 + 2 k) for abs(x)<1

    tanh^(-1)(x) = 1/2 log((1 + x)/2) + 1/2 sum_(k=1)^∞ (2^(-k) (1 + x)^k)/k for abs(1 + x)<2

    tanh^(-1)(x) = log(2)/2 - 1/2 log(1 - x) - 1/2 sum_(k=1)^∞ ((-1/2)^k (-1 + x)^k)/k for abs(-1 + x)<2

    Integral representation

    tanh^(-1)(x) = x integral_0^1 1/(1 - t^2 x^2) dt for (Im(x)>0 or (Im(x) = 0 and -1

    tanh^(-1)(x) = -(i x)/(4 π^(3/2)) integral_(-i ∞ + γ)^(i ∞ + γ) (1 - x^2)^(-s) Γ(1/2 - s) Γ(1 - s) Γ(s)^2 ds for (0<γ<1/2 and abs(arg(1 - x^2))<π)

    tanh^(-1)(x) = -(i x)/(4 π) integral_(-i ∞ + γ)^(i ∞ + γ) ((-x^2)^(-s) Γ(1/2 - s) Γ(1 - s) Γ(s))/Γ(3/2 - s) ds for (0<γ<1/2 and abs(arg(-x^2))<π)

    Back to List | POWERED BY THE WOLFRAM LANGUAGE