Difference between revisions of "Integral of log of inverse erf from 0 to 1"

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(Created page with "==Theorem== The following formula holds: $$\displaystyle\int_0^1 \log(\mathrm{erf}^{-1}(x)) dx = \left( \dfrac{\gamma}{2} + \log(2) \right),$$ where $\mathrm{erf}^{-1}$ denote...")
 
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==Theorem==
 
==Theorem==
 
The following formula holds:
 
The following formula holds:
$$\displaystyle\int_0^1 \log(\mathrm{erf}^{-1}(x)) dx = \left( \dfrac{\gamma}{2} + \log(2) \right),$$
+
$$\displaystyle\int_0^1 \log(\mathrm{erf}^{-1}(x)) \mathrm{d}x = \left( \dfrac{\gamma}{2} + \log(2) \right),$$
 
where $\mathrm{erf}^{-1}$ denotes the [[inverse error function]], $\log$ denotes the [[logarithm]], and $\gamma$ denotes the [[Euler-Mascheroni constant]].
 
where $\mathrm{erf}^{-1}$ denotes the [[inverse error function]], $\log$ denotes the [[logarithm]], and $\gamma$ denotes the [[Euler-Mascheroni constant]].
  

Revision as of 03:41, 3 October 2016

Theorem

The following formula holds: $$\displaystyle\int_0^1 \log(\mathrm{erf}^{-1}(x)) \mathrm{d}x = \left( \dfrac{\gamma}{2} + \log(2) \right),$$ where $\mathrm{erf}^{-1}$ denotes the inverse error function, $\log$ denotes the logarithm, and $\gamma$ denotes the Euler-Mascheroni constant.

Proof

References