TITLE: Evaluating $\int_{-\infty}^0 \log(\frac{1}{2}\operatorname{erfc}(x))\mathrm dx$
QUESTION [8 upvotes]: I am looking to evaluate  

$$\int_{-\infty}^0 \log\left(\frac{1}{2}\operatorname{erfc}(x)\right)\mathrm dx = -0.337~668~477...$$

Both Maple and Mathematica have failed to give a closed-form expression but indicate the value is around $−0.337~668~477...$ which doesn't appear to be close to some well known quantity.

REPLY [7 votes]: $\textbf{Edition of 19.02.19}$
Is known, that

$$\begin{cases}
\operatorname{erf}x = \dfrac2{\sqrt\pi}\int\limits_0^x e^{-t^2}\,\mathrm dt,\quad
\operatorname{erfc}x = \dfrac2{\sqrt\pi}\int\limits_x^\infty e^{-t^2}\,\mathrm dt,\\[4pt]
\operatorname{erf}(-\infty) = -1,\quad 
\operatorname{erf}(0) = 0,\quad 
\operatorname{erf}(\infty) = 1,\\[4pt]
\operatorname{erf}(-x) = - \operatorname{erf}x,\quad 
\operatorname{erfc}x = 1-\operatorname{erf}x,\quad
\operatorname{erfc}(-x)+\operatorname{erfc}(x) = 2,\\[4pt]
\operatorname{erfc}(-\infty) = 2,\quad 
\operatorname{erfc}(0) = 1,\quad 
\operatorname{erfc}(\infty) = 0.\tag1
\end{cases}$$

So
$$I=\int\limits_{-\infty}^0 \log\left(\frac{1}{2}\operatorname{erfc}(x)\right)\mathrm dx 
= \int\limits_{-\infty}^0 \log\left(1-\frac{1}{2}\operatorname{erfc}(-x)\right)\mathrm dx
=\int\limits_0^\infty \log\left(1-\frac{1}{2}\operatorname{erfc}(x)\right)\mathrm dx,$$ 
or, using Maclaurin series for logarithm in the form of

$$\log(1-x)=-\sum\limits_{n=1}^{\infty}\dfrac{x^n}n,\tag2$$

$$I=-\sum\limits_{n=1}^{\infty}\dfrac1{2^n n} I_n,\tag3$$
wherein
\begin{align}
&I_n=\int\limits_0^\infty\operatorname{erfc}^n(x)\,\mathrm dx 
= \operatorname{erfc}^{n}(x)\cdot x\Bigg|_0^\infty
-\dfrac{2n}{\sqrt\pi}\int\limits_0^\infty x e^{-x^2}\operatorname{erfc}^{n-1}(x)\,\mathrm dx,\\[4pt]
\end{align}
\begin{align}
&I_n = \dfrac{2n}{\sqrt\pi}\int\limits_0^\infty x e^{-x^2}\operatorname{erfc}^{n-1}(x)\,\mathrm dx,\tag{4.1}
\end{align}
\begin{align}
&I_1 = \dfrac2{\sqrt\pi}\int\limits_0^\infty\,x e^{-x^2}\,\mathrm dx 
= \dfrac1{\sqrt\pi}\approx0.56418\,95835\,47756,\tag{5.1}
\end{align}
\begin{align}
&I_n= \dfrac{n}{\sqrt\pi}\int\limits_0^\infty \operatorname{erfc}^{n-1}(x)\,\mathrm d e^{-x^2}\\[4pt]
&= \dfrac{n}{\sqrt\pi}\operatorname{erfc}^{n-1}(x)\,e^{-x^2} \Bigg|_0^\infty
- \dfrac{2n(n-1)}\pi \int\limits_0^\infty e^{-2x^2}\operatorname{erfc}^{n-2}(x)\,\mathrm dx\\[4pt]
\end{align}
\begin{align}
&I_n = \dfrac{n}{\sqrt\pi} - \dfrac{2n(n-1)}\pi \int\limits_0^\infty e^{-2x^2}\operatorname{erfc}^{n-2}(x)\,\mathrm dx,\quad n\geq2,\tag{4.2}
\end{align}
\begin{align}
&I_2 = \dfrac{2}{\sqrt\pi} - \dfrac4\pi \int\limits_0^\infty e^{-2x^2}\,\mathrm dx = \dfrac{2}{\sqrt\pi} - \sqrt{\dfrac2\pi} \operatorname{erf}(x\sqrt2)\Bigg|_0^\infty,
\end{align}
\begin{align}
&I_2 = \int\limits_0^\infty\operatorname{erfc}^2(x)\,\mathrm dx = \dfrac{2-\sqrt2}{\sqrt\pi}\approx0.33049\,46062\,92647,\tag{5.2}
\end{align}
\begin{align}
&I_3 = \dfrac{3}{\sqrt\pi} - \dfrac{12}\pi \int\limits_0^\infty e^{-2x^2}\operatorname{erfc}(x)\,\mathrm dx,\\[4pt]
& \int\limits_0^\infty e^{-2x^2}\operatorname{erfc}(x)\,\mathrm dx
= \dfrac1{\sqrt2} \int\limits_0^\infty e^{-2x^2}\operatorname{erfc}(x)\,\mathrm d(x\sqrt2)
=\dfrac1{\sqrt2} \int\limits_0^\infty e^{-x^2}\operatorname{erfc}\left(\dfrac x{\sqrt2}\right)\,\mathrm dx\\[4pt]
& = \sqrt{\dfrac2\pi}\int\limits_0^\infty \int\limits_{\frac x{\sqrt2}}^\infty e^{-(x^2+y^2)}\,\mathrm dy\,\mathrm dx
= \sqrt{\dfrac2\pi}\int\limits_0^\infty \int\limits_{\arctan\frac1{\sqrt2}}^{\large\frac\pi2} e^{-\rho^2}\rho \,\mathrm d\rho\,\mathrm d\varphi= \dfrac1{\sqrt{2\pi}}\arctan\sqrt2
,\\[4pt]
\end{align}
\begin{align}
&I_3 = \dfrac{3}{\sqrt\pi} - \dfrac{6\sqrt2}{\pi\sqrt\pi}\arctan\sqrt2\approx 0.23681\ 21373\ 68199\tag{5.3}
\end{align}
(see also Wolfram Alpha).
Numerical calculations confirm the correctness of the formula $(3).$
But it was not possible to get a closed form of $I_n$ for $n>3.$
At the same time, the integration of the issue integral by parts gives
\begin{align}
&I=\int\limits_0^\infty \log\left(1-\frac12\operatorname{erfc}(x)\right)\mathrm dx 
= \log\left(1-\frac12\operatorname{erfc}(x)\right)\cdot x\Bigg|_0^\infty
-\frac1{\sqrt\pi}\int\limits_0^\infty \dfrac{xe^{-x^2}\,\mathrm dx}{1-\frac12\operatorname{erfc}(x)},\\[4pt]
\end{align}
\begin{align}
&I = -\frac1{\sqrt\pi}\int\limits_0^\infty \dfrac{xe^{-x^2}\,\mathrm dx}{1-\frac12\operatorname{erfc}(x)},\tag6
\end{align}
\begin{align}
&I = -\frac1{\sqrt\pi}\int\limits_0^\infty \left(1 + \frac12\operatorname{erfc}(x) + \frac14\operatorname{erfc}^2(x) + \frac18\operatorname{erfc}^3(x) + \dots+\frac1{2^k}\operatorname{erfc}^k(x)+\dots\right) xe^{-x^2}\,\mathrm dx\\[4pt]
&= -\left(\dfrac12I_1+\dfrac18I_2+\dfrac1{24}I_3 +\frac1{8\sqrt\pi}\int\limits_0^\infty \dfrac{\operatorname{erfc}^3(x)\,xe^{-x^2}\,\mathrm dx}{1-\frac12\operatorname{erfc}(x)}\right)\\[4pt]
& = -\frac1{\sqrt\pi}\left(\frac12+\dfrac{2-\sqrt2}8+\dfrac1{24}\left(3-\dfrac{6\sqrt2\,\arctan\sqrt2}\pi\right) +\dfrac18\int\limits_0^\infty \dfrac{\operatorname{erfc}^3(x)\,xe^{-x^2}\,\mathrm dx}{1-\frac12\operatorname{erfc}(x)}\right),
\end{align}
\begin{align}
&I = -\frac1{8\sqrt\pi}\left(7-\sqrt2\left(1+\dfrac2\pi\arctan\sqrt2\right) +\int\limits_0^\infty \dfrac{\operatorname{erfc}^3(x)\,xe^{-x^2}\,\mathrm dx}{1-\frac12\operatorname{erfc}(x)}\right)\tag7
\end{align}
(see also Wolfram Alpha)
On the other hand,
\begin{align}
&I=-\frac2{\sqrt\pi}\int\limits_0^\infty \dfrac{xe^{-x^2}\,\mathrm dx}{1 +\operatorname{erf}(x)}
=\frac1{\sqrt\pi}\int\limits_0^\infty \dfrac{\mathrm de^{-x^2}}{1+\operatorname{erf}(x)}\\[4pt]  
&=\frac1{\sqrt\pi} \dfrac{e^{-x^2}}{\left(1+\operatorname{erf}(x)\right)}\Bigg|_0^\infty
+\frac2\pi\int\limits_0^\infty \dfrac{\mathrm e^{-2x^2}}{\left(1+\operatorname{erf}(x)\right)^2}\mathrm dx\  
=\frac2{\pi}\int\limits_0^\infty \dfrac{\mathrm e^{-2x^2}}{\left(1+\operatorname{erf}(x)\right)^2}\mathrm dx-\frac1{\sqrt\pi}\\[4pt]
&=\frac1{\sqrt{2\pi}}\int\limits_0^\infty \dfrac{\mathrm d\operatorname{erf}(x\sqrt2)}{\left(1+\operatorname{erf}(x)\right)^2}
-\frac1{\sqrt\pi}\\
&=\frac1{\sqrt{2\pi}}\dfrac{\operatorname{erf}(x\sqrt2)}{\left(1+\operatorname{erf}(x)\right)^2}\Bigg|_0^\infty
+\frac1{\sqrt{2\pi}}\cdot2\cdot\frac2{\sqrt\pi}\int\limits_0^\infty \dfrac{e^{-x^2}\operatorname{erf}(x\sqrt2)}{\left(1+\operatorname{erf}(x)\right)^3}\,\mathrm dx
-\frac1{\sqrt\pi}\\[4pt]
&=\frac{2\sqrt2}\pi\int\limits_0^\infty \dfrac{e^{-x^2}\operatorname{erf}(x\sqrt2)}{\left(1+\operatorname{erf}(x)\right)^3}\,\mathrm dx-\frac1{\sqrt\pi}+\frac1{4\sqrt{2\pi}}.\\[4pt]  
\end{align}
Binomial decomposition in the form of

$$\dfrac1{(1-y)^{m+1}} = \sum_{n=m}^{\infty}\binom nm y^{n-m}\tag8$$

allows the further transformations in the form of
$$\begin{align}
&I = \frac 1{2\pi\sqrt2}\int\limits_0^\infty \dfrac{e^{-x^2}\operatorname{erf}(x\sqrt2)}{\left(1-\frac12\operatorname{erfc} x\right)^3}\,\mathrm dx-\frac1{\sqrt\pi}+\frac1{4\sqrt{2\pi}}\\[4pt]  
& = \frac 1{2\pi\sqrt2}\int\limits_0^\infty e^{-x^2}\operatorname{erf}(x\sqrt2)\sum_{n=2}^{\infty}\dfrac{n(n-1)}{2^{n-2}}\,\operatorname{erfc}^{n-2} x\,\mathrm dx-\frac1{\sqrt\pi}+\frac1{4\sqrt{2\pi}}\\[4pt]  
& = \frac 1{2\pi\sqrt2}\sum_{n=0}^{\infty}\dfrac{(n+1)(n+2)}{2^n}\int\limits_0^\infty e^{-x^2}\operatorname{erf}(x\sqrt2)\,\operatorname{erfc}^n x\,\mathrm dx-\frac1{\sqrt\pi}+\frac1{4\sqrt{2\pi}},
\end{align}$$
$$I = -\frac1{\sqrt\pi}+\frac1{4\sqrt{2\pi}} + \dfrac{\arctan\sqrt2}{2\pi\sqrt{2\pi}}+\frac 1{2\pi\sqrt2}\sum_{n=1}^{\infty}\dfrac{(n+1)(n+2)}{2^n}J_n,\tag9$$
where
$$J_n=\int\limits_0^\infty e^{-x^2}\operatorname{erf}(x\sqrt2)\,\operatorname{erfc}^n x\,\mathrm dx\tag{10}$$
(see also Wolfram Alpha).
Besides, can be used the presentations
$$\begin{align}
&I = \frac1{\sqrt\pi}\int\limits_0^\infty \dfrac{\mathrm d(e^{-x^2})}{1+\operatorname{erfc}(x)}
 = -\frac1{\sqrt\pi}\int\limits_0^1 \dfrac{\mathrm dz}{1+\operatorname{erf}(\sqrt{-\log z})},
\end{align}$$
wherein
$$\dfrac1{1+\operatorname{erf}\sqrt{-\log \dfrac{\sqrt\pi y}2}} =\dfrac1{1+y}\left(1+\dfrac\pi{12}\dfrac{y^3}{1+y}\left(1+\dfrac\pi{120}\dfrac{y^3-9y^2}{1+y}\right)\right)+\dots$$
(see also Wolfram Alpha).
However, closed form was not obtained.