Calculating L moments of a standard normal As part of my research, I am having trouble calculating the 2nd L moment $\lambda_2$ and the third and fourth moment ratios $\tau_3$ and $\tau_4$ of the standard normal distribution, where L moments are defined as they are in this paper (Hosking 1990), page 3. 
Take $\lambda_2$ for example. 
$\lambda_2 = \frac{1}{2} [ EX_{2:2} - EX_{1:2}] \\
\implies \lambda_2 = \int_{-\infty}^{\infty} xF(x)f(x) dx - EX + \int_{-\infty}^{\infty} x F(x) f(x) dx \\
= \ \cdots \\
= \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} x e^{-x^2/2} dx + \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} x \ erf(\frac{x}{\sqrt{2}}) e^{-x^2/2} dx \\
=  \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} x \ erf(\frac{x}{\sqrt{2}}) e^{-x^2/2} dx$
There must be an error in my calculation as this is supposed to be equal to $\frac{1}{\sqrt{ \pi}}$. 
$\lambda_3 = \frac{1}{3}E(X_{3:3} - 2X_{2:3}+X_{1:3}) = \int_0^1 x(F) (6F^2 - 6F + 1) dF$
$\lambda_4 = \frac{1}{4}E(X_{4:4} - 3X_{3:4}+3X_{2:4} -X_{1:4}) = \int_0^1 x(F) (20F^3 - 30F^2 + 12F - 1) dF$
$\tau_3 = \frac{\lambda_3}{\lambda_2}$
$\tau_4 = \frac{\lambda_4}{\lambda_2}$
$\tau_3$'s formula is rather involved.
EDIT: I was able to figure it out. Instead of working through the integral as an ugly function of exponentials, x's and erf, work through it as a function of a power of F multiplied by the inverse erf term. The latter is analytically tractable. I am not sure the former is.
 A: Using the formula (2.4) in the paper you refer to, we get (with $f$ the gaussian pdf and $F$ the cdf)
$$\begin{aligned}
\lambda_2 &= \int_{-\infty}^{+\infty} x (2F(x) - 1) f(x) d x \\
          &= 2\int_{-\infty}^{+\infty} x F(x) f(x) d x\\
\end{aligned}$$
Let $g(x) = x f(x) = {x \over \sqrt{2\pi}} e^{-x^2\over 2}$. This is the derivative of $G(x) = -{1\over \sqrt{2\pi}} e^{-x^2\over 2}$. We have $G(x) = -f(x)$. Integration by parts gives
$$\begin{aligned}
{1\over 2} \lambda_2 &= \int_{-\infty}^{+\infty} F(x) g(x) d x\\
&= \Bigl[ F(x) G(x) \Bigr]_{-\infty}^{+\infty} -  \int_{-\infty}^{+\infty} f(x) G(x) d x \\
&=  \int_{-\infty}^{+\infty} f^2(x) dx \\
&= {1\over 2 \pi} \int_{-\infty}^{+\infty} e^{-x^2} d x \\
&= {1\over 2 \pi} \sqrt{\pi}
\end{aligned}$$
and $\lambda_2 = {1 \over \sqrt{\pi}}$.
Edit I tried to check if the same computation is tractable for higher $L$-moments. I had hard times but after a while I found the following, which this seems ok but a bit cumbersome. Everything boils down to computing 
$$ I_k = \int_{-\infty}^{+\infty} x F^k(x) f(x) dx.$$ 
Integration by part seem to lead nowhere. A change of variable $y = -x$ leads to
$$ I_k = - \int_{-\infty}^{+\infty} y (1 -F(y))^k f(y) dy,$$
hence
$$ I_k = - \sum_{\ell=1}^k (-1)^{\ell} {k \choose \ell} I_\ell.$$
For $k=2, 3$, this leads to $I_2 = I_1$. For $k=4, 5$, this leads to two linear equations in $I_1, I_2, I_3, I_4$, allowing to compute $I_3$ and $I_4$ from $I_1$. This can be continued... I really can’t find anything simpler. I think this trick is well known – but not by me.
