Mean of log of $X$ with normal distribution: $\int^{\infty}_{-\infty} \log(| \alpha x + \beta |) \frac{ 1}{\sqrt{2\pi} }e^{\frac{-x^2}{2} }{\rm d}x$

Question

The expectation of the log of the absolute value of random variable with the standard normal distrbution can be computed based on the result provided here 1.

How can this be computed for an arbitrary normal distrbution. It is equivalent to finding the following expectation:

$$\mathbb E [\log|\alpha X + \beta|]=\int^{\infty}_{-\infty} \log(| \alpha x + \beta |) \frac{1}{\sqrt{2\pi}} e^{\frac{-x^2}{2} }{\rm d}x $$

where $X$ has the standard normal distribution.

For $\beta=0$, the formula is

$$\mathbb E [\log|\alpha X|]=\log|\alpha|-\frac{1}{2} (\gamma +\log (2))$$

where $\gamma$ denotes the Euler's constant.

Answer 1

Notations

• $\displaystyle F(z):=\frac{\sqrt{\pi}}{2}e^{-z^2}\text{erfi}(z)\text{ is the Dawson integral}$
• $\gamma$ is the Eulero-Mascheroni constant
• ${}_pF_q(a_1,...,a_p; b_1,...,b_q|z)$ is the hypergemetric function
• P.V. is the principal value

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You can use again the Feymann trick, but now it's more complex: $$\begin{align}\int_{-\infty}^{\infty}\frac{\ln\left(\left|ax+b\right|\right)}{\sqrt{2\pi}}e^{-\frac{x^{2}}{2}}dx=&\frac{\ln\left(\left|a\right|\right)}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-\frac{x^{2}}{2}}dx+\int_{-\infty}^{\infty}\frac{\ln\left(\left|x+\frac{b}{a}\right|\right)}{\sqrt{2\pi}}e^{-\frac{x^{2}}{2}}dx\\ =&\ln\left(\left|a\right|\right)+\int_{-\infty}^{\infty}\frac{\ln\left(\left|x+\frac{b}{a}\right|\right)}{\sqrt{2\pi}}e^{-\frac{x^{2}}{2}}dx\end{align}$$

Let $$G\left(s\right):=\int_{-\infty}^{\infty}\frac{\ln\left(\left|x+s\right|\right)}{\sqrt{2\pi}}e^{-\frac{x^{2}}{2}}dx$$ $$g\left(s\right):=G'(s)=\frac{1}{\sqrt{2\pi}}\text{P.V.}\int_{-\infty}^{\infty}\frac{e^{-\frac{x^{2}}{2}}}{x+s}dx=\sqrt{2}F\left(\frac{s}{\sqrt{2}}\right)$$

So $$G(s)=\frac{s^2}{2}{}_2F_2\left(\left.{1,1\atop\frac{3}{2},2}\right|-\frac{s^2}{2}\right)+c_0$$ $c_0$ is a constant and is determined by imposing the passage for $0$: $$c_0=\int_{-\infty}^{\infty}\frac{\ln\left(\left|x\right|\right)}{\sqrt{2\pi}}e^{-\frac{x^{2}}{2}}dx=-\frac{\ln\left(2\right)+\gamma}{2}$$

So

$$\color{blue}{\int_{-\infty}^{\infty}\frac{\ln\left(\left|ax+b\right|\right)}{\sqrt{2\pi}}e^{-\frac{x^{2}}{2}}dx=\ln(|a|)+\frac{b^2}{2a^2}\cdot{}_2F_2\left(\left.{1,1\atop\frac{3}{2},2}\right|-\frac{b^2}{2a^2}\right)-\frac{\ln\left(2\right)+\gamma}{2}}$$

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P.S.: To check the correctness of my formula I leave you two Wolfram links: in the first the one with the integral, in the second the one with the solution (you can change the command and put random values, in this case I put $a=6.2$ and $b=- 2.3$)

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Update

In answer to the question if the hypergeometric function can be simplied the answer is no. At most I can give you a numerical method to calculate it yourself:

$${}_2F_2\left(\left.{1,1\atop\frac{3}{2},2}\right|-\frac{z^2}{2}\right)=2\sum_{n=0}^{\infty}\left(-1\right)^{n}\frac{2^{n}n!}{\left(2n+2\right)!}z^{2n}$$