Solve the following integral: $\int_{x_0}^\infty \log(x) e^{-(x-a)^2/b}dx \quad x_0,b>0$ I want to compute this integral preferably in closed-form without expanding the $\log$ function; however, efficiently computable approximations might also solve my problem:
$\int_{x_0}^\infty \log(x) e^{-\frac{(x-a)^2}{b}}dx \quad  x_0,b>0$
Background:
I am trying to compute the expectation of a logarithmic function of random variables from a truncated normal distribution:
$\mathbb{E}_{x\sim \mathcal{N}(\mu,\sigma^2|x\geq x_0>0)}[\log(x)]$
Equivalently, I need to solve the follwing integral:
$\int_{x_0}^\infty \log(x) \frac{e^{-\frac{(x-\mu)^2}{2\sigma^2}}}{\sigma^2\sqrt{2\pi}(1-\Phi(\frac{x_0-\mu}{\sigma}))}dx=\frac{1}{\sigma^2\sqrt{2\pi}(1-\Phi(\frac{x_0-\mu}{\sigma}))}\int_{x_0}^\infty \log(x) e^{-\frac{(x-\mu)^2}{2\sigma^2}}dx$
I'm stuck here.
Thanks for any help.
 A: Here is a suggested solution. Recall that
$$
\frac{\partial}{\partial t} x^t = x^t\log(x)
$$
and notice that the above derivative will be equal to $\log(x)$ when $t = 0$. Therefore,
$$
I = \int^\infty_{x_0} \log(x) \exp\left(-\frac{1}{2} \left[\frac{x - \mu}{\sigma}\right]^2\right)dx = \left.\int^\infty_{x_0} \left[\frac{\partial}{\partial t} x^t\right] \exp\left(-\frac{1}{2} \left[\frac{x - \mu}{\sigma}\right]^2\right)dx\right|_{t = 0}
$$
Using Leibniz integral rule, the above integral can be rewritten as
$$
I = \left.\dfrac{d}{dt} \int^\infty_{x_0} x^t \exp\left(-\frac{1}{2} \left[\frac{x - \mu}{\sigma}\right]^2\right)dx\right|_{t = 0}.
$$
Let $u = \frac{1}{2}\left(\frac{x - \mu}{\sigma}\right)^2$, accordingly, $\frac{\sigma}{\sqrt{2}} u^{-\frac{1}{2}} du = dx$. Hence,
$$
I = \frac{\sigma}{\sqrt{2}} \left.\dfrac{d}{dt} \int^\infty_{\xi_0} u^{1 - \frac{t}{2} - 1} e^{-u}du\right|_{t = 0} = \frac{\sigma}{\sqrt{2}} \left.\dfrac{d}{dt} \Gamma\left(1 - \frac{t}{2}, \xi_0 \right)\right|_{t = 0}
$$
where $\xi_0 = \frac{1}{2}\left(\frac{x_0 - \mu}{\sigma}\right)^2$. Using the definition of the first-order derivative of the upper incomplete gamma function $\Gamma(\cdot, \cdot)$ that was given by Geddes et. al. (1990), the above integral reduces to:
$$
I = -\frac{\sigma}{2\sqrt{2}} \left.\left[\log \xi_0 \Gamma\left(1 - \frac{t}{2}, \xi_0 \right) + \xi_0 T\left(3, 1 - \frac{t}{2}, \xi_0\right) \right]\right|_{t = 0} = -\frac{\sigma}{2\sqrt{2}} \left[\log \xi_0 \Gamma\left(1, \xi_0 \right) + \xi_0 T\left(3, 1, \xi_0\right) \right]
$$
where
$$
T\left(3, 1, \xi_0\right) = G^{3,0}_{2,3} \left(\xi_0 \left|
\begin{matrix}
0 , 0 \\
-1, 0, - 1
\end{matrix} \right.
\right)
$$
is a special case of the Meijer G-function.
Reference:
K.O. Geddes, M.L. Glasser, R.A. Moore and T.C. Scott (1990), Evaluation of Classes of Definite Integrals Involving Elementary Functions via Differentiation of Special Functions. Applicable Algebra in Engineering, Communication and Computing, vol. 1, pp. 149-165 
A: A quick approximation for large $a$ using the saddle point method: 
The integrand can be written as 
$e^{f(x)}$, where 
$$f(x) = \log\log x - \frac{(x-a)^2}{b}.$$
We find 
$$f'(x) = \frac{1}{x\log x} - \frac{2(x-a)}{b}.$$
Thus, for large $a$ we find $f'(a)=\frac{1}{a\log a}\approx 0$. 
We find 
$f(a) = \log\log a$
and 
$$f''(a) = -\frac{2}{b} + O\left(\frac{1}{a^2\log a}\right)
\approx -\frac{2}{b}.$$
Therefore, 
\begin{align*}
\int_{x_0}^\infty \log(x) e^{-(x-a)^2/b}dx
&\approx
\int_{x_0}^\infty
\exp\left(\log\log a -\frac{1}{b}(x-a)^2\right) dx \\
&= \sqrt{\pi b}\log a\,
\Phi\left(\sqrt{\frac{2}{b}}(a-x_0)\right).
%&= \frac{1}{2}\sqrt{\pi b}\log a
%\left(1+\mathrm{erf}\left(\frac{a-x_0}{\sqrt{b}}\right)\right).
\end{align*}
(It can be shown that the same result holds if we assume instead that $b$ is small.)
Addendum: Below we plot the absolute relative error and the absolute error using this approximation for $x_0=1$.


A: Too long for a comment : Even the simple case $\color{#0055AA}{a=x_0}$ yields an expression in terms of  $($ generalized $)$ hypergeometric functions : 

$$
I(a,b)~=~\frac14~\sqrt{\frac\pi b}~\bigg[2a^2\cdot~_2F_2\bigg(\{1,1\},\bigg\{\frac32,~2\bigg\},-\frac{a^2}b\bigg)-b\cdot J(a,b)\bigg],
$$

with 

$$
J(a,b)~=~\gamma~+~\ln\frac4b~+~\text{erf}~\bigg(\frac a{\sqrt b}\bigg)\cdot\bigg(\gamma-2+\ln\frac{a^2}b\bigg),
$$

where $\text{erf}$ represents the error function, and $\gamma$ stands for the Euler-Mascheroni constant. 
It is therefore highly doubtful that the more general definite integral posted in the original question possesses a closed form even in terms of these special functions. Perhaps Meijer's G-function might sweep in and save the day ? Just a thought $\ldots$ 
