I'm trying to evaluate the following integral:
$$A(a)=\int\nolimits_{-\infty}^{\infty}e^{-x^2/2}\log\left(\int_{x-a}^{x+a}e^{-t^2/2}dt\right)dx$$
I'll be satisfied with a reasonable lower bound on it. I've tried the Mean Value Theorem bound on the inner integral as follows:
$$\begin{align} A(a)&=\int_{-\infty}^{0}e^{-x^2/2}\log\left(\int_{x-a}^{x+a}e^{-t^2/2}dt\right)dx+\int_{0}^{\infty}e^{-x^2/2}\log\left(\int_{x-a}^{x+a}e^{-t^2/2}dt\right)dx\\ &=\int_{-\infty}^{0}e^{-x^2/2}\log\left(2ae^{-y_0^2/2}\right)dx+\int_{0}^{\infty}e^{-x^2/2}\log\left(2ae^{-y_1^2/2}\right)dx\\ &~~~~~~~~\text{where }y_0,y_1\in[x-a,x+a]\text{ in respective integrals by MVT}\\ &\geq \sqrt{2\pi}\log(2a)+\int_{-\infty}^{0}e^{-x^2/2}\log\left(e^{-(x-a)^2/2}\right)dx+\int_{0}^{\infty}e^{-x^2/2}\log\left(e^{-(x+a)^2/2}\right)dx\\ \end{align} $$
since $y_0=x-a$ and $y_1=x+a$ achieve minimum value of the respective function on $[x-a,x+a]$.
This lower bound has a great upside of resulting in two easy-to-evaluate integrals. However, I am wondering if there are other lower bounds that are tighter and result in an analytically solvable outer integral.
