Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm trying to evaluate the following integral:

$$A(a)=\int_{-\infty}^{\infty}e^{-x^2/2}\log\left(\int_{x-a}^{x+a}e^{-t^2/2}dt\right)dx$$

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.

share|improve this question
add comment

1 Answer

up vote 1 down vote accepted

Would you be interested in a series expansion? Maple gives

$$ \sqrt{2 \pi} \left(\ln(2a) - \frac{1}{2} - \frac{a^4}{36} + \frac{a^6}{81} - \frac{37 a^8}{5400} + \frac{31 a^{10}}{6075} - \frac{128831 a^{12}}{26790750} + \ldots \right) $$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.