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 am interested in the Taylor series expansion around $t=0$ of the following expression:

$$I(t)=\int_{0}^{\infty}e^{-x^2}\log\left(e^{-(x-t)^2}+e^{-(x+t)^2}\right)dx$$

Normally, I would proceed by taking the derivatives inside the integral using the Leibniz Integral Rule, however, in this case I am not sure if I can do this, since I can't find the dominating function $g(x)$ for the integrand such that $|e^{-x^2}\log\left(e^{-(x-t)^2}+e^{-(x+t)^2}\right)|\leq g(x)$ that is independent of $t$. Such dominating function is a condition for interchanging the order of differential and integration.

Re-arranging the equation above using arithmetic, interchanging the differential and the integral in the following expression would be very useful:

$$\frac{\partial}{\partial t}\int_0^\infty e^{-x^2}\log \operatorname{cosh}(xt)dx$$ where $\operatorname{cosh}(x)=\frac{e^{x}+e^{-x}}{2}$ is the hyperbolic cosine function. However, I cannot find a dominating function for the integrand in this case that does not depend on $t$.

Any help would be appreciated. I think restricting $t$ to $0\leq t \leq t_{\max} <\infty$ would work, but I am wondering if I can do this without the restriction on $t$ (other than being a real number).

share|improve this question
add comment

1 Answer 1

up vote 1 down vote accepted

For every positive $a$ and $b$, $a+b\geqslant2\sqrt{ab}\geqslant\sqrt{ab}$ hence $\log(a+b)\geqslant\frac12(\log a+\log b)$. Thus, considering $f(t,x)=\mathrm e^{-x^2}\log\left(\mathrm e^{-(x-t)^2}+\mathrm e^{-(x+t)^2}\right)$, one sees that $$ -\mathrm e^{-x^2}(x^2+t^2)\leqslant f(t,x)\leqslant\mathrm e^{-x^2}\log2\leqslant\mathrm e^{-x^2}. $$ Thus, for every $|t|\leqslant1$, $|f(t,x)|\leqslant g(x)$ with $$ g(x)=\mathrm e^{-x^2}(1+x^2). $$ Note that $g$ is integrable and that $(-1,1)$ is a neighbourhood of $t=0$. On the other hand, for every $x$, $f(t,x)\to-\infty$ when $|t|\to\infty$ hence $$ \sup_{t\in\mathbb R}|f(t,x)|=+\infty. $$ This proves that no dominating function is valid for every $t$.

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.