# Taylor expansion of an integral

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).

-

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 $t$.