Continuity of a parametric integral (where the integrated function is discontinuous) For all $t\in\mathbb{R}$ consider
$$F(t):=\int_\mathbb{R}e^{-x^2/2}\log|t+e^x|\,dx \;.$$
I managed to show that $F(t)$ is well-defined and finite for every $t$.
I would like to show that $F$ is continuous on $\mathbb{R}$.
Notice that the difficult part is continuity at $t\leq0$.
Anyway I belive the funtion is continuous, because of a numerical plot.
How could I prove it? I tried to apply dominated convergence, but I don't manage.
 A: This should work,
Let 
$$F(t):=\int_\mathbb{R}e^{-x^2/2}\log|t+e^x|\,dx \;.$$
Note first that $f(t,x):=e^{-x^2/2}\log|t+e^x|$ is continuous for almost all $x$. The only singularity we can find is if $|t+e^x|=0$. But as $e^x$ is injective this happens at most in one point, namely if $t<0$. Since singeltons are of Lebesgue measure zero this has no influence on the coninuity of $F(t)$.
Furthermore the follwing holds,
$$|e^{-x^2/2}\log|t+e^x||\leq e^{-x^2/2}|t+e^x|\leq e^{-x^2/2}|t|+e^{-x^2/2}e^x$$
and since continuity is a local property it is sufficent to find a dominating function vor every $t$ in a arbitrary small neighborhood of $t$. Thus let $\epsilon>0$ be arbitrary small and consider $(t-\epsilon,t+\epsilon)$ for some fixed $t\in\mathbb R$. 
Then we have 
$$e^{-x^2/2}|t|+e^{-x^2/2}e^x\leq|t+\epsilon|e^{-x^2/2}+e^{-x^2/2}e^x$$
where the last function is integrable in $\mathbb R$. Consequently by the dominated convergence theorem the continuity follows in $t$ and since $t$ was arbitrary we can conlude that $F(t)$ is indeed continous for every $t\in\mathbb R$.
