I already have derived so far that $|f|=f$, and that $f$ is even. Thus, the integral we're ultimately concerned with is
$$2\int_0^\infty \frac{1+e^x}{1+x^2}dx$$
My current instinct is that this integral is infinite, i.e. $f$ is not absolutely integrable. I just need to prove as much. This is not really a simple integral to evaluate, so I'm a little stumped.
My current approach is through the comparison test. Suppose we have a second function $g$ such that $| g(x) | \leq | f(x) |$. Then if $g$ is not absolutely integrable, neither is $f$.
My goal is to try to get a second function $g$ that satisfies this. I first split up the integral into two integrals, and focus on the one with the exponential in it. (I've already calculated the former to be $\pi/2$.)
$$2\int_0^\infty \frac{1+e^x}{1+x^2}dx = 2\left( \int_0^\infty \frac{1}{1+x^2}dx +\int_0^\infty \frac{e^x}{1+x^2}dx \right ) = \pi +\int_0^\infty \frac{e^x}{1+x^2}dx $$
I've tried a number of functions: I want a function less than $\frac{e^x}{1+x^2}$ that is also not a finite value to make this work. I've got heuristic arguments as for why $f$ should not be absolutely integrable, but I want be more formal.
Any ideas for a theorem I might be overlooking, or for a suitable function that actually has a nice antiderivative that would make this work?
(Sorry for all the edits you might have seen as I made this post. I originally deleted this when I saw that my entire approach in the previous time this was posted was ultimately incorrect owing to a flaw of my own.)
Consider:
$$\int_0^\infty\frac{e^x}{1+x^2}\,\mathrm{d}x$$
This integral diverges, because, following from the power series expansion of $e^x$,
$$ \frac{e^x}{1+x^2}\ge\frac{1+x+\frac12x^2}{1+x^2}\ge\frac12 $$