If a sequence of functions diverges pointwise to infinity, does its integral diverge? Consider a smooth function $f: \mathbb R^2\to \mathbb R$ with the property that:
$$\forall x\in \mathbb R, \; \; \int_{-\infty}^{\infty} |f(x,y)|dy<\infty. $$
Furthermore, suppose that, $\forall y\in \mathbb R$, $\lim_{x \to \infty} f(x,y)= \infty, $ i.e., $f$ diverges to infinity pointwise in $y$.  Then, I am wondering whether it is necessary true that:
$$\lim_{x \to \infty}\int_{-\infty}^{\infty} f(x,y)dy=\infty.  $$
I believe the answer is no, for a similar reason that pointwise convergence does not imply uniform convergence. But, here, we are dealing with divergence. Also, is there a variant of Dini's theorem that would ensure this holds if $f$ is monotonically increasing in $x$?
 A: I do not believe this is the case.  Fatou's lemma does not apply because $f$ is not assumed to be non-negative, or at least I don't see how to apply Fatou's lemma directly.  As a counterexample, we could consider $$f(x,y) := x 1_{-x \le y \le x} - x(1_{-2x \le y < x} + 1_{x < y \le 2x}).$$
Then for any $y \in \mathbb{R}$, we have $f(x,y) = x$ when $x \ge |y|$ so $\lim_{x \rightarrow \infty} f(x,y) = \infty$.
However, for any $x \in \mathbb{R}$, $$ \int_{-\infty}^\infty f(x,y)dy = 2x^2 - 2x^2 = 0$$ so $\lim_{x \rightarrow \infty} \int_{-\infty}^\infty f(x,y)dy = 0.$
This isn't quite exactly what you want because this $f$ is not smooth, but I think that can be fixed without substantially changing the integrals or the pointwise divergence in $y$ by mollifying $f$.
You also asked whether having $f$ be monotone increasing in $x$ would help.
That would make the clam true because then we could apply Fatou's lemma.  Specifically, we would have $f(x,y)-f(0,y)$ is non-negative, so $$ \lim_{x \rightarrow \infty} \int_{-\infty}^\infty (f(x,y)-f(0,y))dy \ge \int_{-\infty}^\infty \lim_{x \rightarrow \infty} (f(x,y)-f(0,y))dy = \infty $$
