If $\int_{R^2} f(x,y)dxdy$ exists, must $\int_R f(a,y)dy$ exist? 
Let $f$ be a smooth real function on $R^2$ such that 
  $$\int_{R^2} f dxdy$$ exists. Let $a\in R$. Must $\int_R f(a,y) dy$ exist?

I believe this is true, but I don't know how to prove it.
 A: For illustration a simple but non-smooth example first:
Let $h:\mathbb R\to \mathbb R$ be the hat function, that's is $h(x) = \max(0, 1 - |x|)$. Now, let
$$ f(x, y) = h(x \cdot \exp(|y|)). $$
Then, we have
$$ \int f(x,y) dx dy = \int \exp(-|y|) dy = 2. $$
But, $f(0, y) = 1$ is obviously not integrable on $\mathbb R$.
For a smooth example:
Let 
$$g(x) = \frac1{\sqrt{2\pi}} \exp(-x^2)$$ be the density of the standard normal distribution. Now, consider 
$$ f(x,y) = g(x /g(y)). $$
For a symmetric example: 
$$ \tilde f (x, y) = f(x,y) + f(y,x). $$
A: Suppose $f(x):\mathbb{R}\to \mathbb{R}$ is an odd, nonzero and smooth function such that the integral exists (In a Riemann sense). Then $F(x,y):\mathbb{R}^2\to \mathbb{R}$ given by $F(x,y)=f(x)$ is such a counter example, just take $a$ to be some value such that $f(a)\neq 0$. 
Edit: so the above conditions on $f$ aren't exactly what I was picturing in my head. What I mean is a smooth function $f$ such that it is positive for $x>0$ and for $x<0$ it is given by $f(x):= -f(-x)$. This way the function cancels itself out when taking the integral and so it is zero.
