Asymptotic scaling of integral Say I have some integral of the form 
$$
I(y)=\int_0^\infty  f(x,y) g(x) dx,
$$
where $\int_0^\infty f(x,y) dx=y$, and  $g(x)>0$. Also assume the integral $I(y)$ exists for every $y \in \mathbb{N}$. I feel that somehow $I(y)$ should scale as $y$ as well, at least asymptotically,
$$
I(y) \overset{?}{\sim} c y, \ (y\to\infty)
$$
for some constant $c$. Is there some theorem that states this?  
Attempt at a solution:
If we make a variable substitution $dz=g(x)dx$, we have (assuming the boundaries of integration do not change) $I=\int_0^{\infty} f(x(z),y) dz$. This starts to look a bit like $\int_0^\infty f(x,y)\, dx$ already. Also, I have gone through some examples that support my belief. These are  $f(x,y)=e^{-x^2/y^2}$, $g(x)=x$, and the same $f(x,y)$ with $g(x)=\cos^2(x)$. (I know $\cos^2(x)=0$ for some $x$, but the set where this is true has measure zero, so I wasn't too worried about it.)
 A: The case $f(x,y) = \frac{2}{\sqrt{\pi}}e^{-x^2/y^2}$, $g(x)=x$ that you claim supports your belief, actualy shows that it is not true. You have
$$ \int_0^\infty f(x,y)  dx = \frac{2}{\sqrt{\pi}}\int_0^\infty e^{-x^2/y^2}dx = \frac{2y}{\sqrt{\pi}}\int_0^\infty e^{-z^2}dz = y$$
$$ \int_0^\infty f(x,y) g(x) dx = \frac{2}{\sqrt{\pi}}\int_0^\infty xe^{-x^2/y^2}dx = \frac{2y^2}{\sqrt{\pi}}\int_0^\infty ze^{-z^2}dz = \frac{y^2}{\sqrt{\pi}} $$
We can consider a special case where there exist limits $$\lim_{y\rightarrow\infty} f(zy,y) = F(z), \qquad \int_0^\infty F(z) dz   =1 $$
$$\lim_{y\rightarrow\infty} \frac{g(z y)}{y^\alpha} = G(z) \text{ for some }\alpha\in\mathbb R$$
and the convergence is sufficiently uniform. We have $$ \lim_{y\rightarrow\infty} \frac{1}{y}\int_0^\infty f(x,y) dx = \lim_{y\rightarrow\infty}\int_0^\infty f(zy,y) dz =\int_0^\infty F(z) dz = 1 $$
$$ \lim_{y\rightarrow\infty} \frac{1}{y^{\alpha+1}}\int_0^\infty f(x,y)g(x) dx = \lim_{y\rightarrow\infty} \int_0^\infty f(zy,y) \frac{g(zy)}{y^\alpha} dz = \int_0^\infty F(z) G(z) dz =: c$$
so $$ \int_0^\infty f(x,y) dx \sim y$$ $$ I(y) \sim c y^{\alpha+1}$$
