Why can't I interchange integration and differentiation here? I think my questions relates to this other: "counterexamples to differentiation under integral sign"
In fact, it provides a counterexample
Consider $f(x,y)=y^3e^{-y^2x}$ and define $F(y) =\int_0^{\infty}f(x,y)dx$
We have that $F'(0)\not = \int_0^{\infty} \frac{\partial f}{\partial y}(x,0)dx$
(We calculate $F'(0)$ essentially using Monotone convergence theorem we can show that,
for $y\in \mathbb{R}\setminus\{0\} $,   $F(y)=y$ moreover $F(0)=0$ so $F'(0)=1$)
Now, I want to understand which hypothesis of Theorem 2 at this page does not hold.
Obviously the third hypothesis does not hold, but I want to consider the case in which we replace it by a weacker condition:
"For each $b \in \mathbb{R}$, there exists an open interval $b\in J$ and an integrable function $g(x)$ over $(0, \infty)$ ,such that $| \frac{\partial f}{\partial y}(x,y)| \leq g(x)$ for every $y\in J$ and $\forall x$"
Now, the first hypothesis certainly holds as
$\forall  y, \ x\rightarrow f(x,y)$ is integrable $(0,\infty)$ by comparison with $e^{-kx}$ for appropriate positive value of $k$
Moreover $ \frac{\partial f}{\partial y}(x,y)$ exists everywhere...
So is the last hypothesis to be problematic but I can't see how as I can bound $y$ in $J$ and then just use some linear combination of $e^{-kx}$ and $ xe^{-lx}$ for suitable $k,l$  as they are both integrable over $(0,\infty)$
Thank you very much!  
 A: You need a convenient locally convex space $E$ of functions in $x$ such that 
$g\mapsto \int_0^\infty g(x)\,dx$ is a bounded linear functional, and such that
$y\mapsto f(\quad,y)$ is a differentiable curve in $E$, at least of Hoelder class $C^{1,\alpha}$. Then the interchange follows by the chain rule. 
The space $\mathcal S(-1,\infty)$ of smooth functions which decrease rapidly on the right  suggests itself. This is a Frechet space with seminorms
$sup \{(1+|x|^2)^m\partial_x^{k} g(x): -1<x<\infty\}$ for fixed $m,k$
Now we look at the paper:
Peter W. Michor and David Mumford: A zoo of diffeomorphism groups on $\mathbb ℝ^n$.  arXiv:1211.5704. (pdf)
By repeating the proof of 3.5 there (which uses the very strong theorem 2.4 of Frölicher and Kriegl)
we see that:
The space $C^\infty(\mathbb R,\mathcal S(-1,\infty))$ of smooth
curves in $\mathcal S(-1,\infty)$ consists of all functions 
$c\in C^\infty(\mathbb R\times(-1,\infty),\mathbb R)$ satisfying the following 
property:
$\bullet$
For all $k,m\in \mathbb N_{\ge0}$ and $\alpha\in \mathbb N_{\ge0}^n$,  the expression
$(1+|x|^2)^m\partial_t^{k}\partial^\alpha_xc(t,x)$ is uniformly bounded in $x\in (-1,\infty)$, locally in $t\in \mathbb R$.
Obviously, your function $f$ satisfies this.
EDIT: The "obviously" is wrong. 
$(1+x^2) y^3 e^{-y^2x}$ is not uniformly bounded for $(x,y)\in (-1,\infty)\times[-1,1]$.
To see this insert the curve $x(t) = 1/(1+t^2)$, $y(t)=t$ to get
$$(1+x^2) y^3 e^{-y^2x}= \Big(1+\frac{1}{(1+t^2)^2}\Big) t^3 e^{-\frac{t^2}{1+t^2}}, \quad t\in[-1.1]$$
which is unbounded. 
Let me try again: $g\mapsto \int_0^\infty g(x)dx$ is a bounded linear functional on $L^1(-1,\infty)$, but $y\mapsto  f(\quad,y)$ is not $Lip^1$ into  (otherwise you could interchange).  
