Help showing that $g(x)$ is differentiable. Let $f:\mathbb{R}^{2}\rightarrow \mathbb{R}$ be a bounded function, continuously differentiable function and with bounded derivative.  Define $g$ as:
$$g(x)=\int_{\mathbb{R}}\frac{f(x,t)}{t^2+x^2}\,dt.$$
I need to show that $g$ is differentiable in $\mathbb{R}-\{0\}$. To do that I proved that $g(x)$ is well defined since the function $$t\to \frac{f(t,x)}{t^2+x^2}$$
is Lebesgue integrable and I tried to calculate the limit
$$\lim_{h\to 0}\frac{g(x+h)-g(x)}{h}=\lim_{h\to 0}\frac{\int_{\mathbb{R}}\frac{f(t,x+h)}{t^2+(x+h)^{2}}-\int_{\mathbb{R}}\frac{f(t,x)}{t^2+x^{2}}}{h}\, dt$$
But I don't know how to use the hypothesis to calculate that limit. Any suggestion or help would be appreciated. 
 A: Let $q(x,t) = {f(x,t) \over t^x+x^2}$, then we want to show that
$g'(x) = \lim_{h \to 0} \int {g(x+h,t)-q(x,y) \over h} dt$ exists.
Note that $\lim_{h \to 0} {q(x+h,t)- q(x,t) \over h} =  {\partial q(x,y) \over \partial x} = { {\partial f(x,y) \over \partial x} \over x^2+t^2} -2x {f(x,t) \over (x^2+t^2)^2} $.
Suppose $f$ and ${\partial f(x,y) \over \partial x}$ are bounded by $M$. 
Suppose $x \neq 0$ and $|h| < {1 \over 4} |x|$. Then ${3 \over 2 } |x| > |x+h| > {1 \over 2 } |x|$.
We have $q(x+h,t)-q(x,t) = \int_x^{x+h} {\partial q(s,y) \over \partial x} ds$.
Then $|q(x+h,t)-q(x,t)| \le \int_x^{x+h} |{\partial q(s,y) \over \partial x}| ds \le ({M \over {x \over 2}^2+t^2}+2 {3 \over 2}|x|{M \over (x^2+t^2)^2}) |h|$, and so
$|{ q(x+h,t)-q(x,t) \over h}| \le {M \over {x \over 2}^2+t^2}+2 {3 \over 2}|x|{M \over (x^2+t^2)^2}$. Note that the right hand side is integrable with respect to $t$.
Hence the dominated convergence theorem gives
$g'(x)  =\lim_{h \to 0} \int {q(x+h,t)-q(x,y) \over h} dt =  \int \lim_{h \to 0}{q(x+h,t)-q(x,y) \over h} dt = \int {\partial q(x,t) \over \partial x} dt$.
A: $\newcommand{\R}{\mathbb{R}}$
So you want to calculate 
$$
\frac{d}{dx} g(x) = \frac{d}{dx} \int_\mathbb{R} \frac{f(x,t)}{x^2 + t^2} dt.
$$
One very powerful tool in the Lebesgue theory of integrals is the fact, that under relative weak assumptions you can actually swap differentiation and integration. That fact is known as dominated convergence and in your case it goes likes this:
You already argued that $t \mapsto \frac{f(x,t)}{x^2+t^2}$ is (absolutely) integrable for all $x \in\R\setminus\{0\}$ and also that for those $x$ the above function is differentiable (and this derivative is also absolute integrable). Now, if one could show that there exist an absolutely integrable function $h : \R \to \R$ which satisfies
$$
\left|\frac{\partial}{\partial x} \frac{f(x,t)}{x^2 + t^2} \right| \leq |h(t)|
\qquad \text{for all} \quad t\in\R, \, x \in \R\setminus\{0\},
$$
then you are "already" allowed to conclude that 
$$
\frac{d}{dx} g(x) = \frac{d}{dx} \int_\mathbb{R} \frac{f(x,t)}{x^2 + t^2} dt
= \int_\mathbb{R} \frac{\partial}{\partial x}  \frac{f(x,t)}{x^2 + t^2} dt
$$
Please notice though, that after further thinking about it I am not sure anymore how to find this $h$ or that it even exists. The twice upvoted comment under your question though is to my knowledge very much related to this issue and if that shall work so should my approach. I thought about deleting this but maybe even though this not really an answer it might be helpful?!
