Under what hypothesis is $F(x) = \int_a^b f(t,x) dt$ is continuous? differentiable? If we consider a sequence of functions $f_n$ there are theorems relating the properties of the functions $f_n$ with the properties of the function $\displaystyle F(x) = \sum_{n=0}^\infty f_n$ (when the series converges).
What about a continuous analog for this?
If $f:[a,b]\times \mathbb{R} \to \mathbb{R}$, what conditions should we impose on $f$ so $F(x)=\int_a^b f(t,x) dt$ to be, lets say, differentiable?
I found this example: if $\displaystyle f(t,x) = \frac{x \sin(t x)}{1+x^2}$, then $\displaystyle F(x) = \int_{-\infty}^\infty f(t,x) dt = \pi\ \mathrm{sgn}(x) e^{-|x|}$ that is not continuous at $x=0$, even though $f$ is analytical in both variables.
This example shows that the hypothesis i'm looking for may be very restrictive.
 A: The standard analysis theorem says that $f(t,x)$ should be uniformly (on $t$) continuous in $x$.
A: Suppose that $F(x)=\int_{a}^{b}f(t,x)dt$. Informally, we expect that
$F'(x)=\int_{a}^{b}f_{x}(t,x)dt$, where $f_{x}$ denotes the partial
derivative $\frac{\partial f}{\partial x}$. Therefore, we should
require that for each $x\in\mathbb{R}$, $f(\cdot,x)$ and $f_{x}(\cdot,x)$
are integrable. In order to employ Dominated Convergence Theorem,
we require that there exists an integrable function $g:[a,b]\rightarrow[0,\infty)$
such that for each $|f_{x}(t,x)|\leq g(t)$. Putting
all conditions together, we formally state and prove the following.

Proposition: Let $U\subseteq\mathbb{R}$ be an open set. Let $f:[a,b]\times U \rightarrow \mathbb{R}$ be a function. Suppose that
(1) For each $x\in U$, $f(\cdot,x)$ is integrable,
(2) For each $(t,x)\in[a,b]\times U$, $f_{x}(t,x)$ exists, and
(3) There exists an integrable function $g:[a,b]\rightarrow[0,\infty)$
such that $|f_{x}(t,x)|\leq g(t)$ for all $(t,x)\in[a,b]\times U$.
Then $F:U\rightarrow\mathbb{R}$ defined by $F(x)=\int_{a}^{b}f(t,x)dt$
is differentiable. Moreover, $F'(x)=\int_{a}^{b}f_{x}(t,x)dt$.

Proof: By condition (1), $F$ is well-defined. We go to show that
$F$ is differentiable. Let $x\in U$ be fixed. Let $(h_{n})$ be
an arbitrary sequence of real numbers such that $h_{n}\neq0$ and
$x+h_{n}\in U$. For each $t\in[a,b]$, applying Mean-Value Theorem
on $f(t,\cdot)$, we have $f(t,x+h_{n})-f(t,x)=f_{x}(t,\xi)h_{n}$,
for some $\xi$ between $x$ and $x+h_{n}$. (Note that $\xi$ also relies on $t$, but this is not important.) Hence,
$$
\left|\frac{f(t,x+h_{n})-f(t,x)}{h_{n}}\right|\leq g(t).
$$
Now, define $g_{n}:[a,b]\rightarrow\mathbb{R}$ by $g_{n}(t)=\frac{f(t,x+h_{n})-f(t,x)}{h_{n}}$.
Then, $|g_{n}|\leq g$ and $g_{n}\rightarrow f_{x}(\cdot,x)$ pointwisely.
By Dominated Convergence Theorem, we have
\begin{eqnarray*}
\lim_{n\rightarrow\infty}\frac{F(x+h_{n})-F(x)}{h_{n}} & = & \lim_{n\rightarrow\infty}\int_{a}^{b}g_{n}(t)dt\\
 & = & \int_{a}^{b}f_{x}(t,x)dt.
\end{eqnarray*}
Since $(h_{n})$ is arbitrary, by Heine Theorem, we conclude that
$$
\lim_{h\rightarrow0}\frac{F(x+h)-F(x)}{h}
$$
exists (i.e., $F$ is differentiable at $x$) and $F'(x)=\int_{a}^{b}f_{x}(t,x)dt.$
