Continuity by the right of a function defined by an improper integral Suppose we have a continuous function defined by $$f:\Bbb R\times(0,\infty)\to\Bbb R,\quad (t,x)\mapsto f(t,x)$$
such that $F(x):=\int_0^\infty f(t,x)\,\mathrm dt$ converges uniformly in $[c,\infty)$ for any chosen $c>a$, where the integral is an improper integral of Riemann. When we can say that
$$\lim_{x\to a^+}F(x)=\int_0^\infty \lim_{x\to a^+}f(t,x)\,\mathrm dt=\int_0^\infty f(t,a)\,\mathrm dt$$
for some $a>0$? That is, when we can say that $F$ is continuous at $a$ by the right?

Background: this question comes just from curiosity, I dont have a clear answer by now.
 A: This is not true in general even if the improper integral $\int_0^\infty f(t,a) \, dt$ converges and $F(a)$ exists.  
The assertion is true, of course, if the improper integral $\int_0^\infty f(t,x) \, dt$ is uniformly convergent on $[a,\infty)$, but it is not enough for the convergence to be uniform on the  interval $[c,\infty)$ for every $c > a.$
For a  counterexample, take $f(t,x) = \sin(xt)/t$ for $t \neq 0$ and $f(0,x) = x$.  For the improper integral  
$$\tag{*} F(x) = \int_0^\infty \frac{\sin (xt)}{t} \, dt$$
we have  $F(0) = 0$ and $F(x) = \frac{\pi}{2}$ for all $x > 0$. Hence, 
$$F(0) = 0 \neq \frac{\pi}{2} = \lim_{x \to 0+} F(x).$$
We show that (*) is uniformly convergent on $[c, \infty)$ with $c >0$  using the Cauchy criterion.  By the second mean value theorem for integrals, for all $a_2 > a_1 >0$ there exists $\xi \in (a_1,a_2)$ such that 
$$\tag{**}\left|\int_{a_1}^{a_2} \frac{\sin (xt)}{t} \, dt\right| = \left|\frac{1}{a_1}\int_{a_1}^{\xi} \sin (xt) \, dt\right|  = \frac{|\cos (xa_1) - \cos(x \xi)|}{xa_1} \leqslant \frac{2}{c a_1}$$.
For any $\epsilon > 0$, the RHS of (**) is less than $\epsilon$ if $a_1 > 2/(c \epsilon)$. 
