What is the domain of the function $F(x)=\int_{0}^{x}\frac{\operatorname{arctan}(t)}{t}dt$? What is the domain of the following function?
$$F(x)=\int_{0}^{x}\frac{\operatorname{arctan}(t)}{t}dt$$
On the one hand, the internal function is not defined at 0, but on the other hand, it's defined on every other point except zero, so the region can be calculated.
 A: $F$ has a well defined integral from $(-\infty, \infty)$. The integral at $0$ is well defined, you can see that by expanding the taylor series of $$\arctan(t)/t = \sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{2n+1},$$ which converges for $|t| < 1$. For regions away from zero, the function is a ratio of continuous function with the denominator nonvanishing. Thus the integral is well defined for all $x$.
A: The function can be extended in $x=0$ to obtain a continuous function defined on the whole real line. Hence such extended function is Riemann integrable and hence its integral on $[0,x]$ is well defined for all $x$. Now remember that if you modify a function in a finite number of points, its integral does not change. Hence you can extended your function with any value in $x=0$ and obtain an Riemann integrable function. 
However, strictly speaking, Riemann integrals on the interval $[a,b]$ is defined only for functions which are defined on the whole interval $[a,b]$. If the function is only defined on the interval $(a,b]$ this should be considered an improper integral i.e.:
$$
F(x) = \int_0^x f(t)\, dt = \begin{cases}\lim_{a\to 0^+}  \int_a^x f(t)\, dt & \text{if } x>0 \\ \lim_{a\to 0^-}\int_a^x f(t)\, dt & \text {if $x<0$ }\\ 0 &\text{if } x=0\end{cases}.
$$
But, again, since the function $f$ can be extended with continuity in $x=0$ the limit exists and is equal to the integral on $[0,x]$ of the extended function.
