Domain of the integral function $F(x)=\int_2^x (1-\frac{\sin^2 t}{t})\, dt$ I have to study the integral function given by $F(x)=\int_2^x (1-\frac{\sin^2 t}{t})\, dt$.
First of all I have to determine the domain of $F(x)$. To to this I can observe that $f(t)=(1-\frac{\sin^2 t}{t})$ is definite and continous in $(-\infty, 0)\cup (0,+\infty)$. 
For the reason why the finite limit of the integral, that is $x_0\in 2$ is in $(0,+\infty)$, I want to study the behaviour of $f$ for $t\to 0^+$ in order to study the convergence of $\int_2^0 f(x) \, dx$.
To do this I have done the following:
$$\left(1-\frac{\sin^2 t}{t}\right) \sim 1-t \text{ for}\, t\to 0^+$$
And now since $\int_2^0 1-t\, dt$ is finite then also $\int_2^0 (1-\frac{\sin^2 t}{t})\, dt$ is convergent. So surely $(0,+\infty)\in \operatorname{Domain}_F$.
$\textbf{1 question}$: do you think my passages are right?
$\textbf{2 question}$: I have also to study the behaviour for $t\to 0^-$?
EDIT:
Someone states that my doubts are not clear. My doubts are summarized on the two question above. If you think that my attempt of work is a bit confused can you help me writing how can I find out the domain of $F$? Thanks in advance.
 A: If you define $f:\mathbb{R}\to \mathbb{R}$ by $f(t)=1-\sin^2(t)/t$ for $t\neq 0$ and $f(t)=1$ for $t=0$, then $f$ is continuous and bounded on $\mathbb{R}$.
Now, for any $x\in\mathbb{R}$, there are two cases: either $x\geq 2$ or $x<2$.  If $x\geq 2$, then $f$ is continuous and bounded on $[2,x]$. If $x<2$, then $f$ is continuous and bounded on $[x,2]$. In both cases, $$\int_2^xf(t)dt\text{ and }\int_x^2f(t)dt=-\int_2^xf(t)dt$$ exist, since continuous functions on compact sets are integrable. It should follow that the domain of $F$ is all of $\mathbb{R}$.
You say you want to study the behavior of $f$ as $t$ approaches $0$ in order to study the convergence of $F(0)$. This is as simple as showing that $\lim\limits_{x\to 0}f(t)=1$. Then you just continuously extend $f$ from $\mathbb{R}\setminus\{0\}$ to all of $\mathbb{R}$ by defining $f(0)=1$, as I have done above.
Hopefully that clears things up and answers your question of how to find the domain of $F$.
A: $$I=\int \Bigg[1-\frac{\sin ^2(t)}{t}\Bigg]\,dt=\int \Bigg[1-\frac {1-\cos(2t)}{2t}\Bigg]\,dt$$
$$I=t+\frac 12{\text{Ci}(2 t)}-\frac 12{\log (t)}+C$$ which is defined for any $t>0$. But, for small $t$, we have
$$I=\frac{1}{2} (\gamma +\log (2))+t-\frac{t^2}{2}+O\left(t^3\right)$$ So, $I$ is defined $\forall t \geq 0$.
But the cosine integral function has the series expansion
$$\text{Ci}(x)=\gamma+\log(x)+\sum_{k=1}^\infty \frac {(-x^2)^k}{2k \,(2k)!}$$ then the logarithms disappear and the domain is $\mathbb{R}$ (just as you wrote it).
A: Let's state some assumptions first. We are dealing here with Riemann integrals and that too proper ones (in contrast to improper Riemann integrals).
With this assumption the symbol $\int_a^b f(x) \, dx$ is defined for functions $f$ which are bounded on $[a, b] $. Further if $a=b$ the symbol is defined to be $0$ irrespective of nature of $f$. And if $a>b$ the symbol is defined as $-\int_b^a f(x) \, dx$.
In most cases the function $f$ is given in terms of a formula. When this is the case it is expected that the formula for $f(x) $ makes sense for all $x\in[a, b] $. If that's not the case then we need to dig deeper. If the formula for $f(x) $ makes sense for all $x\in[a, b] $ except for a finite number of points then there is absolutely no problem and we are at liberty to define $f$ at those points in any arbitrary manner whatsoever. This is because the values of a function at a finite number of points neither affects Riemann integrability of the function nor the value of the integral (if it exists).
Based on this we see that the integrand $f(t) =1-\dfrac{\sin^2t}{t}$ is defined everywhere except at $f=0$. Defining $f(0)$ in any manner is ok and we don't care about it. Then $f$ is defined and bounded in any closed interval and moreover it has only one possible discontinuity, namely at $0$. Thus $f$ is Riemann integrable on any closed interval.
It follows that $F(x) =\int_2^x f(t) \, dt$ is defined for all real values of $x$ and hence the domain of $F$ is $\mathbb {R} $.

In particular you don't need to take limit of integrals here. That procedure is needed for improper Riemann integrals when either the function is unbounded or the interval of integration is unbounded or both are unbounded. For functions which are bounded on bounded intervals we don't need this limiting procedure.
Also for completeness sake one can't have a formula for the integrand undefined at at infinite number of points. If you get something like that then understand that there is no point talking about the Riemann integral of that integrand.
