Does it immediately follow that the sine integral is the antiderivative of $\frac{\sin x}{x}$? The sine integral is commonly defined as $\operatorname{Si}(x) = \int_0^x \frac{\sin t}{t}\,dt$. It seems that by the Fundamental Theorem of Calculus, the derivative of $\operatorname{Si}(x)$ is $\frac{\sin x}{x}$. However, isn't one of the assumptions violated, namely that $\frac{\sin t}{t}$ be defined on $[0, x]$ ($x$ is non-negative)? It's not defined at $0$, which means that the sine integral is actually an improper one. So it becomes $\operatorname{Si}(x) = \lim_{a\to 0^{+}}\int_a^x \frac{\sin t}{t}\,dt$ which you then have to differentiate. Obviously you can avoid the issue by redefining it as $\int_1^x \frac{\sin t}{t}\,dt$ for example. My question is whether the FTC can be immediately applied to the integral when the lower limit is $0$?
 A: Let's understand that when we are dealing with Riemann integral $\int_{a} ^{b} f(x) \, dx$ then it is a prerequisite that $f$ is defined and bounded on $[a, b] $. 
Normally the function $f$ is given by a formula for $f(x) $ in terms of $x$ and in the current case $f(x) =(\sin x) /x$. When the function is specified in such manner there can be a scenario when the defining formula is meaningless for some values of $x$ in the interval $[a, b] $. If there are a finite number of such exceptional values of $x$ in $[a, b] $ for which the formula for $f(x) $ in terms of $x$ does not make any sense then this is not a problem at all. Changing the values of a function at a finite number of points in $[a, b] $ does not affect the boundedness and Riemann integrability of the function and the value of its Riemann integral (if the integral exists). Hence in such scenarios we are at liberty to assign the values of the function at these exceptional points in any manner we wish.
For the current problem we can define $f(0)$ to be any number without any problem. In this current case we are lucky as $0$ turns out to be a removable discontinuity for $f(x)=(\sin x) /x$ and hence defining $f(0)=1$ makes the function $f$ continuous and Fundamental Theorem of Calculus applies to show that $\operatorname {Si} '(x) =\dfrac{\sin x} {x}, x\neq 0$ and $\operatorname {Si}' (0)=1$. Note also that the derivative $\operatorname {Si} '(0)=1$ no matter how we define $f(0)$. 
We may not be always so lucky to have only removable discontinuities. A nice example is $g(x) =\cos(1/x)$ and $G(x) =\int_{0}^{x}\cos(1/t)\,dt$. No matter how we define $g(0)$, $g$ has an essential discontinuity at $0$ and the Fundamental Theorem of Calculus cannot be used directly to compute $G'(0)$. But using some tricky arguments one can show that $G'(0)=0$ and thus if we define $g(0)=0$ then $G$ is an anti-derivative of $g$ on whole of $\mathbb {R} $.
Thus FTC can be used directly in case of the sine integral even when $0$ is the lower limit of integral. Further if you carefully observe the proof of Fundamental Theorem of Calculus you will find that the result being proved is as follows:

Actual FTC: Let $f:[a, b] \to\mathbb{R} $ be Riemann integrable on $[a, b] $ and let $F:[a, b] \to\mathbb {R} $ be defined by $F(x) =\int_{a} ^{x} f(t) \, dt$. Let $c\in[a, b] $ then the left (right) hand derivative of $F$ at $c$ equals the left (right) hand limit of $f$ at $c$ provided the latter exists.

The FTC does not help if the left (right) hand limit of $f$ at $c$ does not exist. Also note that the value $f(c) $ is immaterial when trying to find derivative $F'(c) $ simply because the value of $f$ at a single point (or a finite number of points) has no impact on the integral defining $F$. What is really needed here  is the left (right) limit of $f$ at $c$.
There is no need to invoke improper Riemann integrals in these scenarios because the usual Riemann integral handles bounded functions on bounded intervals. Improper integrals are needed when either the interval of integration is unbounded or the function itself is unbounded. 
A: We can define $\frac{\sin t}t$ at $t=0$ to be 1 such that the function is continuous/differentiable over the whole real line, and then the problem disappears. The undefined point of that function is called a removable singularity.
