I want to find out what the Taylor expansion of
$$F(x) = \int_0^x \frac{\sin(t)}{t} dt .$$
Am I wrong in saying that by the fundamental theorem of calculus, $F'(x) = sin(t)/t$? Should I continue from there? It just doesn't sit well with me for some reason.
Thanks for your time.
As justified by Fantini, your idea is good and you just continue. I supposed that you started using the Taylor series for $sin(t)$, divided by $t$, integrate between $0$ and $x$ and you are done. I suppose you arrived to something looking like $$x-\frac{x^3}{18}+\frac{x^5}{600}-\frac{x^7}{35280}+\frac{x^9}{3265920}-\frac{x^{11}}{ 439084800}+O\left(x^{13}\right)$$
No, you are not. You can continue from there, nothing is wrong with your approach since $\sin(t)/t$ is a continuous function for all $x >0$, therefore by the fundamental theorem of calculus $F(x)$ is differentiable and $F'(x) = \sin(x)/x$.
