I am working through an exercise that will require me to have a workable understanding of improper integrals. I would like to make sure the following concepts are correct:
Suppose I have the function $F(x)=\int_0^x \frac{1}{\sqrt{t}}$. Firstly, because $\frac{1}{\sqrt{\cdot}}$ is a function that is not defined at $t=0$, we implicitly understand that $F(x)$ is actually shorthand for $\displaystyle \lim_{\varepsilon \to 0^+}\int^x_{\varepsilon}\frac{1}{\sqrt{t}}$. From what I know, this amounts to a definitional convention. Next, suppose I am interested in the value $F(0)$ (assuming it exists...i.e. assuming $F$ is definable at $0$).
What is not clear to me is whether or not this, too, is shorthand for something...or just altogether invalid syntax. Would it be appropriate to think that $F(0)$ is just shorthand for $\displaystyle \lim_{x\to 0^+} \left(\lim_{\varepsilon \to 0^+} \int^x_{\varepsilon}\frac{1}{\sqrt{t}}\right)$? Is this convention? Or something I would have to explicitly define for my own purposes?
I bring this case up because $F(0)$ appears to approach $0$ when I use graphing software. Comparatively, in the case of $G(x)=\int_0^x\frac{1}{x}$, the graph does not even appear, which I suppose means that $\lim_{\varepsilon \to 0^+}\int_{\varepsilon^+}^x \frac{1}{x}$ does not exist...but does $G(0)=\lim x \to 0^+ \left(\lim_{\varepsilon \to 0^+} \int_{\varepsilon}^x\frac{1}{x}\right)$ exist? How are such double limits calculated?