How to formalize my argument for existence of $\lim \limits_{\epsilon \to 0^{+}} \int_{\epsilon}^{c} \frac{\sin x}{x} dx$

Question

I know this exists - $\frac{\sin x}{x}$ is not defined on $0$ but it has an asymptote there. However, I'm finding it difficult to formalize my argument in a way that I don't say things which might not be always true. I was trying to do it defining a function $f$ by: $f(x) = \frac{\sin x}{x}$ if $x > 0$, $ f(0) = 1$. Then $f$ is continuous in $[0, +\infty[$, because $$\lim \limits_{x \to 0^{+}} f(x) = 1$$ and so is integrable in an interval $[0,c] \subset [0, +\infty[$. I don't know if this suffices to say that the original integral exists, because $\frac{\sin x}{x}$ is not defined in $0$.

Answer 1

As $\lim_{x\to 0}\frac{\sin x}{x}=1$ we find that the singularity at $x=0$ is removable. As the only discontinuities of $\frac{\sin x}{x}$ on $\left[0,c\right]$ are removable for all $c\geq 0$ we find that the integral exists.

Answer 2

Your title contains the expression $$\int_{0^+}^c \frac{\sin x}{x}\,dx.$$ I expect this has been defined (for $c\gt 0$) as being equal to the following limit, if it exists: $$\lim_{\epsilon \to 0^+}\int_{\epsilon}^c\frac{\sin x}{x}\,dx.\tag{1}$$ So does the limit exist? As $\epsilon$ approaches $0$ through positive values, the integral $\int_{\epsilon}^c\frac{\sin x}{x}\,dx$, as soon as $\epsilon$ reaches a reasonably small value, like $\min(c,1)$, is increasing. It is clearly bounded above, since $\frac{\sin x}{x}$ has absolute value $\lt 1$. So the limit (1) exists.