# Proving $f(x) = \frac{\sin x}{x}$ converges by improper integral test? [duplicate]

So my professor talked about one example of improper integrals and I'm having difficulty understanding the general proof outline for proving convergence.

I was given the problem to prove that $$f(x) = \begin{cases} \frac{\sin x}{x}, & x>0 \\ 1, & x=0 \end{cases}$$ converges . The proof then goes like (proof is in blockquotes and my question/comments follow underneath it):

$$f$$ is continuous at $$x \ \forall \ x >0$$ and $$f$$ is continuous at $$x=0$$.

My question: I know how to show continuity with the epsilon-delta argument, but other than that how can I tell if something is continuous? Specifically for this case, why do we have continuity for all $$x>0$$ and $$x=0$$?

Therefore $$f$$ is RI on $$[a,b] \ \forall \ b>0$$. In particular, $$f$$ is RI on $$[0,1]$$. So $$\displaystyle \int_0^{\infty} f(x) dx$$ converges iff $$\displaystyle \int_1^{\infty} f(x)dx$$ converges.

I'm guessing that this is from the comparison test for improper integrals where if $$|f(x)| \leq g(x)$$ then $$\displaystyle \int_a^{\infty} g(x) dx$$ converges. Clearly the integral from $$1$$ to infinity is smaller than the integral from $$0$$ to infinity.

And so we look at \begin{align*} \lim_{b \rightarrow \infty} \int_1^{b} \frac{\sin x}{x} dx &= \lim_{b \rightarrow \infty} \left[-\frac{1}{x} \cos x\right]_1^{b} - \int_1^{b} \frac{\cos x}{x^2} dx \\ &= \lim_{b \rightarrow \infty} \left(\frac{-\cos b}{b} +\cos(1)\right) - \int_1^{\infty} \frac{\cos x}{x^2} dx. \end{align*}

I understand this is done by integration by parts.

Finally, we have that $$\displaystyle \int_1^{\infty} \frac{\cos x}{x^2} dx$$ converges because $$\displaystyle \left|\frac{\cos x}{x^2}\right| < \frac{1}{x^2}$$ and $$\displaystyle \int_1^{\infty} \frac{1}{x^2}$$ converges.

Does $$\displaystyle \int_1^{\infty} \frac{1}{x^2}$$ converge from the fact that $$\displaystyle \sum \frac{1}{n^p}$$ converges if $$p>1$$?