Show that $\int_{0}^{1}{\frac{\sin{x}}{x}\mathrm dx}$ converges As title says, I need to show that the following integral converges, and I can honestly say I don't really have an idea of where to start. I tried evaluating it using integration by parts, but that only left me with an $I = I$ situation. 
$$\int \limits_{0}^{1}{\frac{\sin{x}}{x} \mathrm dx}$$

Note that $\frac{\sin x}{x} = \sum_{n=1}^\infty (-1)^n \frac{x^{2n}}{(2n+1)!} $.
Then for $|x| \le M$, we have $\left| \frac{\sin x}{x}  \right| \le \sum_{n=1}^\infty \frac{x^{2n}}{(2n+1)!} \le \sum_{n=1}^\infty \frac{M^{2n}}{(2n+1)!} \le \frac{1}{M} e^M$.
