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. 
$$\int \limits_{0}^{1}{\frac{\sin{x}}{x} \mathrm dx}$$

Notice that, for all $0 < x < 1$, we have
\begin{eqnarray*}
\left|\int_0^1 \frac{\sin x}{x} \, \operatorname{d}\!x\right| &\le& \int_0^1 \left|\frac{\sin x}{x} \right| \operatorname{d}\!x \\ \\
&\le& \int_0^1 1 \, \operatorname{d}\!x \\ \\
&\le& 1
\end{eqnarray*} 
