Show that $\int_0^\infty \frac{\sin(x)}{x}\,\mathrm{d}x = \frac{\pi}{2}$, and using that show that $\sum\nolimits_{0<\mid n \mid \leq N} \frac{\mathrm{e}^{inx}}{n}$ is uniformly bounded in N and $x\in[-\pi,\pi]$. Use the fact that $$\frac{1}{2i}\sum\nolimits_{0<\mid n \mid \leq N} \frac{\mathrm{e}^{inx}}{n}= \sum \limits_{n=1}^N \frac{\sin(nx)}{n}=\frac{1}{2}\int_0^x (D_N(t)-1)\,\mathrm{d}t,$$ where $D_N(t)$ is the Dirichlet Kernel.
I was able to show that $\int_0^\infty \frac{\sin(x)}{x}\,\mathrm{d}x = \frac{\pi}{2}$ by using $\int_{-\pi}^\pi D_n(t)\,\mathrm{d}t = 2\pi$. Then using the definition of $D_N(t)=\frac{\sin(N+1/2)x}{\sin(x/2)}$, I showed that the integral results in $\int_{-\pi}^\pi \frac{2\sin(N+1/2)x}{x}\,\mathrm{d}x$ which then by the Riemann Lebesgue Lemma and a change of variable gave me $\int_{0}^{(N+1/2)\pi} \frac{\sin(x)}{x}\,\mathrm{d}x$ which tends to $\pi/2$ as $N\to\infty$. Then using the Mean Value Theorem I arrived at the result that $$\int_0^\infty \frac{\sin(x)}{x}\,\mathrm{d}x = \frac{\pi}{2}.$$ Is this the right method to arrive at the first part of the question?
The second part of the question in how I would go about using that to show that the sum is uniformly bounded is where I'm getting stuck. Am I just over-complicating things and therefore can't get it? Any help will be appreciated. Thank you!