Boundedness of an integral Got stuck on this one:
Show that there is a constant $C>0$ such that $$\left|\int_0^x \frac{\sin (N+1/2)t}{\sin t/2} dt \right|\le C$$ for all
$x\in [-\pi,\pi]$ and integer $N\ge 1$.
I thought this should follow from $\int_0^\infty \frac{\sin t}{t}dt<\infty$, but somehow
don't get the connection.
 A: This is a classic integral from summing Fourier series. I saw the following argument in Katznelson's book I believe. By symmetry we can assume $x > 0$. We divide the integral into $0$ to ${1 \over N}$ and ${1 \over N}$ to $x$ parts. (The second term will be $0$ if $x$ is small enough). Since $|\sin((N + {1 \over 2})t)| \leq (N + {1 \over 2})t$ and $\sin({t \over 2}) \geq C't$ the integrand is bounded by $CN$ and the first term is at most $C$.
For the second term, you can integrate by parts, integrating the $\sin((N + {1 \over 2})t)$ and differentiating the ${1 \over \sin({t \over 2})}$. You get a factor of ${1 \over N + {1 \over 2}}$ from the integration, and the resulting integrand is now bounded by $C{1 \over Nt^2}$. Taking absolute values and integrating from ${1 \over N}$ to $x$ again gives a bound of $C$. The left endpoint term in this integration by parts is bounded by $C{1 \over Nt}$ at $t = {1 \over N}$, so once again you just get a $C$. The right-endpoint term is even smaller. Thus you're done.
A: Consider: 
$$\sin (Nt + \frac12 t) = \frac1{2i} \left[ \exp (i Nt + \frac{i}2 t) - \exp (-Nt - \frac{i}2 t)\right] = \sin \frac{t}2  \sum_{m = -N}^{N} \exp (imt) $$
So your integrand reduces to 
$$ 1 + 2\sum_{m = 1}^N \cos(mt) $$
Now, $\int_0^x \cos(mt) dt = \frac{\sin (mx)}{m}$. To estimate $\sum_1^N \frac{\sin(mx)}{m}$ you can now use your fact that $\int_1^\infty \frac{\sin t}{t} dt < \infty$.  
A: If you expand the numerator you get a sum of two terms.  One doesn't depend on t, one is proportional to cot(t/2)
