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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.

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It should be said that the integrand is the so called Dirichlet kernel which has a central role in the study convergence of Fourier series on the circle since $D_n*f=S_n[f]$, that is the $n^{\text{th}}$ partial sum of the Fourier series of $f$. Also, it is well known that $\|D_n\|_{L^1}\to\infty$ as $\log n$, in particular this problem reflects that $D_n$ is not a positive summation kernel (such as the Fejér kernel $K_n =\frac{1}{n+1}\sum_0^nD_k$). More of $D_n$ can be found here en.wikipedia.org/wiki/Dirichlet_kernel –  AD. Dec 4 '10 at 6:40

3 Answers 3

up vote 4 down vote accepted

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.

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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 $\int_1^\infty \frac{\sin t}{t} dt < \infty$.

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I don't quite see how to use the integral to estimate the sum. Can you elaborate? Thanks. –  TCL Dec 6 '10 at 6:34
    
You are right, I was being too hasty. You have to actually directly estimate the trigonometric sum. You can probably find a proof of the boundedness of that sum in Zygmund's Trigonometric Series. –  Willie Wong Dec 7 '10 at 18:23

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)

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But $C$ has to bound $\int_0^x \sin Nt \cot (\frac{1}{2}t) dt$ for all N and $x\in [-\pi,\pi]$. I don't see how. –  TCL Dec 4 '10 at 4:32
    
Near $x=0, sin(Nt)\approx Nt$ and $\int(t\cos(t))=t sin(t) + cos(t)$. cos(t) is bounded by 1. –  Ross Millikan Dec 4 '10 at 5:44
    
But this argument will need much refining if you want to show that the integral is bounded independently of $N$. –  Willie Wong Dec 4 '10 at 5:48
    
@Willie Wong: You are right. For small x and large N we have $\int_0^x{2Ndt}=2Nx which fails. We need to take advantage of the fact that sin(Nt) turns around before it makes much area. More thinking required. –  Ross Millikan Dec 4 '10 at 6:04

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