Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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 – AD. Dec 4 '10 at 6:40
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.

share|cite|improve this answer


$$\sin (Nt + \frac12 t) = \frac1{2i} \left[ \exp (i Nt + \frac{i}2 t) - \exp (-Nt - \frac{i}2 t)\right] = \sin \frac{t}2 \sum_{m = -N}^{N} \exp (imt) $$

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

share|cite|improve this answer
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)

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.