Approximating an integral

1. May 20, 2005

Zaare

I'm stuck trying to solve the following problem:

If $$D_n$$ is the Dirichlet kernel, show that there exist positive constants $$c_1$$ and $$c_2$$ such that
$$c_1 \log n \le \int\limits_{ - \pi }^\pi {\left| {D_n \left( t \right)} \right|dt} \le c_2 \log n ,$$​
for $$n=2,3,4,...$$. By $$\log$$ they mean the natural logarithm.

I know that
$$D_n \left( t \right) = \frac{1}{\pi }\left( {\frac{1}{2} + \sum\limits_{N = 1}^n {\cos \left( {Nt} \right)} } \right) = \frac{1}{{2\pi }}\sum\limits_{N = - n}^n {e^{iNt} } = \frac{{\sin \left( {nt + \frac{t}{2}} \right)}}{{2\pi \sin \left( {\frac{t}{2}} \right)}}$$​
And it's easy to see that
$$\left| {D_n \left( t \right)} \right| = \frac{1}{\pi }\left( {\frac{1}{2} + \left| {\sum\limits_{N = 1}^n {\cos \left( {Nt} \right)} } \right|} \right) \le \frac{1}{\pi }\left( {\frac{1}{2} + n} \right)$$​
But that is all I can do. I have no idea on how $$\log$$ comes into the picture. Any help, suggestions or tips would be much appreciated.

2. May 20, 2005

saltydog

Zaare, isn't $|D_n(t)|$ always greater than $\frac{1}{2\pi}$? Thus the integral of that over -pi to pi is 1. Thus that integral is always greater than 1. So what does $c_1$ (in terms of n) have to be to make the expression hold? Same dif for the other side right since (according to your results:

$$|D_n(t)|\leq \frac{1}{\pi}(\frac{1}{2}+n)$$

So the integral is always less than 1+2n. Now, what does $c_2$ have to be to satisfy this one?

3. May 20, 2005

HackaB

D_2(t) (according to the forumula involving sines) has a root at t = 2 pi/5. So its absolute value isn't always greater than any positive number.

4. May 21, 2005

saltydog

Thanks for pointing that out HackaB. I see now that the last inequality should be:

$$|D_n(t)|=|\frac{1}{2\pi}+\frac{1}{\pi}\sum_N^n Cos(Nt)|$$

5. May 21, 2005

shmoe

Hi, notice the Dirichlet kernel is an even function, so it's enough to consider the integral on 0 to pi.

Using the

$$\frac{{\sin \left( {nt + \frac{t}{2}} \right)}}{{2\pi \sin \left( {\frac{t}{2}} \right)}}$$

version, you see it changes sign whenever $\sin \left( {nt + \frac{t}{2}} \right)=0$, find all these points and treat seperately the integral on each of these intervals. This removes the absolute value problem.

Now finding the integral on each of these intervals isn't easy, but fortunately a good enough approximation isn't difficult. Find (constant) bounds for $$1/\sin(t/2)$$ seperately on each of these smaller intervals to remove it from the integrals, then integrate, sum and you're nearly done. Nothing fancy for the bounds is needed, a partial hint remember $$\sin{x}\leq x$$.

One more thing, you'll have to treat the interval containing 0 in a different way, but I'll leave that to you.

6. May 27, 2005

Zaare

Finally, I solved this a couple of days ago.
Schmoe, I followed your "recipe". It was of great help.
Thank you all.