How can we show that $D_n(t)$ has a bounded integral over a certain interval?

[Euge]

Here is this week's POTW:

-----
Let $D_n(t) = \sum\limits_{\lvert k\rvert \le n} e^{2\pi i kt}$ for $t\in [-.5, .5]$. Show that if $n \ge 2$, there are positive constants $A$ and $B$ independent of $n$, such that

$$A \le \frac{1}{\log n}\int_{-.5}^{.5} |D_n(t)|\, dt \le B$$

-----

Remember to read the https://mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to https://mathhelpboards.com/forms.php?do=form&fid=2!

 

[Euge]

Seeing that users are having some trouble, I’m giving another week for POTW submissions and corrections. You might find it useful to show first that $|D_n(t)|$ is bounded by a constant times the smaller of $n$ and $\lvert t\rvert^{-1}$ for all nonzero $t\in [-.5,.5]$.

 

[Euge]

Opalg gets honorable mention for his near perfect solution. Since his overall argument is solid, I’ll post his solution below. Just note that in fact, $D_n(t) = \sin[(2n+1)\pi t]/\sin(\pi t)$.

Spoiler

\(\displaystyle D_n(t) = \sum_{k=-n}^n e^{2\pi ikt}\) is a geometric series with sum $$\frac{e^{-2\pi int}(1 - e^{2\pi i(n+1)t})}{1 - e^{2\pi it}} = e^{-\pi i(2n-1)t}\frac{\sin(n+1)\pi t}{\sin\pi t}.$$ So \(\displaystyle |D_n(t)| = \frac{|\sin(n+1)\pi t|}{|\sin\pi t|}\), and $$\int_{-1/2}^{1/2}|D_n(t)|\,dt = \int_{-1/2}^{1/2}\frac{|\sin(n+1)\pi t|}{|\sin\pi t|}\,dt = 2\int_0^{1/2}\frac{|\sin(n+1)\pi t|}{\sin\pi t}\,dt.$$ On the interval $[0,1/2]$, $2t \leqslant \sin\pi t \leqslant \pi t$. It follows that \(\displaystyle \int_{-1/2}^{1/2}|D_n(t)|\,dt\) lies between two positive multiples of\(\displaystyle \int_0^{1/2}\frac{|\sin(n+1)\pi t|}{ t}\,dt\), so it will be sufficient to estimate the size of that integral. The substitution $x = (n+1)\pi t$ gives $$ \int_0^{1/2}\frac{|\sin(n+1)\pi t|}{ t}\,dt = \int_0^{(n+1)\pi/2}\frac{|\sin x|}{x}\,dx = \sum_{r=0}^n \int_{r\pi/2}^{(r+1)\pi/2}\frac{|\sin x|}{x}\,dx.$$ On the interval $[0,\pi/2]$, $\frac{\sin x}x$ lies between $\frac12$ and $1$.On the interval $[r\pi/2,(r+1)\pi/2]$ for $r\geqslant1$, \(\displaystyle \frac{|\sin x|}{x} \leqslant \frac1{r\pi/2}.\) Therefore \(\displaystyle \int_{r\pi/2}^{(r+1)\pi/2}\frac{|\sin x|}{x}\,dx \leqslant \int_{r\pi/2}^{(r+1)\pi/2}\frac1{r\pi/2}\,dx = \frac1r.\)To get an inequality in the opposite direction, notice that \(\displaystyle |\sin x| \geqslant \frac1{\sqrt2}\) throughout one or other of the subintervals $[r\pi/2,(r+0.5)\pi/2]$, $[(r+0.5)\pi/2,(r+1)\pi/2]$ (each of which has length $\pi/4$). On that subinterval, \(\displaystyle \frac{|\sin x|}x \geqslant \frac{1/\sqrt2}{(r+1)\pi/2} = \frac{\sqrt2}{(r+1)\pi}.\) It follows that \(\displaystyle \int_{r\pi/2}^{(r+1)\pi/2}\frac{|\sin x|}{x}\,dx \geqslant \frac{\sqrt2}{4(r+1)} \geqslant \frac{\sqrt2}{8r}\) (when $r\geqslant1$).Thus \(\displaystyle \int_{r\pi/2}^{(r+1)\pi/2}\frac{|\sin x|}{x}\,dx\) lies between two fixed (independent of $r$) multiples of $\frac1r$, and therefore \(\displaystyle \int_{-1/2}^{1/2}|D_n(t)|\,dt\) lies between two positive multiples of \(\displaystyle \sum_{r=1}^n\frac1r\). But that harmonic sum is asymptotically close to $\ln n$.In conclusion (modulo tidying up the little details of what happens when $r=0$, and how to incorporate the Euler–Mascheroni constant), \(\displaystyle \int_{-1/2}^{1/2}|D_n(t)|\,dt\) lies between two fixed (independent of $n$) positive multiples of $\ln n$.