Prove that $\int_0^u\frac{\sin(mt)}{2\sin(t/2)}dt$ is bounded

Question

When I read the Fourier Series theory, a property show in the book but without the details.

That property is:

$\displaystyle\int_0^u\frac{\sin(mt)}{2\sin(t/2)}dt$ is bounded for all $m$ and $0\leq u\leq\pi$

I want to ask how to prove it? Also, what is the situation we need this theorem? Maybe it can simply be found on other books on this topic, but I don't have them.

Answer 1

Since $ \sin x \ge \frac{2}{\pi}x$ for $0 \le x \le \frac{\pi}{2}$, then for $0 \le u \le \pi$,

$$\left\lvert \int_0^u \frac{\sin mt}{2\sin \frac{t}{2}}\, dt\right\rvert \le \int_0^{u} \left\lvert \frac{\sin mt}{2\sin \frac{t}{2}}\right\rvert\, dt \le \frac{\pi}{2}\int_0^\pi \frac{\lvert \sin mt\rvert}{t}\, dt$$

Using the inequality $\lvert \sin mt\rvert \le \lvert m t\rvert$ for all $t$, we estimate

$$\int_0^{\pi} \frac{\lvert \sin mt\rvert}{t}\, dt \le \int_0^\pi \lvert m\rvert\, dt = \lvert m\rvert\pi$$

Thus $$\left\lvert\int_0^u \frac{\sin mt}{2\sin \frac{t}{2}}\, dt\right\rvert \le \frac{\lvert m\rvert\pi^2}{2}$$

Answer 2

Through the substitution $t=2s$ it is enough to prove that

$$ I_m(v)=\int_{0}^{v}\frac{\sin(2ms)}{\sin(s)}\,ds $$ is bounded for any $m\in\mathbb{N}$ and $v\in\left(0,\frac{\pi}{2}\right)$. Since $\frac{s}{\sin(s)}$ is a bounded continuous function over such interval, by the Cauchy-Schwarz inequality $$\begin{eqnarray*} I_m(v)^2 &\leq& \int_{0}^{v}\frac{\sin(2ms)^2}{s^2}\,ds\int_{0}^{v}\frac{s^2}{\sin(s)^2}\,ds\\&\leq&\int_{0}^{+\infty}\frac{\sin(2ms)^2}{s^2}\,ds\int_{0}^{\pi/2}\frac{s^2}{\sin(s)^2}\,ds\\&=&\pi m\cdot \pi\log(2) \end{eqnarray*}$$ hence $$ \left|I_m(v)\right|\leq \left(\pi\sqrt{\log 2}\right)\sqrt{m}.$$ You may also notice that $I_1(v)=2\sin(v)$ and $$ I_{m+1}(v)-I_{m}(v) = \frac{2}{2m+1}\sin((2m+1)v), $$ so $I_m(v)$ is given by the $m$-th partial sum of the Fourier series of a rectangle wave and $|I_m(v)|$ is uniformly bounded by the constant $\int_{0}^{\pi}\frac{\sin t}{t}\,dt = \text{Si}(\pi)$ appearing in Gibbs phenomenon.

Answer 3

Notice that, in general, for $n \ge 0$

$$\cos nx + \Bbb i \sin nx = \Bbb e ^{\Bbb i n x} = (\Bbb e ^{\Bbb i x})^n = (\cos x + \Bbb i \sin x)^n = \sum _{k=0} ^n \binom n k \Bbb i ^k \cos ^{n-k} x \sin ^k x$$

whence equating the imaginary parts in both sides gives us

$$\sin nx = \binom n 1 \cos ^{n-1} x \sin x - \binom n 3 \cos ^{n-3} x \sin ^3 x + \binom n 5 \cos ^{n-5} x \sin ^5 x - \\ \binom n 7 \cos ^{n-7} x \sin ^7 x + \dots = (\sin x) P_{n-1} (\cos x)$$

with $P_{n-1} = a_{n-1,n-1} x^{n-1} + a_{n-1,n-2} x^{n-2} + \dots + a_{n-1,0}$ a polynomial function of degree $n-1$.

It follows that

$$\sin mt = \sin \left( 2m \frac t 2 \right) = \left( \sin \frac t 2 \right) P_{2m-1} \left( \cos \frac t 2 \right) ,$$

whence, using the obvious fact that $\int _a ^b f \le \int _a ^b |f|$,

$$\left| \int \limits _0 ^u \frac {\sin mt} {\sin \frac t 2} \ \Bbb d t \right| = \left| \int \limits _0 ^u \frac {\left( \sin \frac t 2 \right) P_{2m-1} \left( \cos \frac t 2 \right)} {\sin \frac t 2} \ \Bbb d t \right| = \left| \int \limits _0 ^u P_{2m-1} \left( \cos \frac t 2 \right) \ \Bbb d t \right| \le \int \limits _0 ^u \left| P_{2m-1} \left( \cos \frac t 2 \right) \right| \ \Bbb d t \le \\ \sum _{k=0} ^{2m-1} |a_{2m-1,k}| \int \limits _0 ^u \left| \cos ^k \frac t 2 \right| \ \Bbb d t \le \sum _{k=0} ^{2m-1} |a_{2m-1,k}| \int \limits _0 ^u 1 \ \Bbb d t = \sum _{k=0} ^{2m-1} |a_{2m-1,k}| u \le \sum _{k=0} ^{2m-1} |a_{2m-1,k}| \pi$$

which is a constant independent of $u$.

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Of course, the much simpler approach would be to notice that the only point where the integrand might raise problems is $t=0$, but there

$$\lim _{t \to 0} \frac {\sin mt} {\sin \frac t 2} = 2m ,$$

so using the comparison test for improper integrals your integral must converge, i.e. be finite for all $0 \le u \le \pi$. Since the result of the integration is continuous with respect to $u$, and since $[0, \pi]$ is compact, Weiertrass's extreme value theorem tells us that your integral is bounded and attains its bounds.