# Problem of the Week #217 - Jul 26, 2016

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#### Euge

##### MHB Global Moderator
Staff member
Here is this week's POTW:

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Prove

$$\lim_{n\to \infty} \int_0^{\pi/(2n)} \frac{\sin 2nx}{\sin x}\, dx = \int_0^\pi \frac{\sin x}{x}\, dx.$$

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#### Euge

##### MHB Global Moderator
Staff member
No one answered this week's problem. You can read my solution below.

I'll prove this in two ways: (1) by Lebesgue integration and (2) by Riemann integration.

(1) By a change of variable $y = 2n x$, I write

$$\int_0^{\pi/2n} \frac{\sin 2n x}{\sin x}\, dx = \int_0^\pi \frac{\sin y}{2n\sin \frac{y}{2n}}\, dy\tag{*}$$

Since $\sin x \ge \dfrac{2x}{\pi}$ for $0 \le x \le \pi/2$, then $2n \sin \dfrac{y}{2n} \ge \dfrac2\pi y$ for $0 \le y \le \pi$. Thus, the integrand of the right-hand side of (*) is bounded by $\dfrac{\pi}{2}\cdot\dfrac{\sin y}{y}$ on $(0,\pi]$. Furthermore, the integrand converges pointwise to $\dfrac{\sin y}{y}$ over $(0,\pi]$. Hence, by the dominated convergence theorem, the integral converges to $\int_0^\pi \frac{\sin y}{y}\, dy$, as desired.

(2) Note that for every $x$,

$$\sin 2n x = \sum_{k = 1}^n (\sin 2kx - \sin\, [2(k-1)x]) =\sum_{k = 1}^n 2\cos\, [(2k-1)x] \sin x.$$

Thus

$$\int_0^{\pi/2n} \frac{\sin 2nx}{\sin x}\, dx = \int_0^{\pi/2n} \sum_{k = 1}^n 2\cos\,[(2k-1)x]\, dx = \int_0^\pi \frac{1}{n}\sum_{k = 1}^n \cos\, \left[\frac{2k-1}{2n}x\right]\, dx.\tag{**}$$

The sequence of continuous functions

$$f_n(x) = \frac{1}{n}\sum_{k = 1}^n \cos \left[\frac{2k-1}{2n}x\right] \quad (n = 1, 2,3,\ldots)$$

increase monotonically, on the compact interval $[0,\pi]$, to the function $F(x)$, where $F(0) = 1$ and for $x > 0$, $$F(x) = \frac{1}{x}\int_0^x \cos t\, dt = \frac{\sin x}{x}.$$

By Dini's theorem, $f_n$ converges uniformly to $F$ on $[0,\pi]$. Hence,

$$\lim_{n\to \infty} \int_0^\pi f_n(x)\, dx = \int_0^\pi \lim_{n\to \infty} f_n(x)\, dx = \int_0^\pi \frac{\sin x}{x}\, dx,$$

and consequently by (**),

$$\lim_{n\to \infty} \int_0^{\pi/2n} \frac{\sin 2nx}{\sin x}\, dx = \int_0^\pi \frac{\sin x}{x}\, dx.$$

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