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Question

Demostrar que la función de $\csc(x/2)-2/x$ es integrable en a $(0,\pi)$. De hecho, demostrar que es acotada. De hecho, probar que tiende a cero, como se $x\to0$. Use esto para mostrar que $$\lim_{N\to\infty}\int_0^\pi\left(\frac1{\sin\frac{x}2}-\frac2x\right)\sin\left((N+\frac12)x\right)dx=0$$ Entonces demostrar que $$\lim_{N\to\infty}\int_0^\pi\frac{\sin(N+\frac12)x}xdx=\pi/2$$ Finally, prove that $$\int_0^\infty\frac{\sin x}xdx=\pi/2$$

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Una observación es $\sum_{n=-N}^Ne^{inx}=\frac{\sin(N+\frac12)x}{\sin\frac{x}2}$, y la integral de esta de$0$$2\pi$$2\pi$, ya que la integral de la $\int_{0}^{2\pi}e^{inx}dx=0$ si $n\neq0$.

Answer 1

En primer lugar observamos que como $x \to 0^{+}$ la función $$f(x) = \dfrac{1}{\sin\left(\dfrac{x}{2}\right)} - \frac{2}{x}$$ tends to a definite limit $0$ and hence can be assumed continuous in $[0, \pi]$. Therefore $f(x)$ is Riemann-integrable on interval $[0, \pi]$. It now follows by Riemann-Lebesgue Lemma (related to coefficients of Fourier series of $f(x)$, a proof is available in Tom M. Apostol's "Mathematical Analysis", 2nd Ed., Page 313) that $$\lim_{N \to \infty}\int_{0}^{\pi}f(x)\sin(Nx + b)\,dx = 0$$ This settles the hard part of the question. The limit of $f(x)$ as $x \to 0^{+}$ se calcula como sigue:

$\displaystyle \begin{aligned}\lim_{x \to 0^{+}}f(x) &= \lim_{x \to 0^{+}}\dfrac{1}{\sin\left(\dfrac{x}{2}\right)} - \frac{2}{x}\\ &= \lim_{x \to 0^{+}}\dfrac{x - 2\sin(x/2)}{x\sin(x/2)}\\ &= \lim_{x \to 0^{+}}\dfrac{x - 2\sin(x/2)}{x\dfrac{\sin(x/2)}{x/2}\cdot(x/2)}\\ &= 2\lim_{x \to 0^{+}}\dfrac{x - 2\sin(x/2)}{x^{2}}\\ &= 2\lim_{x \to 0^{+}}\dfrac{1 - \cos(x/2)}{2x}\text{ (by L'Hospital's Rule)}\\ &= \lim_{x \to 0^{+}}\dfrac{2\sin^{2}(x/4)}{x}\\ &= 2\lim_{x \to 0^{+}}\dfrac{\sin^{2}(x/4)}{(x/4)^{2}}\cdot\frac{(x/4)^{2}}{x} = 0\end{aligned}$

La observación que se han hecho acerca de la representación de $\sin(N + 1/2)x$ como una suma podemos ayudar en mostrar que $$\int_{0}^{\pi}\dfrac{\sin\left(N + \dfrac{1}{2}\right)x}{\sin(x/2)} = \pi$$ Using this result together with the earlier established limit $$\lim_{N \to \infty}\int_{0}^{\pi}\left(\dfrac{1}{\sin(x/2)} - \frac{2}{x}\right)\sin\left(N + \frac{1}{2}\right)x\,dx = 0$$ gives us $$\lim_{N \to \infty}\int_{0}^{\pi}\dfrac{\sin\left(N + \dfrac{1}{2}\right)x}{x}\,dx = \frac{\pi}{2}$$The last part of the question can be easily deduced by putting $(N + 1/2)x = t$ en la integral anterior.