Riemann-Lebesgue lemma.

Virgil Nicula
Mesaje: 244
Membru din: Sâm Oct 30, 2010 3:55 pm
Localitate: Bradenton, Florida

Riemann-Lebesgue lemma.

Mesaj de Virgil Nicula »

Exercise. If the function $$g$$ is continuous and $$0<T$$-periodic on $$[0,\infty)$$ , then $$\lim_{n \to \infty} \int^{b}_{a} g(nx) \ \mathrm{dx} = \frac{b-a}{T} \cdot\int^{T}_{0} g(x) \ \mathrm{dx}$$ .

Generalization (Riemann-Lebesgue lemma). Let $$f$$ be a continuous function on $$0 \le a < b$$ . If $$g$$ is a continuous

and $$T$$-periodic on $$[0,\infty)$$ , then $$\lim_{n \to \infty} \int^{b}_{a} f(x) g(nx) \ \mathrm{dx} = \frac{1}{T} \cdot\int^{T}_{0} g(x) \ \mathrm{dx}\cdot \int_{a}^{b} f(x) \ \mathrm{dx} $$ .



Example 1. Ascertain $$\lim_{n\to\infty}f(n)$$ , where $$f(n)= \int_0^{\pi} e^{x}|\sin nx|\ \mathrm{dx}\ ,\ n\in\mathbb N^*$$ .

Example 2. For any continuous function $$f$$ following equality holds $$\lim_{n\to\infty}\int_a^b |sin(nx)|f(x)\ \mathrm{dx}dx=\frac{2}{\pi}\int_a^b f(x)\ \mathrm{dx}$$ .
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