Let a_n = int_{-pi}^{pi} |x-1| cos(nx) , dx f | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) By the Riemann-Lebesgue Lemma,if $f(x)$ is an integrable function on $[a, b]$,then $\lim_{n \rightarrow \infty} \int_{a}^{b} f(x) \cos(nx) \, dx =

Vedclass · Accepted Answer

$\lim_{n \rightarrow \infty} a_n = 0$

Vedclass · Answer

(D) By the Riemann-Lebesgue Lemma,if $f(x)$ is an integrable function on $[a, b]$,then $\lim_{n \rightarrow \infty} \int_{a}^{b} f(x) \cos(nx) \, dx = 0$. Here,$f(x) = |x-1|$ is a continuous and therefore integrable function on the interval $[-\pi, \pi]$. Since $f(x) = |x-1|$ is integrable,the integral $a_n = \int_{-\pi}^{\pi} |x-1| \cos(nx) \, dx$ represents the Fourier cosine coefficient (up to a constant factor) of the function $f(x)$. According to the Riemann-Lebesgue Lemma,as $n \rightarrow \infty$,the integral of a function multiplied by $\cos(nx)$ approaches $0$. Therefore,$\lim_{n \rightarrow \infty} a_n = 0$.

Let $a_n = \int_{-\pi}^{\pi} |x-1| \cos(nx) \, dx$ for all natural numbers $n$. Then,the sequence $(a_n)_{n \geq 1}$ satisfies:

Similar Questions

For $n \in N$,let $P_n = \int_1^e (\ln x)^n dx$. Then $(P_{10} - 90P_8)$ is equal to

$\int_0^{\pi /2} \frac{\sqrt{\cot x}}{\sqrt{\cot x} + \sqrt{\tan x}} \, dx = $

Let $A = \int\limits_0^1 \frac{e^t}{1 + t} \, dt$. Then $\int\limits_{a - 1}^a \frac{e^{-t}}{t - a - 1} \, dt$ has the value:

Let $I=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{1}{2-\cos 2 x}\left(\frac{3}{\pi}+\log \left(\frac{4+\sin x}{4-\sin x}\right)\right) d x$. Given that $\int \frac{d x}{1+k x^2}=\frac{1}{\sqrt{k}} \tan ^{-1}(\sqrt{k} x)+c, \tan ^{-1}(0)=0$ and $\tan ^{-1}(\sqrt{3})=\frac{\pi}{3}$. Then $3 I^2=$

$\int_{\pi / 11}^{9 \pi / 22} \frac{d x}{1+\sqrt{\tan x}} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module