∫^(π/2) _0 (dx/(cox(x/2)∙cos(x/2^2 )∙∙∙∙∙cos(x/2^n )))=?


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 183320 by mathlove last updated on 25/Dec/22

${\int_{0}}^{\frac{π}{2}} \frac{d x}{c o x \frac{x}{2} \cdot c o s \frac{x}{2^{2}} \cdot \cdot \cdot \cdot \cdot c o s \frac{x}{2^{n}}} = ?$
	Answered by Vynho last updated on 28/Dec/22

	$\int_{0}^{\frac{π}{2}} \frac{d x}{\underset{k = 1}{\prod^{n}} c o s (\frac{x}{2^{k}})}$ $l n (\underset{k = 1}{\prod^{n}} c o s (\frac{x}{2^{k}})) = \underset{k = 1}{\sum^{n}} l n (c o s (\frac{x}{2^{k}}))$ $\underset{k = 1}{\prod^{n}} c o s (\frac{x}{a^{k}}) + i s i n (\frac{x}{a^{k}}) = \underset{k = 1}{\prod^{n}} e^{i \frac{x}{a^{k}}}$ $U_{n} = \underset{k = 1}{\prod^{n}} e^{i x {(\frac{1}{a})}^{k}}$ $l n (U_{n}) = \underset{k = 1}{\sum^{n}} l n (e^{i x {(\frac{1}{a})}^{k}}) = i x \underset{k = 1}{\sum^{n}} {(\frac{1}{a})}^{k}$ $l n (U_{n}) = i x \frac{1}{a} . \frac{1 - {(\frac{1}{a})}^{n}}{1 - \frac{1}{a}} = i x . \frac{1 - {(\frac{1}{a})}^{n}}{a - 1}$ $U_{n} = e^{i x \frac{1 - {(\frac{1}{a})}^{n}}{a - 1}}$ $\int_{0}^{\frac{π}{2}} e^{- i x \frac{1 - {(\frac{1}{a})}^{n}}{a - 1}} d x = \int_{0}^{\frac{π}{2}} e^{- i x θ} d x = \frac{1}{- i θ} {[e^{- i x θ}]}_{0}^{\frac{π}{2}}$ $I = \frac{i}{θ} [e^{- i \frac{π}{2} θ} - 1] = \frac{1}{θ} [i c o s (\frac{π}{2} θ) + s i n (\frac{π}{2} θ)]$ $\int_{0}^{\frac{π}{2}} \frac{d x}{\underset{k = 1}{\prod^{n}} c o s (\frac{x}{2^{k}})} = \frac{1}{θ} s i n (\frac{π}{2} θ)$ $θ = \frac{1 - {(\frac{1}{a})}^{n}}{a - 1}$
	Commented by Vynho last updated on 28/Dec/22

	$a = 2$
	Commented by mathlove last updated on 03/Jan/23

	$t h a n k s$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum