let give 0≤x≤1 calculate ∫_0 ^∞ ((arctan((x/t)))/(1+t^2 )) dt


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 32737 by caravan msup abdo. last updated on 01/Apr/18

$l e t g i v e 0 ⩽ x ⩽ 1 c a l c u l a t e \int_{0}^{\infty} \frac{a r c t a n (\frac{x}{t})}{1 + t^{2}} d t$
	Answered by hknkrc46 last updated on 09/Apr/18

	$t = x c o t u \Rightarrow d t = - {x c s c}^{2} u d u \land (t \to 0 \Rightarrow u \to \frac{π}{2} \land t \to \infty \Rightarrow u \to 0)$ $\int_{\frac{π}{2}}^{0} \frac{- {u x c s c}^{2} u}{1 + x^{2} {c o t}^{2} u} d u = \int_{\frac{π}{2}}^{0} \frac{- u x}{{s i n}^{2} u + x^{2} {c o s}^{2} u} d u = \int_{\frac{π}{2}}^{0} \frac{- u x}{\frac{1 - c o s 2 u + x^{2} (1 + c o s 2 u)}{2}} d u$ $\int_{\frac{π}{2}}^{0} \frac{- 2 u x}{(x^{2} - 1) c o s 2 u + x^{2} + 1} d u = \frac{- 2 x}{x^{2} - 1} \int_{\frac{π}{2}}^{0} \frac{u}{c o s 2 u + \frac{x^{2} + 1}{x^{2} - 1}} d u$ $= \frac{- 2 x}{x^{2} - 1} \int_{\frac{π}{2}}^{0} \frac{u}{2 {c o s}^{2} u + \frac{2}{x^{2} - 1}} d u = \frac{- x}{x^{2} - 1} \int_{\frac{π}{2}}^{0} \frac{u}{\frac{1}{x^{2} - 1} + {c o s}^{2} u} d u$ $I D O N O T K N O W : ($
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum