let-give-0-x-1-calculate-0-arctan-x-t-1-t-2-dt-

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 : (

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