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Question Number 77886 by mathmax by abdo last updated on 11/Jan/20

c a l c u l a t e \int_{0}^{\infty} \frac{a r c t a n (x^{2} + x^{- 2})}{x^{2} + a^{2}} d x w i t h a > 0

2) f i n d t h e v a l u e o f \int_{0}^{\infty} \frac{a r c t a n (x^{2} + x^{- 2})}{x^{2} + 1} d x

Commented by mathmax by abdo last updated on 12/Jan/20

1) l e t A = \int_{0}^{\infty} \frac{a r c t a n (x^{2} + \frac{1}{x^{2}})}{x^{2} + a^{2}} d x t h e c o n v e r g e n c e o f A i s a s d u r e d

b e c a u s e ∣ a r c t a n (x^{2} + x^{- 2}) ∣< \frac{π}{2} f o r x \neq 0

w e h a v e 2 A = \int_{- \infty}^{+ \infty} \frac{a r c t a n (x^{2} + x^{- 2})}{x^{2} + a^{2}} d x l e t W (z) = \frac{a r c t a n (z^{2} + z^{- 2})}{z^{2} + a^{2}}

\int_{- \infty}^{+ \infty} W (z) d z = 2 i π R e s (W, i a) = 2 i π \frac{∣ a r c t a n (- a^{2} - \frac{1}{a^{2}}) ∣}{2 i a}

= \frac{π}{a} a r c t a n (a^{2} + \frac{1}{a^{2}}) \Rightarrow A = \frac{π}{2 a} a r c t a n (a^{2} + \frac{1}{a^{2}})

2) a = 1 \Rightarrow \int_{0}^{\infty} \frac{a r c t a n (x^{2} + x^{- 2})}{x^{2} + 1} d x = \frac{π}{2} a r c t a n (2) .