find-the-value-of-0-x-arctan-2x-2-x-2-2-dx-

Question Number 28035 by abdo imad last updated on 18/Jan/18

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

Commented by abdo imad last updated on 23/Jan/18

l e t i n t e g r a t r b y p a r t s

Missing \left or extra \right

= \int_{0}^{\infty} \frac{d x}{(2 + x^{2}) (1 + 4 x^{2})}

= \frac{1}{2} \int_{R} \frac{d x}{(2 + x^{2}) (1 + 4 x^{2})} l e t i n t r o d u c e

t h e c o m p l e x f u n c t i o n

f (z) = \frac{1}{(z^{2} + 2) (4 z^{2} + 1)} p o l e s o f f ?

f (z) = \frac{}{4 (z - \sqrt{2} i) (x + \sqrt{2} i) (z - \frac{i}{2}) (z + \frac{i}{2})}

t h e p o l e s o f f a r e \sqrt{2} i, - \sqrt{2} i, \frac{i}{2} a n d \frac{- i}{2}

\int_{R} f (x) d z = 2 i π (R e s (f, \sqrt{2} i) + R e s (f, \frac{i}{2}))

R e s (f, \sqrt{2} i) = \frac{1}{4 (2 \sqrt{2} i)) ({(\sqrt{2} i)}^{2} + \frac{1}{4})}

= \frac{1}{8 \sqrt{2} i (- 2 + \frac{1}{4})} = \frac{1}{8 \sqrt{2} i . \frac{- 7}{4}}

= \frac{- 1}{14 \sqrt{2} i}

R e s (f, \frac{i}{2}) = \frac{1}{4 (\frac{i}{2} - \sqrt{2} i) (\frac{i}{2} + \sqrt{2} i) i}

= \frac{1}{4 (- \frac{1}{4} + 2) i} = \frac{1}{7 i}

\int_{R}^{} f (z) d z = 2 i π (\frac{- 1}{14 \sqrt{2} i} + \frac{1}{7 i})

= \frac{- π}{7 \sqrt{2}} + \frac{2 π}{7} = \frac{2 π \sqrt{2} - π}{7 \sqrt{2}} .

Commented by abdo imad last updated on 23/Jan/18

I = \frac{1}{2} \int_{R} f (z) d z .