find-0-x-2-1-x-2-3-x-2-dx-

Question Number 29855 by abdo imad last updated on 13/Feb/18

f i n d \int_{0}^{\infty} \frac{x^{2}}{(1 + x^{2}) (3 + x^{2})} d x .

Commented by abdo imad last updated on 13/Feb/18

l e t p u t I = \int_{0}^{\infty} \frac{x^{2}}{(1 + x^{2}) (3 + x^{2})}

I = \frac{1}{2} \int_{- \infty}^{+ \infty} \frac{x^{2}}{(1 + x^{2}) (3 + 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{z^{2}}{(1 + z^{2}) (3 + z^{2})} t h e p o l e s o f f a r e i, - i

\sqrt{3} i a n d - \sqrt{3} i a n d

\int_{- \infty}^{+ \infty} f (z) d z = 2 i π (R e s (f, i) + R e s (f, i \sqrt{3}))

R e s (f, i) = {l i m}_{z \to i} (z - i) f (z) = \frac{- 1}{(2 i) (3 - 1)} = \frac{- 1}{4 i}

R e s (f, i \sqrt{3}) = {l i m}_{z \to i \sqrt{3}} (z - i \sqrt{3}) f (z) = \frac{- 3}{2 i \sqrt{3} (- 2)} = \frac{3}{4 i \sqrt{3}}

\int_{- \infty}^{+ \infty} f (z) d z = 2 i π (\frac{- 1}{4 i} + \frac{3}{4 i \sqrt{3}}) = - \frac{π}{2} + \frac{3 π}{2 \sqrt{3}}

= \frac{π}{2} (\sqrt{3} - 1) \Rightarrow I = \frac{π}{4} (\sqrt{3} - 1) .

Answered by ajfour last updated on 13/Feb/18

= - \frac{1}{2} \int \frac{d x}{1 + x^{2}} + \frac{3}{2} \int \frac{d x}{3 + x^{2}}

= - \frac{\tan^{- 1} x}{2} + \frac{\sqrt{3}}{2} \tan^{- 1} (\frac{x}{\sqrt{3}}) + c .

w i t h l i m i t s 0 t o \infty

I = - \frac{π}{4} + \frac{\sqrt{3} π}{4} = \frac{(\sqrt{3} - 1) π}{4} .

Commented by abdo imad last updated on 14/Feb/18

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