find-the-value-of-0-dx-1-x-2-a-2-x-2-2-find-the-value-of-A-0-dx-1-x-2-x-2-1-sin-2-0-lt-lt-pi-2-

Question Number 33341 by prof Abdo imad last updated on 14/Apr/18

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

2) f i n d t h e v a l u e o f

A (θ) = \int_{0}^{\infty} \frac{d x}{(1 + x^{2}) (x^{2} + 1 - {s i n}^{2} θ)}

0 < θ < \frac{π}{2} .

Commented by prof Abdo imad last updated on 18/Apr/18

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

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

t h e c o m p l e x f u n c t i o n φ (z) = \frac{1}{(z^{2} + 1) (z^{2} + a^{2})}

t h e p o l e s o f φ i s i, - i, i a, - i a

c a s e 1 a > 0

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π (R e s (φ, i) + R e s (φ, i a)) b u t

φ (z) = \frac{1}{(z - i) (z + i) (z - i a) (z + i a)} \Rightarrow

R e s (φ, i) = \frac{1}{(2 i) (a^{2} - 1)}

R e s (φ, i a) = \frac{1}{(2 i a) (1 - a^{2})} \Rightarrow

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π (\frac{1}{(- 2 i) (1 - a^{2})} + \frac{1}{(2 i a) (1 - a^{2})})

= \frac{2 π}{1 - a^{2}} (\frac{1}{2 a} - \frac{1}{2}) = \frac{π}{1 - a^{2}} (\frac{1}{a} - 1)

= \frac{π}{1 - a^{2}} \frac{1 - a}{a} = \frac{π}{a (1 + a)} \Rightarrow I = \frac{π}{2 a (1 + a)}

c a s e 2 a < 0

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

R e s (φ, - i a) = \frac{1}{(- 2 i a) (1 - a^{2})} \Rightarrow

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π (\frac{- 1}{2 i (1 - a^{2})} - \frac{1}{2 i a (1 - a^{2})})

= \frac{- π}{1 - a^{2}} (1 + \frac{1}{a}) = \frac{- π (1 + a)}{a (1 - a^{2})} = \frac{- π}{a (1 - a)}

= \frac{π}{a (a - 1)} \Rightarrow I = \frac{π}{2 a (a - 1)} a n d w e m u s t

s t u d y t h e s p e c i a l c a s e a = \bar{+} 1 .

Commented by prof Abdo imad last updated on 18/Apr/18

2) w e h a v e A (θ) = \int_{0}^{\infty} \frac{d x}{(1 + x^{2}) (x^{2} + {c o s}^{2} θ)}

= f (c o s θ) = \frac{π}{2 c o s θ (1 + c o s θ)} .