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Question Number 66466 by mathmax by abdo last updated on 15/Aug/19

f i n d f (a, b) = \int_{0}^{\infty} \frac{c o s (a x) c o s (b x)}{(x^{2} + a^{2}) (x^{2} + b^{2})} d x w i t h a > 0 a n d b > 0

2) c a l c u l a t e \int_{0}^{\infty} \frac{c o s (x) c o s (2 x)}{(x^{2} + 1) (x^{2} + 4)} d x

Commented by mathmax by abdo last updated on 17/Aug/19

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

= \frac{1}{4} \int_{- \infty}^{+ \infty} \frac{c o s (a + b) x}{(x^{2} + a^{2}) (x^{2} + b^{2})} d x + \frac{1}{4} \int_{- \infty}^{+ \infty} \frac{c o s (a - b) x}{(x^{2} + a^{2}) (x^{2} + b^{2})} d x \Rightarrow

4 f (, b) = H + K

H = R e (\int_{- \infty}^{+ \infty} \frac{e^{i (a + b) x}}{(x^{2} + a^{2}) (x^{2} + b^{2})} d x) l e t φ (z) = \frac{e^{i (a + b) z}}{(x^{2} + a^{2}) (x^{2} + b^{2})}

φ (z) = \frac{e^{i (a + b) z}}{(x - i a) (x + i a) (x - i b) (x + i b)} s o t h e p o l e s o f φ a r e

\bar{+} i a a n d \bar{+} i b r e s i d u s t h e o r e m g i v e

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

R e s (φ, i a) = \frac{e^{i (a + b) i a}}{(2 i a) (b^{2} - a^{2})} (i f a \neq b)

= \frac{e^{- a (a + b)}}{(2 i a) (b^{2} - a^{2})}

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

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

= \frac{π}{a^{2} - b^{2}} {\frac{e^{- b^{2} - a b}}{b} - \frac{e^{- a^{2} - a b}}{a}} = H (t h e i n t e g r a l i s r e a l) f o r k

w e c h a n g e b b y - b w e g e t K = \frac{π}{a^{2} - b^{2}} {\frac{e^{- b^{2} + a b}}{- b} - \frac{e^{- a^{2} + a b}}{a}} \Rightarrow

f (a, b) = \frac{π}{4 (a^{2} - b^{2})} {\frac{e^{- b^{2} - a b} - e^{- b^{2} + a b}}{b} - \frac{e^{- a^{2} - a b} + e^{- a^{2} + a b}}{a}}

a n d w e m u s t s t u d y t h e c a s e a = b \dots

2) \int_{0}^{\infty} \frac{c o s x c o s (2 x)}{(x^{2} + 1) (x^{2} + 4)} d x = f (1, 2)

= \frac{π}{4 (- 3)} {\frac{e^{- 6} - e^{- 2}}{2} - \frac{e^{- 3} + e^{1}}{1}} = - \frac{π}{12} {\frac{e^{- 6} - e^{- 2} - 2 e^{- 3} - 2 e}{2}}

= \frac{π}{24} {2 e^{- 3} + e^{- 2} + 2 e - e^{- 6}} .