find f(a,b) =∫_0 ^∞ ((cos(ax)cos(bx))/((x^2 +a^2 )(x^2 +b^2 )))dx with a>0 and b>0 2)calculate ∫_0 ^∞ ((cos(x)cos(2x))/((x^2 +1)(x^2 +4)))dx


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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 bymathmax by abdo last updated on 17/Aug/19
$1) we have f(a,b) =(1/2)∫_0 ^∞ ((cos(a+b)x+cos(a−b)x)/((x^2 +a^2 )(x^2 +b^(2)) ))dx =(1/4)∫_(−∞) ^(+∞) ((cos(a+b)x)/((x^2 +a^2 )(x^2 +b^2 )))dx +(1/4)∫_(−∞) ^(+∞) ((cos(a−b)x)/((x^2 +a^2 )(x^2 +b^2 )))dx ⇒ 4f(,b) =H +K H =Re( ∫_(−∞) ^(+∞) (e^(i(a+b)x) /((x^2 +a^2 )(x^2 +b^2 )))dx)let ϕ(z)=(e^(i(a+b)z) /((x^2 +a^2 )(x^2 +b^2 ))) ϕ(z) =(e^(i(a+b)z) /((x−ia)(x+ia)(x−ib)(x+ib))) so the poles of ϕ are +^− ia and +^− ib residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ {Res(ϕ,ia)+Res(ϕ,ib)} Res(ϕ,ia) =(e^(i(a+b)ia) /((2ia)(b^2 −a^2 ))) (ifa≠b) =(e^(−a(a+b)) /((2ia)(b^2 −a^2 ))) Res(ϕ,ib) = (e^(i(a+b)ib) /((2ib)(a^2 −b^2 ))) =(e^(−b(a+b)) /(2ib(a^2 −b^2 ))) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{ (e^(−a^2 −ab) /(2ia(b^2 −a^2 ))) +(e^(−b^2 −ab) /(2ib(a^2 −b^2 )))} =(π/(a^2 −b^2 )){ (e^(−b^2 −ab) /b)−(e^(−a^2 −ab) /a)} =H (the integral is real) for k we change b by −b we get K =(π/(a^2 −b^2 )){(e^(−b^2 +ab) /(−b))−(e^(−a^2 +ab) /a)} ⇒ f(a,b) =(π/(4(a^2 −b^2 ))){ ((e^(−b^2 −ab) −e^(−b^2 +ab) )/b)−((e^(−a^2 −ab) +e^(−a^2 +ab) )/a)} and we must study the case a=b... 2)∫_0 ^∞ ((cosx cos(2x))/((x^2 +1)(x^2 +4)))dx =f(1,2) =(π/(4(−3))){ ((e^(−6) −e^(−2) )/2) −((e^(−3) +e^1 )/1)} =−(π/(12)){((e^(−6) −e^(−2) −2e^(−3) −2e)/2)} =(π/(24)){2e^(−3) +e^(−2) +2e −e^(−6) } .$
$1) w e h a v e f (a, b) = \frac{1}{2} \int_{0}^{\infty} \frac{c o s (a + b) x + c o s (a - b) x}{(x^{2} + a^{2}) (x^{2} + b^{2)}} d x$ $= \frac{1}{4} \int_{- \infty}^{+ \infty} \frac{c o s (a + b) x}{(x^{2} + a^{2}) (x^{2} + b^{2})} d x + \frac{1}{4} \int_{- \infty}^{+ \infty} \frac{c o s (a - b) x}{(x^{2} + a^{2}) (x^{2} + b^{2})} d x \Rightarrow$ $4 f (, b) = H + K$ $H = R e (\int_{- \infty}^{+ \infty} \frac{e^{i (a + b) x}}{(x^{2} + a^{2}) (x^{2} + b^{2})} d x) l e t φ (z) = \frac{e^{i (a + b) z}}{(x^{2} + a^{2}) (x^{2} + b^{2})}$ $φ (z) = \frac{e^{i (a + b) z}}{(x - i a) (x + i a) (x - i b) (x + i b)} s o t h e p o l e s o f φ a r e$ $\bar{+} i a a n d \bar{+} i b r e s i d u s t h e o r e m g i v e$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {R e s (φ, i a) + R e s (φ, i b)}$ $R e s (φ, i a) = \frac{e^{i (a + b) i a}}{(2 i a) (b^{2} - a^{2})} (i f a \neq b)$ $= \frac{e^{- a (a + b)}}{(2 i a) (b^{2} - a^{2})}$ $R e s (φ, i b) = \frac{e^{i (a + b) i b}}{(2 i b) (a^{2} - b^{2})} = \frac{e^{- b (a + b)}}{2 i b (a^{2} - b^{2})} \Rightarrow$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {\frac{e^{- a^{2} - a b}}{2 i a (b^{2} - a^{2})} + \frac{e^{- b^{2} - a b}}{2 i b (a^{2} - b^{2})}}$ $= \frac{π}{a^{2} - b^{2}} {\frac{e^{- b^{2} - a b}}{b} - \frac{e^{- a^{2} - a b}}{a}} = H (t h e i n t e g r a l i s r e a l) f o r k$ $w e c h a n g e b b y - b w e g e t K = \frac{π}{a^{2} - b^{2}} {\frac{e^{- b^{2} + a b}}{- b} - \frac{e^{- a^{2} + a b}}{a}} \Rightarrow$ $f (a, b) = \frac{π}{4 (a^{2} - b^{2})} {\frac{e^{- b^{2} - a b} - e^{- b^{2} + a b}}{b} - \frac{e^{- a^{2} - a b} + e^{- a^{2} + a b}}{a}}$ $a n d w e m u s t s t u d y t h e c a s e a = b . . .$ $2) \int_{0}^{\infty} \frac{c o s x c o s (2 x)}{(x^{2} + 1) (x^{2} + 4)} d x = f (1, 2)$ $= \frac{π}{4 (- 3)} {\frac{e^{- 6} - e^{- 2}}{2} - \frac{e^{- 3} + e^{1}}{1}} = - \frac{π}{12} {\frac{e^{- 6} - e^{- 2} - 2 e^{- 3} - 2 e}{2}}$ $= \frac{π}{24} {2 e^{- 3} + e^{- 2} + 2 e - e^{- 6}} .$
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