let m>0 and 0<a<b 1) calculate ∫_0 ^∞ ((cos(mx))/((x^2 +a^2 )(x^2 +b^2 )))dx 2)find the value of ∫


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Question Number 35438 by prof Abdo imad last updated on 19/May/18

$l e t m > 0 a n d 0 < a < b$ $1) c a l c u l a t e \int_{0}^{\infty} \frac{c o s (m x)}{(x^{2} + a^{2}) (x^{2} + b^{2})} d x$ $2) f i n d t h e v a l u e o f \int_{0}^{\infty} \frac{c o s (2 x)}{(x^{2} + 1) (x^{2} + 3)} d x$
Commented by prof Abdo imad last updated on 21/May/18
$let put I = ∫_0 ^∞ ((cos(mx))/((x^2 +a^2 )(x^(2 ) +b^2 )))dx 2I = ∫_(−∞) ^(+∞) ((cos(mx))/((x^2 +a^2 )(x^2 +b^2 )))dx =Re( ∫_(−∞) ^(+∞) (e^(imx) /((x^2 +a^2 )(x^2 +b^2 )))dx) let consider the complex function ϕ(z) = (e^(imz) /((x^2 +a^2 )(x^2 +b^2 ))) ϕ(z) = (e^(imz) /((x−ia)(x+ia)(x−ib)(x+ib))) thepoles of ϕ are ia,−ia,ib,−ib ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ { Res(ϕ,ia) +Res(ϕ,ib)} Res(ϕ,ia) = (e^(im(ia)) /(2ia(b^2 −a^2 ))) = (e^(−ma) /(2ia(b^2 −a^2 ))) Res(ϕ,ib) = (e^(im(ib)) /(2ib(a^2 −b^2 ))) = (e^(−mb) /(2ib(a^2 −b^2 ))) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{ (e^(−ma) /(2ia(b^2 −a^2 ))) + (e^(−mb) /(2ib(a^2 −b^2 )))} = (π/a) (e^(−ma) /(b^2 −a^2 )) −(π/b) (e^(−mb) /(b^2 −a^2 )) = (π/(b^2 −a^2 )){ (e^(−ma) /a) − (e^(−mb) /b)} ⇒ I = (π/(2(a^2 −b^2 ))){ (e^(−mb) /b) −(e^(−ma) /a)} 2)let take m=2 , a =1, b=(√3) we get ∫_0 ^∞ ((cos(2x))/((x^2 +1)(x^2 +3))) = (π/(2(1 −3))){ (e^(−2(√3)) /(√3)) −(e^(−2) /1) } =−(π/4){ (e^(−2(√3)) /(√3)) − e^(−2) } = (π/4){ e^(−2) − (e^(−2(√3)) /(√3))}.$
$l e t p u t I = \int_{0}^{\infty} \frac{c o s (m x)}{(x^{2} + a^{2}) (x^{2} + b^{2})} d x$ $2 I = \int_{- \infty}^{+ \infty} \frac{c o s (m x)}{(x^{2} + a^{2}) (x^{2} + b^{2})} d x$ $= R e (\int_{- \infty}^{+ \infty} \frac{e^{i m x}}{(x^{2} + a^{2}) (x^{2} + b^{2})} d x) l e t c o n s i d e r$ $t h e c o m p l e x f u n c t i o n$ $φ (z) = \frac{e^{i m z}}{(x^{2} + a^{2}) (x^{2} + b^{2})}$ $φ (z) = \frac{e^{i m z}}{(x - i a) (x + i a) (x - i b) (x + i b)} t h e p o l e s$ $o f φ a r e i a, - i a, i b, - i b$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {R e s (φ, i a) + R e s (φ, i b)}$ $R e s (φ, i a) = \frac{e^{i m (i a)}}{2 i a (b^{2} - a^{2})} = \frac{e^{- m a}}{2 i a (b^{2} - a^{2})}$ $R e s (φ, i b) = \frac{e^{i m (i b)}}{2 i b (a^{2} - b^{2})} = \frac{e^{- m b}}{2 i b (a^{2} - b^{2})}$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {\frac{e^{- m a}}{2 i a (b^{2} - a^{2})} + \frac{e^{- m b}}{2 i b (a^{2} - b^{2})}}$ $= \frac{π}{a} \frac{e^{- m a}}{b^{2} - a^{2}} - \frac{π}{b} \frac{e^{- m b}}{b^{2} - a^{2}}$ $= \frac{π}{b^{2} - a^{2}} {\frac{e^{- m a}}{a} - \frac{e^{- m b}}{b}} \Rightarrow$ $I = \frac{π}{2 (a^{2} - b^{2})} {\frac{e^{- m b}}{b} - \frac{e^{- m a}}{a}}$ $2) l e t t a k e m = 2, a = 1, b = \sqrt{3} w e g e t$ $\int_{0}^{\infty} \frac{c o s (2 x)}{(x^{2} + 1) (x^{2} + 3)} = \frac{π}{2 (1 - 3)} {\frac{e^{- 2 \sqrt{3}}}{\sqrt{3}} - \frac{e^{- 2}}{1}}$ $= - \frac{π}{4} {\frac{e^{- 2 \sqrt{3}}}{\sqrt{3}} - e^{- 2}} = \frac{π}{4} {e^{- 2} - \frac{e^{- 2 \sqrt{3}}}{\sqrt{3}}} .$
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