find the value of ∫_0 ^∞ ((x^2 −1)/(x^2 +1)) ((sin(x))/x)dx


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Question Number 36204 by prof Abdo imad last updated on 30/May/18

$f i n d t h e v a l u e o f \int_{0}^{\infty} \frac{x^{2} - 1}{x^{2} + 1} \frac{s i n (x)}{x} d x$
Commented by math khazana by abdo last updated on 17/Aug/18
$let I = ∫_0 ^∞ ((x^2 −1)/(x^2 +1)) ((sinx)/x) dx we hsve2I = ∫_(−∞) ^(+∞) ((x^2 −1)/(x^2 +1)) ((sinx)/x)?dx =Im( ∫_(−∞) ^(+∞) ((x^2 −1)/(x(x^2 +1))) e^(ix) dx) let?consider the complex function ϕ(z) = ((z^2 −1)/(z(z^2 +1))) e^(iz) the poles of ϕ are 0,i and −i ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{ Res(ϕ,0)+Res(ϕ,i)} but ϕ(z) = (((z^2 −1)e^(iz) )/(z(z−i)(z+i))) ⇒ Res(ϕ,0) =lim_(z→0) zϕ(z) = (1/1) =1 Res(ϕ,i) = lim_(z→i) (z−i)ϕ(z) =((−2 e^(−1) )/(i(2i))) = e^(−1) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{1+e^(−1) } ⇒ 2I =Im( ∫_(−∞) ^(+∞) ϕ(z)dz) =2π{1+e^(−1) } ⇒ I = π +(π/e) .$
$l e t I = \int_{0}^{\infty} \frac{x^{2} - 1}{x^{2} + 1} \frac{s i n x}{x} d x$ $w e h s v e 2 I = \int_{- \infty}^{+ \infty} \frac{x^{2} - 1}{x^{2} + 1} \frac{s i n x}{x} ? d x$ $= I m (\int_{- \infty}^{+ \infty} \frac{x^{2} - 1}{x (x^{2} + 1)} e^{i x} d x) l e t ? c o n s i d e r t h e$ $c o m p l e x f u n c t i o n φ (z) = \frac{z^{2} - 1}{z (z^{2} + 1)} e^{i z}$ $t h e p o l e s o f φ a r e 0, i a n d - i$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {R e s (φ, 0) + R e s (φ, i)}$ $b u t φ (z) = \frac{(z^{2} - 1) e^{i z}}{z (z - i) (z + i)} \Rightarrow$ $R e s (φ, 0) = {l i m}_{z \to 0} z φ (z) = \frac{1}{1} = 1$ $R e s (φ, i) = {l i m}_{z \to i} (z - i) φ (z) = \frac{- 2 e^{- 1}}{i (2 i)}$ $= e^{- 1} \Rightarrow \int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {1 + e^{- 1}} \Rightarrow$ $2 I = I m (\int_{- \infty}^{+ \infty} φ (z) d z) = 2 π {1 + e^{- 1}} \Rightarrow$ $I = π + \frac{π}{e} .$
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