calculate ∫_0 ^∞ ((x sin(((πx)/2)))/({1+(x+1)^2 }{1+(x−1)^2 }))dx


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Question Number 43058 by maxmathsup by imad last updated on 06/Sep/18
$calculate ∫_0 ^∞ ((x sin(((πx)/2)))/({1+(x+1)^2 }{1+(x−1)^2 }))dx$
$c a l c u l a t e \int_{0}^{\infty} \frac{x s i n (\frac{π x}{2})}{{1 + {(x + 1)}^{2}} {1 + {(x - 1)}^{2}}} d x$
Commented by maxmathsup by imad last updated on 09/Sep/18
$2I = ∫_(−∞) ^(+∞) ((xsin(((πx)/2)))/({1+(x+1)^2 }{1+(x−1)^2 }))dx=Im (∫_(−∞) ^(+∞) ((x e^((iπx)/2) )/({1+(x+1)^2 }{1+(x−1)^2 }))dx) let ϕ(z) = ((z e^((iπz)/2) )/({1+(z+1)^2 }{1+(z−1)^2 })) we have ϕ(z) = ((z e^((iπz)/2) )/({1+z^2 +2z+1}{1+z^2 −2z +1})) = ((z e^((iπz)/2) )/({z^2 +2z+2}{z^2 −2z +2})) poles of ϕ ? roots of z^2 +2z +2→ Δ^′ = 1−2 =−1 ⇒ z_1 =−1+i and z_2 =−1−i roots of z^2 −2z +2 → Δ^′ =1−2=−1 ⇒z_3 =1+i and z_4 =1−i ⇒ ϕ(z) = ((z e^((iπz)/2) )/((z−z_1 )(z−z_2 )(z−z_3 )(z−z_4 ))) = ((z e^((iπz)/2) )/((z−z_1 )(z−z_2 )(z+z_2 )(z+z_1 ))) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ { Res(ϕ,z_1 )+Res(ϕ,z_3 )} Res(ϕ,z_1 ) =lim_(z→z_1 ) (z−z_1 )ϕ(z) = ((z_1 e^((iπz_1 )/2) )/((z_1 −z_2 )(z_1 +z_2 )(2z_1 ))) = (e^((iπz_1 )/2) /(2(2i)(−2))) = (e^((iπz_1 )/2) /(−8i)) Res(ϕ, z_3 ) = Res(ϕ, −z_2 ) =lim_(z→−z_2 ) (z+z_2 )ϕ(z) = ((−z_2 e^((−iπz_2 )/2) )/((−z_2 −z_1 )(−2z_2 )(z_1 −z_2 ))) = (e^(−((iπz_2 )/2)) /(−2(z_1 +z_2 )(z_1 −z_2 ))) = (e^(−((iπz_2 )/2)) /((−2)(−2)(2i))) = (e^(−((iπz_2 )/2)) /(8i)) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{ −(e^((iπz_1 )/2) /(8i)) + (e^(−((iπz_2 )/2)) /(8i))}=−(π/4){ e^((iπz_1 )/2) −e^(−((iπ z_1 ^− )/2)) } =−(π/4) {2i Im(e^((iπz_1 )/2) )} =−((iπ)/2) Im( e^((iπz_1 )/2) ) but e^((iπz_1 )/2) =e^((iπ(−1+i))/2) = e^((−iπ−π)/2) = e^(−(π/2)) (−i) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =−((iπ)/2) (−e^(−(π/2)) ) = ((iπ)/2) e^(−(π/2)) ⇒ 2I = (π/2) e^(−(π/2)) ⇒ I = (π/4) e^(−(π/2)) .$
$2 I = \int_{- \infty}^{+ \infty} \frac{x s i n (\frac{π x}{2})}{{1 + {(x + 1)}^{2}} {1 + {(x - 1)}^{2}}} d x = I m (\int_{- \infty}^{+ \infty} \frac{x e^{\frac{i π x}{2}}}{{1 + {(x + 1)}^{2}} {1 + {(x - 1)}^{2}}} d x)$ $l e t φ (z) = \frac{z e^{\frac{i π z}{2}}}{{1 + {(z + 1)}^{2}} {1 + {(z - 1)}^{2}}} w e h a v e$ $φ (z) = \frac{z e^{\frac{i π z}{2}}}{{1 + z^{2} + 2 z + 1} {1 + z^{2} - 2 z + 1}} = \frac{z e^{\frac{i π z}{2}}}{{z^{2} + 2 z + 2} {z^{2} - 2 z + 2}}$ $p o l e s o f φ ?$ $r o o t s o f z^{2} + 2 z + 2 \to Δ^{^{'}} = 1 - 2 = - 1 \Rightarrow z_{1} = - 1 + i a n d z_{2} = - 1 - i$ $r o o t s o f z^{2} - 2 z + 2 \to Δ^{^{'}} = 1 - 2 = - 1 \Rightarrow z_{3} = 1 + i a n d z_{4} = 1 - i \Rightarrow$ $φ (z) = \frac{z e^{\frac{i π z}{2}}}{(z - z_{1}) (z - z_{2}) (z - z_{3}) (z - z_{4})} = \frac{z e^{\frac{i π z}{2}}}{(z - z_{1}) (z - z_{2}) (z + z_{2}) (z + z_{1})}$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {R e s (φ, z_{1}) + R e s (φ, z_{3})}$ $R e s (φ, z_{1}) = {l i m}_{z \to z_{1}} (z - z_{1}) φ (z) = \frac{z_{1} e^{\frac{i π z_{1}}{2}}}{(z_{1} - z_{2}) (z_{1} + z_{2}) (2 z_{1})}$ $= \frac{e^{\frac{i π z_{1}}{2}}}{2 (2 i) (- 2)} = \frac{e^{\frac{i π z_{1}}{2}}}{- 8 i}$ $R e s (φ, z_{3}) = R e s (φ, - z_{2}) = {l i m}_{z \to - z_{2}} (z + z_{2}) φ (z)$ $= \frac{- z_{2} e^{\frac{- i π z_{2}}{2}}}{(- z_{2} - z_{1}) (- 2 z_{2}) (z_{1} - z_{2})} = \frac{e^{- \frac{i π z_{2}}{2}}}{- 2 (z_{1} + z_{2}) (z_{1} - z_{2})}$ $= \frac{e^{- \frac{i π z_{2}}{2}}}{(- 2) (- 2) (2 i)} = \frac{e^{- \frac{i π z_{2}}{2}}}{8 i} \Rightarrow$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {- \frac{e^{\frac{i π z_{1}}{2}}}{8 i} + \frac{e^{- \frac{i π z_{2}}{2}}}{8 i}} = - \frac{π}{4} {e^{\frac{i π z_{1}}{2}} - e^{- \frac{i π {\bar{z}}_{1}}{2}}}$ $= - \frac{π}{4} {2 i I m (e^{\frac{i π z_{1}}{2}})} = - \frac{i π}{2} I m (e^{\frac{i π z_{1}}{2}}) b u t$ $e^{\frac{i π z_{1}}{2}} = e^{\frac{i π (- 1 + i)}{2}} = e^{\frac{- i π - π}{2}} = e^{- \frac{π}{2}} (- i) \Rightarrow \int_{- \infty}^{+ \infty} φ (z) d z = - \frac{i π}{2} (- e^{- \frac{π}{2}})$ $= \frac{i π}{2} e^{- \frac{π}{2}} \Rightarrow 2 I = \frac{π}{2} e^{- \frac{π}{2}} \Rightarrow I = \frac{π}{4} e^{- \frac{π}{2}} .$
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