Question Number 70237 by mathmax by abdo last updated on 02/Oct/19

calculate ∫_0 ^∞ ((xsin(αx))/(1+x^4 ))dx with α real

Commented bymind is power last updated on 02/Oct/19

nice i deleat my post mistack[ ((zsin(z))/(1+z^4 )) when ∣z∣→∞ is not defind z=re^(ia) a∈[0,π[ sin(z)=((e^(ire^(ia) ) −e^(ire^(ia) ) )/(2i)) the factor e^(−ire^(ia) ) =e^(−irvos(a)) .e^(rsin(a)) ∣e^(−ire^(ia) ) ∣=e^(rsin(a)) witch is get big as we want ⇒lim zf(z)⇏0 jordan lemma dont not applie ∫_(−∞) ^(+∞) f(z)dz≠2iπresid(.....) i have done it quickly without seeing what going on Sorry sir

Commented bymathmax by abdo last updated on 02/Oct/19

let A =∫_0 ^∞ ((xsin(αx))/(x^4 +1))dx ⇒2A =∫_(−∞) ^(+∞) ((xsin(αx))/(x^4 +1))dx =Im(∫_(−∞) ^(+∞) ((x e^(iαx) )/(x^4 +1))dx) let ϕ(z)=((z e^(iαz) )/(z^4 +1)) poles of ϕ? ϕ(z) =((z e^(iαz) )/((z^2 −i)(z^2 +i))) =((z e^(iαz) )/((z−(√i))(z+(√i))(z−(√(−i)))(z+(√(−i))))) =((z e^(iαz) )/((z−e^((iπ)/4) )(z+e^((iπ)/4) )(z−e^(−((iπ)/4)) )(z+e^(−((iπ)/4)) ))) so the poles of ϕ are +^− e^((iπ)/4) and +^− e^(−((iπ)/4)) and ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{Res(ϕ,e^((iπ)/4) )+Res(ϕ,−e^(−((iπ)/4)) )} Res(ϕ,e^((iπ)/4) )=lim_(z→e^((iπ)/4) ) (z−e^((iπ)/4) )ϕ(z) =((e^((iπ)/4) e^(iαe^((iπ)/4) ) )/(2e^((iπ)/4) (2i))) =(e^(iα((1/((√2) ))+(i/(√2)))) /(4i)) =(1/(4i)) e^((iα)/(√2)) e^(−(α/(√2))) Res(ϕ,−e^(−((iπ)/4)) ) =lim_(z→−e^(−((iπ)/4)) ) (z+e^(−((iπ)/4)) )ϕ(z) =((−e^(−((iπ)/4)) e^(iα(−e^(−((iπ)/4)) )) )/((−2i)(−2e^(−((iπ)/4)) ))) =(1/(4i)) e^(−iα((1/(√2))−(i/(√2)))) =−(1/(4i)) e^(−((iα)/(√( 2)))) e^(−(α/(√2))) ⇒∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{(1/(4i)) e^(−(α/(√2))) e^((iα)/(√2)) −(1/(4i)) e^(−(α/(√2))) e^(−((iα)/(√( 2)))) } =(π/2) e^(−(α/(√2))) { 2i sin((α/(√2)))} =iπ e^(−(α/(√2))) sin((α/(√2))) ⇒ 2A =π e^(−(α/(√2))) sin((α/(√2))) ⇒ A =(π/2) e^(−(α/(√2))) sin((α/(√2))) .

Commented bymathmax by abdo last updated on 03/Oct/19

nevermind sir.