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.