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0-sin-x-sin-x-x-2-dx-




Question Number 130417 by Lordose last updated on 25/Jan/21
∫_0 ^( ∞) ((sin(αx)sin(βx))/x^2 )dx
0sin(αx)sin(βx)x2dx
Answered by mathmax by abdo last updated on 25/Jan/21
I =∫_0 ^∞  ((sin(αx)sin(βx))/x^2 )dx ⇒I=_(αx=t)   ∫_0 ^∞  ((sintsin((β/α)t))/(t^2 /α^2 ))×(dt/α)  =∫_0 ^∞    ((sin(t)sin((β/α)t))/t^2 )dt (we suppose α>0 and β>0 )let λ=(β/α) ⇒  I =∫_0 ^∞  ((sintsin(λt))/t^2 )dt  by parts we get   I =[−(1/t)sint sin(λt)]_0 ^∞  +∫_0 ^∞ (1/t)(costsin(λt)+λsint cos(λt))dt  =∫_0 ^∞  ((costsin(λt))/t)dt +λ∫_0 ^∞  ((sintcos(λt))/t)dt =u(λ)+λv(λ)  we have sin(x+y)=sinx cosy +cosx siny  sin(x−y)=sinxcosy−cosxsiny ⇒sinx cosy =(1/2)(sin(x+h)+sin(x−y)) ⇒  u(λ)=∫_0 ^∞ (1/t)(sin(λ+1)t +sin(λ−1)t)dt  =∫_0 ^∞  ((sin(λ+1)t)/t) +∫_0 ^∞  ((sin(λ−1)t)/t)dt  we know λ>0 ⇒  ∫_0 ^∞  ((sin(λ+1)t)/t)dt =_((λ+1)t=y) (λ+1) ∫_0 ^∞  ((siny)/y)(dy/(λ+1)) =∫_0 ^∞  ((siny)/y)dy=(π/2)  if λ>1 ∫_0 ^∞  ((sin(λ−1)t)/t)dt =(π/2) ⇒I=(π/2)+λ(π/2)=(π/2)(λ+1)=(π/2)((β/α)+1)  if 0<λ<1 ⇒∫_0 ^∞  ((sin(λ−1)t)/t)dt =−∫_0 ^∞ ((sin(1−λ)t)/t)dt=−(π/2) ⇒  I =(π/2)−λ(π/2)=(π/2)(1−λ) =(π/2)(1−(β/α))
I=0sin(αx)sin(βx)x2dxI=αx=t0sintsin(βαt)t2α2×dtα=0sin(t)sin(βαt)t2dt(wesupposeα>0andβ>0)letλ=βαI=0sintsin(λt)t2dtbypartswegetI=[1tsintsin(λt)]0+01t(costsin(λt)+λsintcos(λt))dt=0costsin(λt)tdt+λ0sintcos(λt)tdt=u(λ)+λv(λ)wehavesin(x+y)=sinxcosy+cosxsinysin(xy)=sinxcosycosxsinysinxcosy=12(sin(x+h)+sin(xy))u(λ)=01t(sin(λ+1)t+sin(λ1)t)dt=0sin(λ+1)tt+0sin(λ1)ttdtweknowλ>00sin(λ+1)ttdt=(λ+1)t=y(λ+1)0sinyydyλ+1=0sinyydy=π2ifλ>10sin(λ1)ttdt=π2I=π2+λπ2=π2(λ+1)=π2(βα+1)if0<λ<10sin(λ1)ttdt=0sin(1λ)ttdt=π2I=π2λπ2=π2(1λ)=π2(1βα)
Commented by mathmax by abdo last updated on 25/Jan/21
sorry I =α∫_0 ^∞ ((sintsin(λt))/t^2 ) ⇒  I =(π/2)(α+β) or I =(π/2)(α−β)....
sorryI=α0sintsin(λt)t2I=π2(α+β)orI=π2(αβ).

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