Menu Close

1-find-the-value-of-f-x-0-1-cos-xt-t-2-e-t-dt-2-calculate-0-1-cos-t-t-2-e-t-dt-

Tinku Tara 0 Comments
Pin




Question Number 35630 by abdo mathsup 649 cc last updated on 21/May/18
1) find the value of f(x)=∫_0 ^∞ ((1−cos(xt))/t^2 ) e^(−t) dt 2) calculate ∫_0 ^∞ ((1−cos(t))/t^2 ) e^(−t) dt .
1)findthevalueoff(x)=∫0∞1−cos(xt)t2e−tdt2)calculate∫0∞1−cos(t)t2e−tdt.
Commented by abdo mathsup 649 cc last updated on 24/May/18
by parts u^′ = (1/t^2 ) and v =(1−cos(xt))e^(−t) ⇒ v^′ = xsin(xt) e^(−t) −(1−cos(xt))e^(−t) =( x sin(xt) +cos(xt)−1) e^(−t) f(x) = [ −(1/t) (1−cos(xt))e^(−t) ]_0 ^(+∞) −∫_0 ^∞ ((−1)/t) ( xsin(xt) +cos(xt) −1)e^(−t) dt but 1−cos(xt) ∼ ((x^2 t^2 )/2) ⇒ ((1−cos(xt))/t) ∼ ((tx^2 )/2) ⇒ lim_(t→0) (((1−cos(xt))e^(−t) )/t) =0 ⇒ f(x) = x ∫_0 ^∞ ((sin(xt))/t) e^(−t) dt +∫_0 ^∞ ((1−cos(xt))/t) e^(−t) dt let calculate h(x)= ∫_0 ^∞ ((sin(xt))/t) e^(−t) dt h^′ (x) = ∫_0 ^∞ cos(xt) e^(−t) dt =Re(∫_0 ^∞ e^(ixt−t) dt) ∫_0 ^∞ e^((−1 +ix)t) dt = [ (1/(−1+ix)) e^((−1+ix)t) ]_0 ^(+∞) = (1/(−1+ix)) = ((−1)/(1−ix)) = ((−1−ix)/(1+x^2 )) ⇒ h^′ (x) = ((−1)/(1+x^2 )) ⇒ h(x) =λ −arctan(x) but λ =h(0)=0 ⇒ h(x) =−arctan(x)
bypartsu′=1t2andv=(1−cos(xt))e−t⇒v′=xsin(xt)e−t−(1−cos(xt))e−t=(xsin(xt)+cos(xt)−1)e−tf(x)=[−1t(1−cos(xt))e−t]0+∞−∫0∞−1t(xsin(xt)+cos(xt)−1)e−tdtbut1−cos(xt)∼x2t22⇒1−cos(xt)t∼tx22⇒limt→0(1−cos(xt))e−tt=0⇒f(x)=x∫0∞sin(xt)te−tdt+∫0∞1−cos(xt)te−tdtletcalculateh(x)=∫0∞sin(xt)te−tdth′(x)=∫0∞cos(xt)e−tdt=Re(∫0∞eixt−tdt)∫0∞e(−1+ix)tdt=[1−1+ixe(−1+ix)t]0+∞=1−1+ix=−11−ix=−1−ix1+x2⇒h′(x)=−11+x2⇒h(x)=λ−arctan(x)butλ=h(0)=0⇒h(x)=−arctan(x)
Commented by abdo mathsup 649 cc last updated on 24/May/18
let calculate K(x) = ∫_0 ^∞ ((1 −cos(xt))/t) e^(−t) dt K^′ (x) = ∫_0 ^∞ sin(xt) e^(−t) dt =Im( ∫_0 ^∞ e^(ixt −t) dt) = Im( ∫_0 ^∞ e^((−1+ix)t) dt) = ((−x)/(1+x^2 )) ⇒ K(x)= λ −(1/2)ln(1+x^2 ) but λ =K(0)=0 ⇒ K(x) = −(1/2)ln(1+x^2 ) ⇒ f(x) = −x arctan(x) −(1/2)ln(1+x^2 )
letcalculateK(x)=∫0∞1−cos(xt)te−tdtK′(x)=∫0∞sin(xt)e−tdt=Im(∫0∞eixt−tdt)=Im(∫0∞e(−1+ix)tdt)=−x1+x2⇒K(x)=λ−12ln(1+x2)butλ=K(0)=0⇒K(x)=−12ln(1+x2)⇒f(x)=−xarctan(x)−12ln(1+x2)
Commented by abdo mathsup 649 cc last updated on 24/May/18
error in the final line f(x)= −x arctan(x) +(1/2)ln(1+x^2 ).
errorinthefinallinef(x)=−xarctan(x)+12ln(1+x2).
Commented by abdo mathsup 649 cc last updated on 24/May/18
2) we have proved that ∫_0 ^∞ ((1−cos(xt))/t^2 ) e^(−t) dt =(1/2)ln(1+x^2 ) −x arctan(x) let take x=1 ⇒ ∫_0 ^∞ ((1−cos(t))/t^2 ) e^(−t) = (1/2)ln(2) −(π/4) .
2)wehaveprovedthat∫0∞1−cos(xt)t2e−tdt=12ln(1+x2)−xarctan(x)lettakex=1⇒∫0∞1−cos(t)t2e−t=12ln(2)−π4.
Commented by abdo mathsup 649 cc last updated on 24/May/18
error in line 8 f(x)= x ∫_0 ^∞ ((sin(xt))/t) e^(−t) dt −∫_0 ^∞ ((1−cos(xt))/t)e^(−t) dt
errorinline8f(x)=x∫0∞sin(xt)te−tdt−∫0∞1−cos(xt)te−tdt

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *