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Question Number 88415 by abdomathmax last updated on 10/Apr/20
find L(((1−cosx)/x^2 )) with L lsplace transform
findL(1−cosxx2)withLlsplacetransform
Commented by jagoll last updated on 10/Apr/20
L(1−cos x) = 1−(x/(x^2 +1)) = ((x^2 −x+1)/(x^2 +1)) L(((1−cos x)/x)) = (d/dx) [((x^2 −x+1)/(x^2 +1))] = (((2x−1)(x^2 +1)−2x(x^2 −x+1))/((x^2 +1)^2 )) = ((2x^3 −x^2 +2x−1−2x^3 +2x^2 −2x)/((x^2 +1)^2 )) = ((x^2 −1)/((x^2 +1)^2 )) L(((1−cos x)/x^2 )) = (d/dx)[((x^2 −1)/((x^2 +1)^2 ))] = ((2x(x^2 +1)^2 −4x(x^2 +1)(x^2 −1))/((x^2 +1)^4 )) = ((2x(x^2 +1){x^2 +1−2(x^2 −1)})/((x^2 +1)^4 )) = ((2x(−x^2 +3))/((x^2 +1)^3 )) = ((−2x^3 +6x)/((x^2 +1)^3 ))
L(1−cosx)=1−xx2+1=x2−x+1x2+1L(1−cosxx)=ddx[x2−x+1x2+1]=(2x−1)(x2+1)−2x(x2−x+1)(x2+1)2=2x3−x2+2x−1−2x3+2x2−2x(x2+1)2=x2−1(x2+1)2L(1−cosxx2)=ddx[x2−1(x2+1)2]=2x(x2+1)2−4x(x2+1)(x2−1)(x2+1)4=2x(x2+1){x2+1−2(x2−1)}(x2+1)4=2x(−x2+3)(x2+1)3=−2x3+6x(x2+1)3
Commented by mathmax by abdo last updated on 10/Apr/20
L(((1−cosx)/x^2 )) =∫_0 ^∞ f(t)e^(−xt) dt =∫_0 ^∞ ((1−cost)/t^2 ) e^(−xt) dt (x>0) =ϕ(x) ⇒ϕ^′ (x)=−∫_0 ^∞ ((1−cost)/t) e^(−xt) dt ⇒ ϕ^(′′) (x) =∫_0 ^∞ (1−cost)e^(−xt ) dt =∫_0 ^∞ e^(−xt) dt −∫_0 ^∞ e^(−xt) cost dt =[−(1/x)e^(−xt) ]_0 ^(+∞) −Re(∫_0 ^∞ e^(−xt+it) dt) =(1/x) −Re(∫_0 ^∞ e^((−x+i)t) dt) ∫_0 ^∞ e^((−x+i)t) dt =[(1/(−x+i))e^((−x+i)t) ]_0 ^(+∞) =((−1)/(−x+i)) =(1/(x−i)) =((x+i)/(x^2 +1)) ⇒ ϕ^(′′) (x)=(1/x)−(x/(x^2 +1)) ⇒ϕ^′ (x)=lnx −(1/2)ln(x^2 +1) +λ ⇒ ϕ(x)=xlnx−x −(1/2)∫ ln(x^2 +1)dx +λx +α ∫ ln(x^2 +1)dx =x ln(x^2 +1)−∫ x×((2x)/(x^2 +1))dx =xln(x^2 +1)−2 ∫ ((x^2 +1−1)/(x^2 +1))dx =xln(x^2 +1)−2x +2arctanx +c⇒ ϕ(x) =xlnx −x −(x/2)ln(x^2 +1) +x −arctanx +λx +c ⇒ϕ(x)=xlnx −(x/2)ln(x^2 +1)−arctanx +λx +c lim_(x→0) ϕ(x) =∫_0 ^∞ ((1−cost)/t^2 )dt =_(by parts) [−(1/t)(1−cost)]_0 ^∞ +∫_0 ^∞ (1/t)sint dt =(π/2) =c ⇒ ϕ(x)=xlnx −(x/2)ln(x^2 +1)−arctanx +λx +(π/2) rest to find λ .be continued...
L(1−cosxx2)=∫0∞f(t)e−xtdt=∫0∞1−costt2e−xtdt(x>0)=φ(x)⇒φ′(x)=−∫0∞1−costte−xtdt⇒φ″(x)=∫0∞(1−cost)e−xtdt=∫0∞e−xtdt−∫0∞e−xtcostdt=[−1xe−xt]0+∞−Re(∫0∞e−xt+itdt)=1x−Re(∫0∞e(−x+i)tdt)∫0∞e(−x+i)tdt=[1−x+ie(−x+i)t]0+∞=−1−x+i=1x−i=x+ix2+1⇒φ″(x)=1x−xx2+1⇒φ′(x)=lnx−12ln(x2+1)+λ⇒φ(x)=xlnx−x−12∫ln(x2+1)dx+λx+α∫ln(x2+1)dx=xln(x2+1)−∫x×2xx2+1dx=xln(x2+1)−2∫x2+1−1x2+1dx=xln(x2+1)−2x+2arctanx+c⇒φ(x)=xlnx−x−x2ln(x2+1)+x−arctanx+λx+c⇒φ(x)=xlnx−x2ln(x2+1)−arctanx+λx+climx→0φ(x)=∫0∞1−costt2dt=byparts[−1t(1−cost)]0∞+∫0∞1tsintdt=π2=c⇒φ(x)=xlnx−x2ln(x2+1)−arctanx+λx+π2resttofindλ.becontinued…

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