Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 106691 by Laplace last updated on 06/Aug/20

determine using  laplce transformation this  integrale     ∫_0 ^(+∞) ((tsin(tx))/(a^2 +t^(2 ) ))dt

determineusinglaplcetransformationthisintegrale0+tsin(tx)a2+t2dt

Answered by mathmax by abdo last updated on 06/Aug/20

f_a (x) =∫_0 ^∞  ((tsin(tx))/(a^2  +t^2 )) dt ⇒f_a (x) =∫_0 ^∞  tsin(tx)∫_0 ^∞ e^(−(a^2 +t^2 )x) dx dt  =∫_0 ^∞  (∫_0 ^∞ sin(tx)e^(−(a^2  +t^2 )x) dx)tdt (by use of fubini)  ∫_0 ^∞  sin(tx) e^(−(a^2  +t^2 )x)  dx =Im(∫_0 ^∞  e^(itx−(a^2  +t^2 )x)  dx) and  ∫_0 ^∞  e^((it−(a^2  +t^2 ))x) dx =[(1/(it−(a^2 +t^2 ))) e^((it−(a^2  +t^2 ))x) ]_0 ^∞   =((−1)/(it−(a^2  +t^2 ))) =(1/((a^2  +t^2 )−it)) =((a^2 +t^2 +it)/((a^2  +t^2 )^2  +t^2 )) ⇒  Im(...) =(t/((a^2  +t^2 )^2  +t^2 )) ⇒  f_a (x) =∫_0 ^∞   (t^2 /((a^2  +t^2 )^2 +t^2 )) dt =∫_0 ^∞  (t^2 /((a^4 +2a^2 t^2  +t^4  +t^2 ))dt  =∫_0 ^∞  (t^2 /(t^4  +(2a^2  +1)t^2  +a^4 )) dt =(1/2)∫_(−∞) ^(+∞)  (t^2 /(t^4  +(2a^2  +1)t^2  +a^4 )) dt  let ϕ(z) =(z^2 /(z^4  +(2a^2  +1)z^2  +a^4 ))  poles of ϕ?  Δ =(2a^2  +1)^2 −4a^4  =4a^4 +4a^2  +1−4a^4  =4a^2  +1  (z^2  =u)  u_1 =((−(2a^2  +1)+(√(4a^2  +1)))/2)  and u_2 =((−(2a^2  +1)−(√(4a^2  +1)))/2)  ⇒ϕ(z) =(z^2 /((z^2 −u_1 )(z^2 −u_2 ))) =(z^2 /((z−(√u_1 ))(z+(√u_1 ))(z−(√u_2 ))(z+(√u_2 ))))  (√u_1 )=i(√((2a^(2 ) +1−(√(4a^2  +1)))/2)) and (√u_2 )=i(√((2a^2  +1+(√(4a^2  +1)))/2))  ...be continued...

fa(x)=0tsin(tx)a2+t2dtfa(x)=0tsin(tx)0e(a2+t2)xdxdt=0(0sin(tx)e(a2+t2)xdx)tdt(byuseoffubini)0sin(tx)e(a2+t2)xdx=Im(0eitx(a2+t2)xdx)and0e(it(a2+t2))xdx=[1it(a2+t2)e(it(a2+t2))x]0=1it(a2+t2)=1(a2+t2)it=a2+t2+it(a2+t2)2+t2Im(...)=t(a2+t2)2+t2fa(x)=0t2(a2+t2)2+t2dt=0t2(a4+2a2t2+t4+t2dt=0t2t4+(2a2+1)t2+a4dt=12+t2t4+(2a2+1)t2+a4dtletφ(z)=z2z4+(2a2+1)z2+a4polesofφ?Δ=(2a2+1)24a4=4a4+4a2+14a4=4a2+1(z2=u)u1=(2a2+1)+4a2+12andu2=(2a2+1)4a2+12φ(z)=z2(z2u1)(z2u2)=z2(zu1)(z+u1)(zu2)(z+u2)u1=i2a2+14a2+12andu2=i2a2+1+4a2+12...becontinued...

Commented by Ar Brandon last updated on 06/Aug/20

��

Answered by abdomathmax last updated on 07/Aug/20

residus method  let I =∫_0 ^∞  ((tsin(tx))/(t^2 +a^2 ))dt  ⇒ I =_(t=∣a∣u)     ∫_0 ^∞  ((∣a∣usin(∣a∣ux))/(a^2 (u^2  +1)))∣a∣ du  =∫_0 ^∞   ((u sin(x∣a∣u))/(u^2  +1)) =(1/2)∫_(−∞) ^(+∞)  ((u sin(x∣a∣u))/(u^2  +1))du  =(1/2)Im(∫_(−∞) ^(+∞)  ((u e^(ix∣a∣u) )/(u^2  +1))du) let  ϕ(z)=((z e^(ix∣a∣z) )/(z^2  +1))  ∫_(−∞) ^(+∞)  ϕ(z)dx =2iπ Res(ϕ,i) =2iπ×((ie^(−x∣a∣) )/(2i))  =iπ e^(−x∣a∣)  ⇒I =(π/2) e^(−x∣a∣)

residusmethodletI=0tsin(tx)t2+a2dtI=t=∣au0ausin(aux)a2(u2+1)adu=0usin(xau)u2+1=12+usin(xau)u2+1du=12Im(+ueixauu2+1du)letφ(z)=zeixazz2+1+φ(z)dx=2iπRes(φ,i)=2iπ×iexa2i=iπexaI=π2exa

Terms of Service

Privacy Policy

Contact: info@tinkutara.com