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Question Number 106691 by Laplace last updated on 06/Aug/20
determineusinglaplcetransformationthisintegrale∫0+∞tsin(tx)a2+t2dt
Answered by mathmax by abdo last updated on 06/Aug/20
fa(x)=∫0∞tsin(tx)a2+t2dt⇒fa(x)=∫0∞tsin(tx)∫0∞e−(a2+t2)xdxdt=∫0∞(∫0∞sin(tx)e−(a2+t2)xdx)tdt(byuseoffubini)∫0∞sin(tx)e−(a2+t2)xdx=Im(∫0∞eitx−(a2+t2)xdx)and∫0∞e(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+t2⇒Im(...)=t(a2+t2)2+t2⇒fa(x)=∫0∞t2(a2+t2)2+t2dt=∫0∞t2(a4+2a2t2+t4+t2dt=∫0∞t2t4+(2a2+1)t2+a4dt=12∫−∞+∞t2t4+(2a2+1)t2+a4dtletφ(z)=z2z4+(2a2+1)z2+a4polesofφ?Δ=(2a2+1)2−4a4=4a4+4a2+1−4a4=4a2+1(z2=u)u1=−(2a2+1)+4a2+12andu2=−(2a2+1)−4a2+12⇒φ(z)=z2(z2−u1)(z2−u2)=z2(z−u1)(z+u1)(z−u2)(z+u2)u1=i2a2+1−4a2+12andu2=i2a2+1+4a2+12...becontinued...
Commented by Ar Brandon last updated on 06/Aug/20
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Answered by abdomathmax last updated on 07/Aug/20
residusmethodletI=∫0∞tsin(tx)t2+a2dt⇒I=t=∣a∣u∫0∞∣a∣usin(∣a∣ux)a2(u2+1)∣a∣du=∫0∞usin(x∣a∣u)u2+1=12∫−∞+∞usin(x∣a∣u)u2+1du=12Im(∫−∞+∞ueix∣a∣uu2+1du)letφ(z)=zeix∣a∣zz2+1∫−∞+∞φ(z)dx=2iπRes(φ,i)=2iπ×ie−x∣a∣2i=iπe−x∣a∣⇒I=π2e−x∣a∣
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