0-sin-x-sin-x-x-2-dx- Tinku Tara June 4, 2023 Integration 0 Comments FacebookTweetPin Question Number 130417 by Lordose last updated on 25/Jan/21 ∫0∞sin(αx)sin(βx)x2dx Answered by mathmax by abdo last updated on 25/Jan/21 I=∫0∞sin(αx)sin(βx)x2dx⇒I=αx=t∫0∞sintsin(βαt)t2α2×dtα=∫0∞sin(t)sin(βαt)t2dt(wesupposeα>0andβ>0)letλ=βα⇒I=∫0∞sintsin(λt)t2dtbypartswegetI=[−1tsintsin(λt)]0∞+∫0∞1t(costsin(λt)+λsintcos(λt))dt=∫0∞costsin(λt)tdt+λ∫0∞sintcos(λt)tdt=u(λ)+λv(λ)wehavesin(x+y)=sinxcosy+cosxsinysin(x−y)=sinxcosy−cosxsiny⇒sinxcosy=12(sin(x+h)+sin(x−y))⇒u(λ)=∫0∞1t(sin(λ+1)t+sin(λ−1)t)dt=∫0∞sin(λ+1)tt+∫0∞sin(λ−1)ttdtweknowλ>0⇒∫0∞sin(λ+1)ttdt=(λ+1)t=y(λ+1)∫0∞sinyydyλ+1=∫0∞sinyydy=π2ifλ>1∫0∞sin(λ−1)ttdt=π2⇒I=π2+λπ2=π2(λ+1)=π2(βα+1)if0<λ<1⇒∫0∞sin(λ−1)ttdt=−∫0∞sin(1−λ)ttdt=−π2⇒I=π2−λπ2=π2(1−λ)=π2(1−βα) Commented by mathmax by abdo last updated on 25/Jan/21 sorryI=α∫0∞sintsin(λt)t2⇒I=π2(α+β)orI=π2(α−β)…. Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: pi-e-sin-1-pi-2-2e-2-sin-2-pi-3-3e-3-sin-3-pi-4-4e-4-sin-4-Next Next post: find-singular-point-for-each-1-f-z-e-z-z-2-2-f-z-sinz-z-3-f-z-1-cosz-sinz-2-4-f-z-ln-z-how-can-solve-this-help-me-sir- Leave a Reply Cancel replyYour email address will not be published. Required fields are marked *Comment * Name * Save my name, email, and website in this browser for the next time I comment.