Evaluate-sinx-x-2-2x-2-dx- Tinku Tara June 3, 2023 Integration 0 Comments FacebookTweetPin Question Number 131957 by rs4089 last updated on 10/Feb/21 Evaluate∫−∞∞sinxx2+2x+2dx Answered by Dwaipayan Shikari last updated on 10/Feb/21 ∫−∞∞sinx(x+1)2+1x+1=u=∫−∞∞sinucos1−sin1cosuu2+1du=cos1∫−∞∞sinuu2+1−sin1∫−∞∞cosuu2+1du=0−πesin(1)=−πesin(1) Commented by mnjuly1970 last updated on 10/Feb/21 verynicemrpayan… Answered by mathmax by abdo last updated on 10/Feb/21 letf(λ)=∫−∞+∞sin(λx)x2+2x+2dxwithλ>0⇒f(λ)=Im(∫−∞+∞eiλxx2+2x+2dx)Φ(z)=eiλzz2+2z+2wehavez2+2z+2=0⇒(z+1)2+1=0⇒(z+1)2=−1⇒z+1=+−i⇒z=−1+−isithepolesofΦarez1=−1+iandz2=−1−i∫RΦ(z)dz=2iπRes(Φ,z1)=2iπ.eiλz12i=πeiλ(−1+i)=πe−iλ−λ=πe−λ{cos(λ)−isin(λ)}⇒f(λ)=−πe−λsin(λ)∫−∞+∞sinxx2+2x+2dx=f(1)=−πesin(1) Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: show-that-for-a-given-complex-number-z-z-n-r-n-cosn-isinn-Next Next post: x-9-2-3-what-is-x- Leave a Reply Cancel replyYour email address will not be published.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.