find-value-of-0-1-ln-1-ix-2-dx-and-0-1-ln-1-ix-2-dx-with-i-1- Tinku Tara June 4, 2023 Integration 0 Comments FacebookTweetPin Question Number 79758 by mathmax by abdo last updated on 27/Jan/20 findvalueof∫01ln(1+ix2)dxand∫01ln(1−ix2)dxwithi=−1 Commented by mathmax by abdo last updated on 29/Jan/20 letzfromCandf(z)=∫01ln(1+zx2)dx⇒f′(z)=∫01x21+zx2dx=1z∫01zx2+1−11+zx2dx=1z−1z∫01dx1+zx2and∫01dx1+zx2=xz=t∫0zdtz(1+t2)=1zarctan(z)⇒f′(z)=1z−arctan(z)zz⇒f(z)=lnz+∫1zarctan(u)uudu+c(u=t2)=lnz+∫1zarctan(t)t3(2t)dt+c=ln(z)+2∫1zarctan(t)t2dt+cf(1)=∫01ln(1+x2)dx=c⇒f(z)=lnz+2∫1zarctan(t)t2dt+∫01ln(1+x2)dxwehavebyparts∫1zarctan(t)t2dt=[−arctantt]1z+∫1z1t(1+t2)dt=π4−arctan(z)z+∫1z(1t−tt2+1)dt=π4−arctan(z)z+ln(z)−[12ln(t2+1)]1z=π4−arctan(z)z+12ln(z)−12{ln(z+1)−ln(2)}⇒f(z)=ln(z)+π2−2arctan(z)z+ln(z)−ln(z+1)+ln(2)+∫01ln(1+x2)dxf(z)=2ln(z)+π2−2arctan(z)z−ln(z+1)+ln(2)+∫01ln(1+x2)dxbyparts∫01ln(1+x2)dx=[xln(1+x2)]01−∫012x21+x2dx=ln(2)−2∫011+x2−11+x2dx=ln(2)−2+2×π4=ln(2)−2+π2⇒f(z)=2ln(z)+2ln(2)+π−2−2arctan(z)z−ln(z+1)=∫01ln(1+zx2)dx⇒∫01ln(1+ix2)dx=f(i)=2ln(i)+2ln(2)+π−2−2×arctan(i)i−ln(i+1)=2×iπ2+2ln(2)+π−2−2×arctan(eiπ4)eiπ4−ln(2eiπ4)=iπ+2ln(2)+π−2−2e−iπ4arctan(eiπ4)−12ln(2)−iπ4 Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: Given-a-polynomial-p-x-x-4-4x-3-2p-2-x-2-2p-5q-2-x-3q-2r-If-p-x-x-3-2x-2-8x-6-Q-x-then-what-the-value-of-p-2q-r-Next Next post: x-2-2-y-xy-0- Leave a Reply Cancel replyYour email address will not be published.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.