calculate-1-ln-lnx-1-x-2-dx- Tinku Tara June 4, 2023 Integration 0 Comments FacebookTweetPin Question Number 60727 by Mr X pcx last updated on 25/May/19 calculate∫1+∞ln(lnx)1+x2dx Commented by maxmathsup by imad last updated on 26/May/19 letA=∫1+∞ln(lnx)1+x2dxchangementln(x)=tgiveA=∫0+∞ln(t)1+e2tetdt=∫0∞e−tln(t)1+e−2tdt=∫0∞e−tln(t)(∑n=0∞(−1)ne−2nt)dt=∑n=0∞(−1)n∫0∞e−(2n+1)tln(t)dt=∑n=0∞(−1)nAnAn=∫0∞e−(2n+1)tln(t)dt=(2n+1)t=u∫0∞e−uln(u2n+1)du2n+1=12n+1∫0∞e−u{ln(u)−ln(2n+1))du=12n+1{∫0∞e−uln(u)du−ln(2n+1)∫0∞e−udu}∫0∞e−uln(u)du=−γ(resultproved)∫0∞e−udu=[−e−u]0∞=1⇒An=−γ2n+1−ln(2n+1)2n+1⇒A=∑n=0∞(−1)n{−γ2n+1−ln(2n+1)2n+1}=−γ∑n=0∞(−1)n2n+1−∑n=0∞(−1)n2n+1ln(2n+1)=−γπ4−∑n=0∞(−1)nln(2n+1)2n+1withγconstantofEuler…becontinued… Commented by maxmathsup by imad last updated on 26/May/19 lettryanotherwaybyuseofchang.x=1t⇒I=∫1+∞ln(lnx)1+x2dx=−∫01ln(ln(1t))1+1t2(−dtt2)=∫01ln(ln(1t))1+t2dt=∫01ln(ln(1t))(1+it)(1−it)dt=12∫01(11+it+11−it)ln(ln(1t))dt=12∫01ln{ln(1t)}1+itdt+12∫01ln{ln(1t)}1−itdtletW=∫01ln{ln(1t)}1+itdtchang.ln(1t)=ugive1t=eu⇒t=e−u⇒W=−∫0∞lnu1+ie−u(−e−u)du=∫0∞e−uln(u)1+ie−udu=∫0∞e−uln(u)(∑n=0∞(−ie−u)n)du=∑n=0∞(−i)n∫0∞e−(n+1)uln(u)du∫0∞e−(n+1)uln(u)du=(n+1)u=t∫0∞e−tln(tn+1)dtn+1=1n+1∫0∞{e−tln(t)−e−tln(n+1)}dt=1n+1∫0∞e−tln(t)−ln(n+1)n+1∫0∞e−tdt=−γn+1−ln(n+1)n+1⇒W=∑n=0∞−γ(−i)nn+1−∑n=0∞(−i)nln(n+1)n+1H=∫01ln{ln(1t)}1−itdt=ln(1t)=u−∫0∞ln(u)1−ie−u(−e−u)du=∫0∞e−uln(u)1−ie−udu=∫0∞e−ulnu{∑n=0∞ine−nu}du=∑n=0∞in∫0∞e−(n+1)uln(u)du=….becontinued…. Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: x-0-i-j-2019-j-2019-i-j-Next Next post: calculate-1-ln-lnx-x-2-x-1-dx- Leave a Reply Cancel replyYour email address will not be published.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.