Question

请问这道题广义积分怎么积分？

Answer 1

令t=x/2，则x=2t，dx=2dt

原式=∫(0,+∞) [ln(2t)/(4t^2+4)]*2dt
=(1/2)*∫(0,+∞) (lnt+ln2)/(t^2+1)dt
=(1/2)*∫(0,+∞) lnt/(t^2+1)dt+(1/2)*∫(0,+∞) ln2/(t^2+1)dt
=(1/2)*∫(0,1) lnt/(t^2+1)dt+(1/2)*∫(1,+∞) lnt/(t^2+1)dt+(ln2/2)*arctant|(0,+∞)
令第二项的t=1/u，则dt=-du/u^2
原式=(1/2)*∫(0,1) lnt/(t^2+1)dt+(1/2)*∫(1,0) (-lnu)/(u^2+1)(-du)+(ln2/4)*π
=(1/2)*∫(0,1) lnt/(t^2+1)dt-(1/2)*∫(0,1) lnu/(u^2+1)du+(ln2/4)*π
=(ln2/4)*π