Question

设f(X)=∫ lnT/1+T dT ,求f(X)+f(1/X)

Answer 1

令y = 1/t、dt = - 1/y² dy
∫(1→x) lnt/(1 + t) dt
= ∫(1→1/x) ln(1/y)/(1 + 1/y) * (- 1/y² dy)
= ∫(1→1/x) (- lny) * y/(1 + y) * (- 1/y²) dy
= ∫(1→1/x) lny/[y(1 + y)] dy
= ∫(1→1/x) lnt/[t(1 + t)] dt
f(x) + f(1/x) = ∫(1→x) lnt/(1 + t) dt + ∫(1→1/x) lnt/(1 + t) dt
= ∫(1→1/x) lnt/[t(1 + t)] dt + ∫(1→1/x) lnt/(1 + t) dt
= ∫(1→1/x) (lnt + t * lnt)/[t(1 + t)] dt
= ∫(1→1/x) (1 + t)lnt/[t(1 + t)] dt
= ∫(1→1/x) (lnt)/t dt
= ∫(1→1/x) lnt dlnt
= (1/2)(lnt)² |[1→1/x]
= (1/2){[ln(1/x)]² - [ln(1)]²}
= (1/2)[- ln(x)]²
= (1/2)(lnx)²