試求下列定積分 ∫ln(1+x)/(1+x^2) dx

Ans: (pi*ln2)/8

2010-05-08 22:25:58 補充：
method1: substitute x=(1-t)/(1+t)
∫[0~1] ln(1+x)/(1+x^2) dx
=∫[1~0] ln[2/(1+t)] / [ 1+(1-t)^2/(1+t)^2] * (-2)/(1+t)^2 dt
=∫[0~1] [ln(2)- ln(1+t)] / (1+t^2) dt
=∫[0~1] ln(2)/(1+t^2) dt - ∫[0~1] ln(1+t)/ (1+t^2) dt
= ln(2)*arctan(t) |[0~1] - ∫[0~1] ln(1+x)/(1+x^2) dx (dummy variable)
= ln(2)*(π/4) - ∫[0~1] ln(1+x)/(1+x^2) dx
so, 2∫[0~1] ln(1+x)/(1+x^2) dx= πln(2) /4
then ∫[0~1] ln(1+x)/(1+x^2) dx = πln(2) /8

method2: substitute x=tan(t)
∫[0~1] ln(1+x)/(1+x^2) dx
=∫[0~π/4] ln(1+tant) dt
=∫[0~π/4] ln(sint+cost) dt - ∫[0~π/4] ln(cost) dt
=∫[0~π/4] ln[√2 sin(t+π/4)] dt - ∫[0~π/4] ln(cost) dt
(sub. x=π/4- t into the first integral)
=∫[π/4~0] ln[√2 sin(π/2-x)] (-dx) - ∫[0~π/4] ln(cost) dt
=∫[0~π/4] ln[√2 cosx] dx - ∫[0~π/4] ln(cosx) dx
=∫[0~π/4] ln(√2) dx
=πln(2)/8

我不會解, 但是想知道如何解, 故贊助 2 點聊表心意