int[(lnx)^2/(1+x^2)]

(-1)^n*4/(2n+1)^3, n=0,1,2,3,....

2010-01-30 22:45:21 補充：
∫[1~∞] (lnx)^2/(1+x^2) dx (令 x= 1/t)
=∫[1~0] (-lnt)^2/(1+ 1/t^2) *(- dt/t^2)
=∫[0~1] (lnt)^2/(t^2+ 1) dt
=∫[0~1] (lnx)^2/(x^1+1) dx
so,
∫[0~∞] (lnx)^2/(1+x^2) dx
=∫[0~1] (lnx)^2/(1+x^2) dx+∫[1~∞] (lnx)^2/(1+x^2) dx
= 2∫[0~1] (lnx)^2/(1+x^2) dx
= 2∫[0~1] Σ[0~∞] (-1)^n* (lnx)^2*x^(2n) dx
= 2Σ[0~∞] (-1)^n*∫[0~1] (lnx)^2*x^(2n) dx
(by parts)= -2Σ[0~∞] (-1)^n*2/(2n+1) ∫[0~1] (lnx)*x^(2n)dx
(by parts)= Σ[0~∞] (-1)^n *4/(2n+1)^2 *∫[0~1]x^(2n)dx
=Σ[0~∞] (-1)^n*4/(2n+1)^3
= 4/1^3 - 4/3^3 + 4/5^3 - 4/7^3 + ...+ (-1)^n* 4/(2n+1)^3+...

使用上半圓包住+i pole做積分
後面還要配合int[1/(1+x^2)]=pi/2,x:0~infinity
即可得解析解

其實最後的那條infinite series是可以繼續化簡至π³/8。詳見http://www.wolframalpha.com/input/?i=sum+4%28-1%29%5En%2F%282n%2B1%29%5E3%2Cn%3D0..inf。

但是，該如何繼續化簡至π³/8？