Integration

寫成infinite series型式不具太大意義! 本題應是定積分x=0~pi/2 or 0~pi
或堅持indefite integral?

2010-01-10 01:05:35 補充：
(1) 1/√(1-x) =Σ[n=0~∞] (2n)!*x^n/[2^n*n!]^2 =Σ[n=0~∞] a(n)x^n
so,1/√(1-x^2)=Σ[0~∞] (2n)!*x^n/[2^n*n!]^2 =Σ[n=0~∞] a(n)x^(2n)
integrating both sides w.r.t. x, then
(2) arcsin(x)=Σ[n=0~∞] a(n)x^(2n+1)/(2n+1)
(3) Sub. -t=ln(sinx), then
∫ln(sinx)dx=∫ t exp(-t)/√[1-exp(-2t)] dt
=∫Σ(n=0~∞) a(n)*t*exp[-(2n+1)t] dt
(uniformly conv. for t>0, so ∫ and summation can be exchanged)
(by parts)
= -Σ[n=0~∞] a(n)*[t/(2n+1)+ 1/(2n+1)^2] /exp[(2n+1)t]+C
= Σ[n=0~∞] a(n)*[ln(sinx)/(2n+1)+ 1/(2n+1)^2] (sinx)^(2n+1)+C
= ln(sinx)*arcsin(sinx) +Σ[0~∞] a(n)(sinx)^(2n+1)/(2n+1)^2+C
= ln(sinx)*x +Σ[0~∞] a(n)(sinx)^(2n+1)/(2n+1)^2+C
Note:
(i) the first term can be obtained from(2)
(ii)the second term is an integral of (arcsinu)/u and then set u=sinx.
(iii)∫[0~π] ln(sinx)dx= 2∫[0~π/2] ln[sin(2x)] dx
=2∫[0~π/2] [ ln2+ ln(sinx)+ ln(cosx)] dx
= 2∫[0~π/2] [ ln2+ 2ln(sinx)] dx
=πln2+ 4∫[0~π/2] ln(sinx) dx
so, ∫[0~π/2] ln(sinx)dx= (-πln2)/2
∫[0~π] ln(sinx)dx= -πln2

2010-01-13 16:20:31 補充：
插播一下:前幾天有題"培正數學邀請賽"的問題,怎麼不見了呢?

2010-01-13 18:01:52 補充：
不是這題,原題問第七屆培正...,第18題,第20題的作法
謝謝您的熱心!

Supplementary answer:

I believe the integral cannot be expressed in elementary functions. A solution involving a convergent series was already presented by a previous answerer. The power series turns out to have a mathematical name.

The polylogarithm

Li (s , z) = Sum_{k=0}^{infinity} z^k / k^s = G (s , z) / Gamma(s)

where G is the Bose-Einstein function, Gamma is the gamma function. [see wikipedia for more. Polylog is also called the de Jonquiere's function] Polylog satisfies

z d Li (s , z) / dz = Li (s-1 , z)
Li (1 , z) = Log(1-z)

One can check by simple differentiation that

Integrate[ log( Sin(x) ) ]
= x log sin x - x log ( 1 - exp(2ix) ) + (i/2) (x^2 + Li (2 , exp(2ix) ))

堅持indefite integral>

2010-01-13 17:55:00 補充：
先感謝你的回答!
你是說這一題嗎?? - http://hk.knowledge.yahoo.com/question/question?qid=7009121801379

2010-01-13 19:29:27 補充：
嗯...我想是發問者移除了問題~