[積分,極限]求積分的極限值

Ans: 2/pi

2010-06-11 13:57:19 補充：
∫[0~nπ] exp(-x) |sin(nx)| dx
=(1/n)∫[0~n^2π] exp(-x/n) |sinx| dx
=(1/n)Σ[k=0~n^2-1] ∫[kπ~(k+1)π] exp(-x/n) |sinx| dx
=n/(n^2+1)*[exp(-π/n)+1]Σ[k=0~n^2-1]exp(-kπ/n) [geometric series r=exp(-π/n)]
=n/(n^2+1)*[1+exp(-π/n)]*[1-exp(-nπ)]/[1-exp(-π/n)]
=n^2/(n^2+1)*[1+exp(-π/n)]*[1-exp(-nπ)]*{[1/n]/[1-exp(-π/n)]}
-->1*[1+1]*[1-0]*{1/π} (Apply L' rule to the last term)
=2/π

圖片參考：http://imgcld.yimg.com/8/n/AD04686329/o/161006100233313872535900.jpg

相似題:

http://tw.knowledge.yahoo.com/question/question?qid=1106083114274

這題不是那麼的簡單吶

Is lim(n→∞)∫(0 to nπ)e^(-x)|sin nx| dx = lim(n→∞)∫(0 to ∞)e^(-x)|sin nx| dx? Why?

感覺上做得出來，不過我用的方法很爛
還請天助大幫忙詳解