Re: [積分]今年政大微積分考題~
※ 引述《fattybb (我愛小肚寶~)》之銘言:
: 1 X-1
: ∫ ----- dx = ???
: 0 lnX
: 拜託各位高手了~~
有兩種作法
1
1. Let F(t) = ∫ (x^t - 1)/lnx dx
0
then it's obvious that F(0) = 0
1
∫(x-1)/lnx dx = F(1) - 0 = F(1) - F(0)
0
1
F(1) - F(0) 可以寫成 ∫(dF/dt) dt = F(1) - F(0)
0
1 1
dF/dt = (d/dt)∫ (x^t - 1)/lnx dx = ∫ (d/dt)((x^t - 1)/lnx) dx
0 0
利用指數微分的公式 f^g --> (f^g) * ( (f'/f)g + (g')ln|f| )
1 x=1
= ∫ x^t dx = (x^(t+1))/(t+1) | = 1/(1+t)
0 x=0
1
代回去, 得到 F(1) - F(0) = ∫ 1/(1+t) dt = ln2
0 #
2. 作法二, 令 u = ln(x) , e^u = x , (e^u)du = dx
1 0
∫ (x-1)/lnx dx = ∫((e^(2u) - e^u)/u )du
0 -∞
將 e^(2u) - e^u 視作是另一個積分, 範圍 [u,2u]
0 2u
= ∫ ∫(e^t)/u dtdu
-∞ u
利用積分順序互換
0 t
= ∫ ∫ (e^t)/u dudt
-∞ t/2
= ln2
#
--
僅供參考
--
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 125.232.3.188
推
07/08 17:41, , 1F
07/08 17:41, 1F
→
07/08 17:42, , 2F
07/08 17:42, 2F
推
07/08 17:46, , 3F
07/08 17:46, 3F
推
07/08 18:39, , 4F
07/08 18:39, 4F
→
07/08 18:39, , 5F
07/08 18:39, 5F
→
07/08 18:41, , 6F
07/08 18:41, 6F
→
07/08 19:41, , 7F
07/08 19:41, 7F
推
07/08 19:52, , 8F
07/08 19:52, 8F
→
07/08 19:52, , 9F
07/08 19:52, 9F
推
07/08 22:10, , 10F
07/08 22:10, 10F
推
07/09 00:27, , 11F
07/09 00:27, 11F
推
07/09 00:56, , 12F
07/09 00:56, 12F
→
07/09 00:57, , 13F
07/09 00:57, 13F
推
07/09 01:34, , 14F
07/09 01:34, 14F
→
07/09 01:58, , 15F
07/09 01:58, 15F
推
07/09 02:24, , 16F
07/09 02:24, 16F
推
07/09 02:34, , 17F
07/09 02:34, 17F
→
07/09 02:34, , 18F
07/09 02:34, 18F
→
07/09 02:35, , 19F
07/09 02:35, 19F
討論串 (同標題文章)