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10樓: Originally posted by pippi6 at 2013-07-26 16:21:54
對(duì)不起，是我看錯(cuò)了。向你道歉！...

沒(méi)事沒(méi)事，說(shuō)不上啥道不道歉的。解釋清了就好。
我只是急著問(wèn)這題，
第一題二樓的老師大概說(shuō)了一下，第二題求解答，謝謝大家

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11樓2013-07-26 18:01:18

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freejx: 金幣+26, ★★★★★最佳答案, 非常感謝 2013-07-27 05:22:41
freejx: 金幣+2, ★★★★★最佳答案 2013-07-27 16:35:57

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11樓: Originally posted by freejx at 2013-07-26 18:01:18
沒(méi)事沒(méi)事，說(shuō)不上啥道不道歉的。解釋清了就好。
我只是急著問(wèn)這題，
第一題二樓的老師大概說(shuō)了一下，第二題求解答，謝謝大家...

I would like to reply in english, as slow in typing chinese.

all of these will be the application of Lebesgue’s Dominated Convergence Theorem.

1)
Int[ (1+x/n)^n * e^{-2x}, 0, n ] = Int[ f_n(x), 0, \infty]
where f_n(x)= Character(x, [0,n]) *  (1+x/n)^n * e^{-2x} , which is pointwise convergent to e^{-x}.  Now show that f_n(x) <=  e^{-0.5 x}. Or, you only need to show that when t=x/n, 1+t < e^{1.5t} for all t>0.
Then since F(x)=e^{-0.5 x} is integrable on (0, infty), so the limit can be moved into the integral and the answer is int( e^{-x} , 0, \infty)=1.

2)let t=x/n. then it becomes int[sin(t)*n/ (1+t)^n; 0, infty].
int[sin(t)*n/ (1+t)^n] = (integral by parts) = -n* sint /(1+t)^{n-1} /(n-1)
+ n /(n-1) Int[ cos(t)/(1+t)^{n-1}] The first part, the function -n* sint /(1+t)^{n-1} /(n-1) evaluates to 0 at 0 and infty. The second integral Int[ cos(t)/(1+t)^{n-1}; 0, infty] is absolutely less than  Int[ 1/(1+t)^{n-1}; 0, infty]= 1/(n-2).
So combine together, and let n go to infty, the answer is 0.

3). Notice that the function sin(t)/t is continuous over R and is bounded by 1. therefore, the integral is directly bounded by 1/(1+x^2). Therefore, the limit can be taken inside the integral, and it becomes Int[ 1/(1+x^2); 0, infty] = pi/2.

4) it has anti-derivative arctan(nx). arctan(nx) at infty is pi/2. arctan(nx) at a depends on a. If a is positive, then the value is pi/2. If a=0, then the value is 0; if a is negative, then the value is -pi/2.
So by subtraction, you will see that there are three answers depending the sign of a: 0, pi/2, and pi.

答案: (1): 1. (2): 0.  (3): pi/2. (4): o or pi/2 or pi.

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12樓2013-07-27 03:03:15

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12樓: Originally posted by hank612 at 2013-07-27 03:03:15
I would like to reply in english, as slow in typing chinese.

all of these will be the application of Lebesgue’s Dominated Convergence Theorem.

1)
Int = Int
where f_n(x)= Character(x, ) * ...

謝謝，我還有個(gè)問(wèn)題，第一題的（7）怎么證明？

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13樓2013-07-27 05:24:29

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13樓: Originally posted by freejx at 2013-07-27 05:24:29
謝謝，我還有個(gè)問(wèn)題，第一題的（7）怎么證明？...

Int [ x e^{-x^2} ] = 1/2 * e^{-x^2} + C
這個(gè)不定積分有原函數(shù)的，自然可積啦。

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14樓2013-07-27 13:24:23

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14樓: Originally posted by hank612 at 2013-07-27 13:24:23
Int = 1/2 * e^{-x^2} + C
這個(gè)不定積分有原函數(shù)的，自然可積啦。...

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15樓2013-07-27 16:35:27

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