K*****Y
发帖数: 629 | | s********r
发帖数: 529 | 2 我想到一个办法，但是不知道是不是最简单，可能也有纰漏：
这个积分是乘以\exp(-2\sqrt{ac})后，变成\exp{-(ax+\dfrac{c}{x})^2}dx
进行变量代换，令x=\dfrac{c}{au},u是新的积分变量，则原来的积分=\dfrac{-c}{au^
2}\exp{-(au+\dfrac{c}{u})^2},再把u换成x，这个积分和原积分相等
令原来的积分等于C，则2C=\int{-\infty}{+\infty}(1-\dfrac{c}{ax^2}\exp{-(ax+\d
frac{c}{x})^2})dx,可以将e指数前面的一项提出1/a后放入dx内，则积分问题退化为\i
nt_{-\infty}^{+\infty}exp{-x^2}dx
我这个方法可能存在积分上下限的问题，明天还要考试，就当抛砖引玉吧
【在 K*****Y 的大作中提到】
: 见附件。请不吝赐教。多谢！
| p*****k
发帖数: 318 | 3 note the symmetry, this is just twice of int from 0 to infty.
the trick then is to set x = (a/c)^(1/4) * exp(t) and use hyperbolic cosine
and sine. up to a constant factor, it's just a Gaussian integral again.
double check if the answer is exp[-2*sqrt(a*c)] * sqrt(pi/a)
more details can be found at:
http://www.wilmott.com/messageview.cfm?catid=26&threadid=72024 | k*****y
发帖数: 744 | 4 Sorry, I was wrong below. There should not be an sqrt{2} in front. pcasnik's answer is right.
【在 K*****Y 的大作中提到】
: 见附件。请不吝赐教。多谢！
| d*j
发帖数: 13780 | 5 大牛， 不仅牛在你每次都知道答案
最佩服的就是每次都能贴出link
cosine
【在 p*****k 的大作中提到】
: note the symmetry, this is just twice of int from 0 to infty.
: the trick then is to set x = (a/c)^(1/4) * exp(t) and use hyperbolic cosine
: and sine. up to a constant factor, it's just a Gaussian integral again.
: double check if the answer is exp[-2*sqrt(a*c)] * sqrt(pi/a)
: more details can be found at:
: http://www.wilmott.com/messageview.cfm?catid=26&threadid=72024
| s********r
发帖数: 529 | 6 嗯，我也觉得很神
【在 d*j 的大作中提到】
: 大牛， 不仅牛在你每次都知道答案
: 最佩服的就是每次都能贴出link
:
: cosine
| G******r
发帖数: 76 | 7 zan!
cosine
【在 p*****k 的大作中提到】
: note the symmetry, this is just twice of int from 0 to infty.
: the trick then is to set x = (a/c)^(1/4) * exp(t) and use hyperbolic cosine
: and sine. up to a constant factor, it's just a Gaussian integral again.
: double check if the answer is exp[-2*sqrt(a*c)] * sqrt(pi/a)
: more details can be found at:
: http://www.wilmott.com/messageview.cfm?catid=26&threadid=72024
| L*********Z
发帖数: 52 | 8 +1 主要是link。。
【在 d*j 的大作中提到】
: 大牛， 不仅牛在你每次都知道答案
: 最佩服的就是每次都能贴出link
:
: cosine
| K*****Y
发帖数: 629 | | k*******n
发帖数: 180 | 10 对于如下的积分，有个简单的解法
I = int_0^\infty exp(-x^2 - a^2/x^2) dx
对a求导，然后用x=a/y做变量替换，得到
dI/da = Const×I ==> I=Const exp(a)
另a=0，得到Const其实就是Erf(infty)。结果就很简单了。
你这个问题只要变换一下就可以了。这个是教科书上的解法，不过我实在想不起来是哪本教科书了，等
想起来了再补上。 | | | r******g
发帖数: 13 | 11 very confused about the Straight algebraic substitution by MatthewM in the
link, tried to do the same things, not sure is there anything I missed,
1) how to get equation (8) using the two roots? why
2) both equations (1), (2) and (6) show equation between (2) and (3) is 0,
what's wrong with it? but equation (10) gives (12)
thought the algebraic substitution was easy and straight forward, scratching my head to find out what's I did wrong, wasting several papers, still not clue, many thanks for the help.
cosine
【在 p*****k 的大作中提到】
: note the symmetry, this is just twice of int from 0 to infty.
: the trick then is to set x = (a/c)^(1/4) * exp(t) and use hyperbolic cosine
: and sine. up to a constant factor, it's just a Gaussian integral again.
: double check if the answer is exp[-2*sqrt(a*c)] * sqrt(pi/a)
: more details can be found at:
: http://www.wilmott.com/messageview.cfm?catid=26&threadid=72024
| k*****y
发帖数: 744 | 12 One should be careful when the substitution is not monotonic.
In your calculation, the two roots cover the range of u in (1, \infty) and
(1, 0) for v in (2, \infty), so the the integral should be the sum/
difference of the ones from the two substitutions.
scratching my head to find out what's I did wrong, wasting several papers,
still not clue, many thanks for the help.
【在 r******g 的大作中提到】
: very confused about the Straight algebraic substitution by MatthewM in the
: link, tried to do the same things, not sure is there anything I missed,
: 1) how to get equation (8) using the two roots? why
: 2) both equations (1), (2) and (6) show equation between (2) and (3) is 0,
: what's wrong with it? but equation (10) gives (12)
: thought the algebraic substitution was easy and straight forward, scratching my head to find out what's I did wrong, wasting several papers, still not clue, many thanks for the help.
:
: cosine
| r******g
发帖数: 13 | 13 Thank you very much, kinecty, zan
didn't realize the two roots in different range
then it should be the sum of the two roots substitution, it will cancel out
the second parts, get equation (2), why is the difference? in both roots,
the range for v is (2, \infty)? equation (7) gives that v is in (-\infty, -2), then u is negative
what's wrong with equation (3)? related to "the substitution is not
monotonic"?
【在 k*****y 的大作中提到】
: One should be careful when the substitution is not monotonic.
: In your calculation, the two roots cover the range of u in (1, \infty) and
: (1, 0) for v in (2, \infty), so the the integral should be the sum/
: difference of the ones from the two substitutions.
:
:
: scratching my head to find out what's I did wrong, wasting several papers,
: still not clue, many thanks for the help.
| k*****y
发帖数: 744 | 14 Because the second one is from 1 to 0 when v is from 2 to \infty, if you swap them you will have a minus sign. You need to double check whether (2) is the difference.
Sorry. You also need to swap the 0 and \infty in (3). That is why we want to use u - 1/u instead of u + 1/u.
out
-2), then u is negative
【在 r******g 的大作中提到】
: Thank you very much, kinecty, zan
: didn't realize the two roots in different range
: then it should be the sum of the two roots substitution, it will cancel out
: the second parts, get equation (2), why is the difference? in both roots,
: the range for v is (2, \infty)? equation (7) gives that v is in (-\infty, -2), then u is negative
: what's wrong with equation (3)? related to "the substitution is not
: monotonic"?
| r******g
发帖数: 13 | 15 多谢！多谢！
forgot the basic, the order of the range in the integral, attach the correct
for future reference.
swap them you will have a minus sign. You need to double check whether (2)
is the difference.
to use u - 1/u instead of u + 1/u.
【在 k*****y 的大作中提到】
: Because the second one is from 1 to 0 when v is from 2 to \infty, if you swap them you will have a minus sign. You need to double check whether (2) is the difference.
: Sorry. You also need to swap the 0 and \infty in (3). That is why we want to use u - 1/u instead of u + 1/u.
:
: out
: -2), then u is negative
| f****u
发帖数: 50 | 16 Kinecty, don't understand one point in your derivation.
Please see my question in the picture attached.
btw, your solution is elegant.
There should not be an sqrt{2} in front. So pcasnik's answer is right.
【在 k*****y 的大作中提到】
: Sorry, I was wrong below. There should not be an sqrt{2} in front. pcasnik's answer is right.
| k*****y
发帖数: 744 | 17 Thanks for noting that.
The integral on (-infty, infty) can be divided into two parts, (-infty, 0)
and (0, infty), which will be the same. That is where the 2 comes from.
When x is in (0, infty), then y = (x-1/x) covers the range of (-infty, infty
).
【在 f****u 的大作中提到】
: Kinecty, don't understand one point in your derivation.
: Please see my question in the picture attached.
: btw, your solution is elegant.
:
: There should not be an sqrt{2} in front. So pcasnik's answer is right.
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