z*****n
发帖数: 165 | 1 \int_-\inf^\inf Sin(x)/x dx |
J*****n
发帖数: 4859 | |
p****u
发帖数: 2596 | 3 尽是这种８辈子用不到的题目。
【在 z*****n 的大作中提到】
: \int_-\inf^\inf Sin(x)/x dx
|
J****8
发帖数: 117 | 4 请问楼主二面都问了什么？
C++？ 算法？ 随机 ？ BT ？
多谢？ |
r*******y
发帖数: 1081 | 5 我怎么还没有想出来？ 只知道 \int_0^\inf sin(x) / x dx 是收敛的，
可以用 alternating series theorem证明收敛性。
其他的难道能求出 antiderivative?
【在 z*****n 的大作中提到】
: \int_-\inf^\inf Sin(x)/x dx
|
b*******e
发帖数: 86 | 6 Dirichlet integral
【在 r*******y 的大作中提到】
: 我怎么还没有想出来？ 只知道 \int_0^\inf sin(x) / x dx 是收敛的，
: 可以用 alternating series theorem证明收敛性。
: 其他的难道能求出 antiderivative?
|
r*******y
发帖数: 1081 | 7 sorry what is dirichlet integral ? I know I can google. but I just want to
think about it.
【在 b*******e 的大作中提到】
: Dirichlet integral
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z*****n
发帖数: 165 | 8 上来问了几个Derivative pricing的问题，后来主要是一些概率的小题；版上都出现过
；还问了几
个积分、C++的一些基本概念。
【在 J****8 的大作中提到】
: 请问楼主二面都问了什么？
: C++？ 算法？ 随机 ？ BT ？
: 多谢？
|
z*****n
发帖数: 165 | 9 我当时是用复积分算的：计算 e^(ix)/x 积分的虚部；积分线在实轴上；由于实轴过奇
点，所以要shift: x-> x-iu, 留数定理后让u->0 |
e**********n
发帖数: 359 | 10 The answer depends on the sign of u if you use the residue theorem. However
thinking of real numbers only,
use the substitution
1/x -> \int_0^\infty e^{-xy} dy
the new 2D integral is easy to evaluate.
【在 z*****n 的大作中提到】
: 我当时是用复积分算的：计算 e^(ix)/x 积分的虚部；积分线在实轴上；由于实轴过奇
: 点，所以要shift: x-> x-iu, 留数定理后让u->0
|
|
|
p****q
发帖数: 149 | |
a*********r
发帖数: 139 | 12 Many ways to evaluate this integral. I'd love to use complex analysis. It's
a typical example in complex analysis. Just deform the contour integral. |
t*******g
发帖数: 373 | 13 仅此一题？！？！
【在 z*****n 的大作中提到】
: \int_-\inf^\inf Sin(x)/x dx
|
B****n
发帖数: 11290 | 14 說真的 這個要是沒看過類似的做法 決大多數人是想破頭也想不出來的
但如果你是數學系的 大學里至少就看過兩三種不同的做法
【在 r*******y 的大作中提到】
: 我怎么还没有想出来？ 只知道 \int_0^\inf sin(x) / x dx 是收敛的，
: 可以用 alternating series theorem证明收敛性。
: 其他的难道能求出 antiderivative?
|
r*******y
发帖数: 1081 | 15 Still something confusing. Using residual theorem we can show that
\int_{-inf}^{inf} e^{ix} / x dx = i* PI so we can have
\int_{-inf}^{inf} sin(x) / x dx =PI
but we also can have
\int_{-inf}^{inf} cos(x) / x dx = 0 which is very confusing
since \int_{-inf}^{inf} cos(x) / x dx will diverge
Just look at \int_{0}^{1} cos(x) / x dx which is infinity
【在 z*****n 的大作中提到】
: 我当时是用复积分算的：计算 e^(ix)/x 积分的虚部；积分线在实轴上；由于实轴过奇
: 点，所以要shift: x-> x-iu, 留数定理后让u->0
|
a*********r
发帖数: 139 | 16 Based on your answer, your contour is obviously wrong. Note 0 is a
singularity of the function f(z)=\frac{e^{iz}}{z}, you have to play a
typical trick to around 0 to get around this difficulty. If you know Jordan'
s lemma, then the remaining is trivial. Good luck!
【在 r*******y 的大作中提到】
: Still something confusing. Using residual theorem we can show that
: \int_{-inf}^{inf} e^{ix} / x dx = i* PI so we can have
: \int_{-inf}^{inf} sin(x) / x dx =PI
: but we also can have
: \int_{-inf}^{inf} cos(x) / x dx = 0 which is very confusing
: since \int_{-inf}^{inf} cos(x) / x dx will diverge
: Just look at \int_{0}^{1} cos(x) / x dx which is infinity
|
r*******y
发帖数: 1081 | 17 So what is the answer to
\int_{-inf}^{inf} e^{ix} / x dx ?
thanks.
Jordan'
【在 a*********r 的大作中提到】
: Based on your answer, your contour is obviously wrong. Note 0 is a
: singularity of the function f(z)=\frac{e^{iz}}{z}, you have to play a
: typical trick to around 0 to get around this difficulty. If you know Jordan'
: s lemma, then the remaining is trivial. Good luck!
|
z*****n
发帖数: 165 | 18 Due to the ambiguity, the notation in the problem really means:
\int_\-inf^\-epsilon + \int_\epsilon^\inf
and then take limit of \epsilon -> 0+
Which is the "v.p." value of integration.
Under this definition, there is no problem of cos(x)/x, because it's an
odd function, so both parts cancel out after integration.
【在 r*******y 的大作中提到】
: Still something confusing. Using residual theorem we can show that
: \int_{-inf}^{inf} e^{ix} / x dx = i* PI so we can have
: \int_{-inf}^{inf} sin(x) / x dx =PI
: but we also can have
: \int_{-inf}^{inf} cos(x) / x dx = 0 which is very confusing
: since \int_{-inf}^{inf} cos(x) / x dx will diverge
: Just look at \int_{0}^{1} cos(x) / x dx which is infinity
|
x******a
发帖数: 6336 | 19 sinx/x没问题
cosx/x问题很大
【在 z*****n 的大作中提到】
: Due to the ambiguity, the notation in the problem really means:
: \int_\-inf^\-epsilon + \int_\epsilon^\inf
: and then take limit of \epsilon -> 0+
: Which is the "v.p." value of integration.
: Under this definition, there is no problem of cos(x)/x, because it's an
: odd function, so both parts cancel out after integration.
|
r*******y
发帖数: 1081 | 20 I agree on this.
【在 x******a 的大作中提到】
: sinx/x没问题
: cosx/x问题很大
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r*******y
发帖数: 1081 | 21 a little more clear now.
using residual theorem, we can show \int_{-inf}^{-\epsilon} sin(x) / x dx
+ \int_{\epsilon}{inf} sin(x) / x dx (in the sense of v.p.)---> PI as
\epsilon ---> 0. But we also can show \int_{-inf}{inf} sin(x) / x dx
converge
so we can claim \int_{-inf}{inf} sin(x) / x dx = PI.
【在 z*****n 的大作中提到】
: Due to the ambiguity, the notation in the problem really means:
: \int_\-inf^\-epsilon + \int_\epsilon^\inf
: and then take limit of \epsilon -> 0+
: Which is the "v.p." value of integration.
: Under this definition, there is no problem of cos(x)/x, because it's an
: odd function, so both parts cancel out after integration.
|
a*********r
发帖数: 139 | 22 The question should be stated precisely as an improper integral. If you read
my last post, you have a removable singularity at 0, you cann't take that
contour.
For evaluation of the well-known integral, you can consult any standard
textbook on complex analysis, e.g. Ahlfors p158 or Conway 115.
Good luck!
【在 r*******y 的大作中提到】
: So what is the answer to
: \int_{-inf}^{inf} e^{ix} / x dx ?
: thanks.
:
: Jordan'
|
w******i
发帖数: 503 | 23 what is the purpose this kind of question ? what if you have not
studied complex analysis? |
s****p
发帖数: 19 | 24 Laplace transformation.
Let F(y)=\int_0^\inf e^{-yx} sin(x)/x dx then
-F'(y)=\int_0^\inf e^{-yx}sin(x) dx, which has close form.
【在 z*****n 的大作中提到】
: \int_-\inf^\inf Sin(x)/x dx
|
L*****k
发帖数: 327 | 25 Good post.
另外，complex analysis有结论：对于coutour积分具有形式f(z)/(z-a)(contour上有
一个奇点a)，可以直接转化为计 f(a)*\pi*j，在满足一些regular条件下，比如f(z)
is analytic inside the contour
read
【在 a*********r 的大作中提到】
: The question should be stated precisely as an improper integral. If you read
: my last post, you have a removable singularity at 0, you cann't take that
: contour.
: For evaluation of the well-known integral, you can consult any standard
: textbook on complex analysis, e.g. Ahlfors p158 or Conway 115.
: Good luck!
|
k***n
发帖数: 997 | 26 查了一下，是书上例题
\int_-inf^inf \frac{sin x}{x} converges not absolute
\int_c^inf \frac{cos x}{x} converges not absolute, where c>0
c <= 0 diverges. |
