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】![](/moin_static193/solenoid/img/up.png) : \int_-\inf^\inf Sin(x)/x dx
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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】![](/moin_static193/solenoid/img/up.png) : \int_-\inf^\inf Sin(x)/x dx
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b*******e 发帖数: 86 | 6 Dirichlet integral
【在 r*******y 的大作中提到】![](/moin_static193/solenoid/img/up.png) : 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?
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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】![](/moin_static193/solenoid/img/up.png) : 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】![](/moin_static193/solenoid/img/up.png) : 请问楼主二面都问了什么? : C++? 算法? 随机 ? BT ? : 多谢?
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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】![](/moin_static193/solenoid/img/up.png) : 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
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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】![](/moin_static193/solenoid/img/up.png) : \int_-\inf^\inf Sin(x)/x dx
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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】![](/moin_static193/solenoid/img/up.png) : 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?
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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】![](/moin_static193/solenoid/img/up.png) : 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
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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】![](/moin_static193/solenoid/img/up.png) : 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
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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】![](/moin_static193/solenoid/img/up.png) : 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!
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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】![](/moin_static193/solenoid/img/up.png) : 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
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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】![](/moin_static193/solenoid/img/up.png) : 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.
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r*******y 发帖数: 1081 | 20 I agree on this.
【在 x******a 的大作中提到】![](/moin_static193/solenoid/img/up.png) : 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】![](/moin_static193/solenoid/img/up.png) : 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.
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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】![](/moin_static193/solenoid/img/up.png) : So what is the answer to : \int_{-inf}^{inf} e^{ix} / x dx ? : thanks. : : Jordan'
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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】![](/moin_static193/solenoid/img/up.png) : \int_-\inf^\inf Sin(x)/x dx
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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】![](/moin_static193/solenoid/img/up.png) : 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!
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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. |