watchmath
  • watchmath
Compute integral sin x /x from 0 to infinity.
Mathematics
jamiebookeater
  • jamiebookeater
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myininaya
  • myininaya
http://www.physicsforums.com/showthread.php?t=64859
M
  • M
doesn't have answer
myininaya
  • myininaya
no but it could help

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anonymous
  • anonymous
You have to memorize the solutions to these. Don't try to actually solve it. The answer is just \pi/2\
watchmath
  • watchmath
I know the result. I want to know how :D.
M
  • M
you'll have hard time getting help here for anything above pre-calculus
M
  • M
and elementary statistics :)
anonymous
  • anonymous
If you are currently in cal 1, you probably won't do the series stuff until the end of cal 2
anonymous
  • anonymous
I'm going to upload an attachment soon, just hold on. Integration by parts SUCKS
myininaya
  • myininaya
i think this involves advance calculus
anonymous
  • anonymous
This does involve advanced calculus... this is going to take longer than I thought
watchmath
  • watchmath
can you tell me the idea?
myininaya
  • myininaya
you can use laplace transform
myininaya
  • myininaya
http://answers.yahoo.com/question/index?qid=20080129204110AAnGtsz
anonymous
  • anonymous
This is going to use Taylor Series convergence
anonymous
  • anonymous
Honestly, I don't feel like proving that the Taylor Series converges. That would take about a week by hand for me (because I don't actually know how to do it yet). However, I'm assuming that it converges to pi/2
watchmath
  • watchmath
That's ok. I found it :D.
anonymous
  • anonymous
we can do this using complex analysis:\[\frac{1}{2}\int\limits_{-\infty}^{\infty}\frac{\sin(x)}{x}dx=\frac{1}{2}I \left( \int\limits_{-\infty}^{\infty}\frac{e^{ix}}{x}dx\right)\] this is the same as: \[I(\frac{1}{2} \pi i ( res_{0}\frac{e^{ix}}{x}))=I(\frac{\pi i}{2})=\frac{\pi}{2}\]
anonymous
  • anonymous
we need the residue at zero since we have a simple pole there

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