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watchmath

  • 5 years ago

Compute integral sin x /x from 0 to infinity.

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  1. myininaya
    • 5 years ago
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    http://www.physicsforums.com/showthread.php?t=64859

  2. M
    • 5 years ago
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    doesn't have answer

  3. myininaya
    • 5 years ago
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    no but it could help

  4. anonymous
    • 5 years ago
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    You have to memorize the solutions to these. Don't try to actually solve it. The answer is just \pi/2\

  5. watchmath
    • 5 years ago
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    I know the result. I want to know how :D.

  6. M
    • 5 years ago
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    you'll have hard time getting help here for anything above pre-calculus

  7. M
    • 5 years ago
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    and elementary statistics :)

  8. anonymous
    • 5 years ago
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    If you are currently in cal 1, you probably won't do the series stuff until the end of cal 2

  9. anonymous
    • 5 years ago
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    I'm going to upload an attachment soon, just hold on. Integration by parts SUCKS

  10. myininaya
    • 5 years ago
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    i think this involves advance calculus

  11. anonymous
    • 5 years ago
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    This does involve advanced calculus... this is going to take longer than I thought

  12. watchmath
    • 5 years ago
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    can you tell me the idea?

  13. myininaya
    • 5 years ago
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    you can use laplace transform

  14. myininaya
    • 5 years ago
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    http://answers.yahoo.com/question/index?qid=20080129204110AAnGtsz

  15. anonymous
    • 5 years ago
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    This is going to use Taylor Series convergence

  16. anonymous
    • 5 years ago
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    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

  17. watchmath
    • 5 years ago
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    That's ok. I found it :D.

  18. anonymous
    • 5 years ago
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    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: \[I(\frac{1}{2} \pi i ( res_{0}\frac{e^{ix}}{x}))=I(\frac{\pi i}{2})=\frac{\pi}{2}\]

  19. anonymous
    • 5 years ago
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    we need the residue at zero since we have a simple pole there

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