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experimentX Group Title

Evaluate: \[ \displaystyle \int_0^{\infty} \dfrac{(\log x)^2}{x^2 + 1} dx \]

  • 2 years ago
  • 2 years ago

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  1. experimentX Group Title
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    http://tinyurl.com/9csdea3

    • 2 years ago
  2. experimentX Group Title
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    the problem hasn't been solved http://math.stackexchange.com/questions/191736/help-with-integrating-displaystyle-int-0-infty-dfrac-log-x2x2-1

    • 2 years ago
  3. mukushla Group Title
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    *

    • 2 years ago
  4. mukushla Group Title
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    going with \[t=\log x\]\[\displaystyle \int_{-\infty}^{\infty} \dfrac{t^2e^t}{1+ e^{2t}} \text{d}t=\int_{-\infty}^{0} \dfrac{t^2e^t}{1+ e^{2t}} \text{d}t+\int_{0}^{\infty} \dfrac{t^2e^t}{1+ e^{2t}} \text{d}t\]first one for example\[\int_{-\infty}^{0} \dfrac{t^2e^t}{1+ e^{2t}} \text{d}t=\int_{-\infty}^{0} t^2e^t \sum_{n=0}^{\infty}(-1)^ne^{2nt} \text{d}t= \sum_{n=0}^{\infty} (-1)^n\int_{-\infty}^{0} t^2e^te^{2nt} \text{d}t\\=\sum_{n=0}^{\infty} (-1)^n \frac{2}{(2n+1)^3}\]

    • 2 years ago
  5. mukushla Group Title
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    but i'd like to know how can we go with complex integration

    • 2 years ago
  6. experimentX Group Title
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    the series looks like Fourier expansion.

    • 2 years ago
  7. mukushla Group Title
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    yeah

    • 2 years ago
  8. experimentX Group Title
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    hold on .. is that 3?

    • 2 years ago
  9. mukushla Group Title
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    lol .. yes

    • 2 years ago
  10. mukushla Group Title
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    this one is reachable by fourier series i think

    • 2 years ago
  11. experimentX Group Title
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    you made things a bit more complicated http://www.wolframalpha.com/input/?i=sum[1%2F%282n%2B1%29^3%2C+{n%2C+0%2C+Infinity}]

    • 2 years ago
  12. experimentX Group Title
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    but this seems nice http://www.wolframalpha.com/input/?i=sum[%28-1%29^n%2F%282n%2B1%29^3%2C+{n%2C+0%2C+Infinity}]

    • 2 years ago
  13. mukushla Group Title
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    i made it worser

    • 2 years ago
  14. experimentX Group Title
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    No not really ,,, this is nice ... and interesting http://www.wolframalpha.com/input/?i=sum[%28-1%29^n%2F%282n%2B1%29^3%2C+{n%2C+1%2C+Infnity}]

    • 2 years ago
  15. mukushla Group Title
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    ahh yeah so this is reachable man

    • 2 years ago
  16. experimentX Group Title
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    probably some nasty Fourier analysis.

    • 2 years ago
  17. eliassaab Group Title
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    *

    • 2 years ago
  18. eliassaab Group Title
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    Hae you tried using the residue theorem?

    • 2 years ago
  19. experimentX Group Title
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    tried ... but stuck. I need picture of contour .... that t^2 term is bugging me badly http://openstudy.com/users/experimentx#/updates/50476f64e4b0c3bb09860ba6

    • 2 years ago
  20. eliassaab Group Title
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    See http://en.wikipedia.org/wiki/Methods_of_contour_integration Example (V)

    • 2 years ago
  21. experimentX Group Title
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    thanks ... i this is helpful ...

    • 2 years ago
  22. experimentX Group Title
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    *think

    • 2 years ago
  23. experimentX Group Title
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    i'll try it ... if it get answer i'll post solution.

    • 2 years ago
  24. mukushla Group Title
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    |dw:1347643651001:dw|

    • 2 years ago
  25. mukushla Group Title
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    man this contour works

    • 2 years ago
  26. experimentX Group Title
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    did you try it?

    • 2 years ago
  27. mukushla Group Title
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    yes

    • 2 years ago
  28. experimentX Group Title
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    oh great ... but I have QM exam six days later.

    • 2 years ago
  29. mukushla Group Title
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    \[\oint \frac{z^2 e^z}{1+e^{2z}} dz=a_{-1}\]

    • 2 years ago
  30. mukushla Group Title
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    man try it after exams

    • 2 years ago
  31. experimentX Group Title
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    sure ... 14 days to go.

    • 2 years ago
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