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zepdrix

Calc 3 Problem - Tricky Integral \[\int\limits_{0}^{\infty}\frac{ \arctan \pi x -\arctan x }{ x } dx\]

  • one year ago
  • one year ago

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  1. johnsonshelby
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    n=0?

    • one year ago
  2. zepdrix
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    n...? :o

    • one year ago
  3. johnsonshelby
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    Nevermind, haha The equation wasn't showing up correctly. It is now, hang on a minute

    • one year ago
  4. johnsonshelby
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    Are you getting imaginary numbers with this?

    • one year ago
  5. zepdrix
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    Hmmm I don't think so :D And for the life of me, I can't remember the hint that the teacher gave us lol. Something aboutttttt try to rewrite it as the derivative of something else that is easier to integrate, orrrrrr maybe, rewrite it as a double integral somehow, blah i forget what he said XD lolol

    • one year ago
  6. johnsonshelby
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    I'm getting something crazy like -1/2 (l1 i)... I don't think that is right, merp ... sorry :(

    • one year ago
  7. dpaInc
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    when you think about it, all problems are tricky... otherwise they wouldn't be problems... :)

    • one year ago
  8. zepdrix
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    Hmm I found the problem on physicsforum.com. I might be able to get through this one if I just try hard enough c:

    • one year ago
  9. johnsonshelby
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    Good luck young grasshopper :)

    • one year ago
  10. zepdrix
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    http://www.physicsforums.com/showthread.php?t=610439 Hmmm trying to make sense of that...

    • one year ago
  11. johnsonshelby
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    Oh, I love physicsforums :) Did you get it?

    • one year ago
  12. zepdrix
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    Nah it's hurting by head XD maybe ill take a look at it later :3

    • one year ago
  13. Zarkon
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    \[\int\limits_{0}^{\infty}\frac{ \arctan (\pi x) -\arctan (x) }{ x } dx\] \[=\int\limits_{0}^{\infty}\int\limits_{x}^{\pi x}\frac{1}{1+y^2}\frac{1}{x}dydx\] \[=\int\limits_{0}^{\infty}\int\limits_{y/\pi}^{y}\frac{1}{1+y^2}\frac{1}{x}dxdy=\cdots\]

    • one year ago
  14. klimenkov
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    By the way, @zepdrix , can you speak Russian?

    • one year ago
  15. zepdrix
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    No sorry c:

    • one year ago
  16. klimenkov
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    Don't read the text, just see the formula. http://ru.wikipedia.org/wiki/%D0%A4%D0%BE%D1%80%D0%BC%D1%83%D0%BB%D1%8B_%D0%A4%D1%80%D1%83%D0%BB%D0%BB%D0%B0%D0%BD%D0%B8

    • one year ago
  17. zepdrix
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    This problem is starting to make sense, I'm trying to understand how Zarkon switched the integrals though. From the second to the third step, we are integrating with resepect to X first now, but how did he come up with the limits? :( hmm

    • one year ago
  18. zepdrix
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    Google translating the text helped a little bit :D heh thanks

    • one year ago
  19. klimenkov
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    The second formula is your case, \(f(x)=\arctan x\).\[\int\limits_{0}^{\infty}\frac{ \arctan \pi x -\arctan x }{ x } dx=(0-\frac{\pi}2)\ln\frac1\pi=\frac\pi2\ln\pi\]

    • one year ago
  20. zepdrix
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    Ya i need to try and show the steps though :d not just a shortcut. And the proof on that page is rather involved :3 heh

    • one year ago
  21. klimenkov
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    You can ask me anything, because russian mathematicians use sometimes another designation for popular objects in math.

    • one year ago
  22. zepdrix
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    |dw:1351978258980:dw|

    • one year ago
  23. zepdrix
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    |dw:1351978595163:dw|

    • one year ago
  24. Zarkon
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    \[\int\limits_{0}^{\infty}\frac{1}{1+y^2}dy=\lim_{t\to\infty}\int\limits_{0}^{t}\frac{1}{1+y^2}dy\] \[=\left.\lim_{t\to\infty}\arctan(y)\right| _{0}^{t}=\lim_{t\to\infty}\arctan(t)-\arctan(0)=\frac{\pi}{2}-0=\frac{\pi}{2}\] combine with the above

    • one year ago
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