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zepdrix Group Title

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

  • 2 years ago
  • 2 years ago

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  1. johnsonshelby Group Title
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    n=0?

    • 2 years ago
  2. zepdrix Group Title
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    n...? :o

    • 2 years ago
  3. johnsonshelby Group Title
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    Nevermind, haha The equation wasn't showing up correctly. It is now, hang on a minute

    • 2 years ago
  4. johnsonshelby Group Title
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    Are you getting imaginary numbers with this?

    • 2 years ago
  5. zepdrix Group Title
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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

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

    • 2 years ago
  7. dpaInc Group Title
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    when you think about it, all problems are tricky... otherwise they wouldn't be problems... :)

    • 2 years ago
  8. zepdrix Group Title
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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:

    • 2 years ago
  9. johnsonshelby Group Title
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    Good luck young grasshopper :)

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

    • 2 years ago
  11. johnsonshelby Group Title
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    Oh, I love physicsforums :) Did you get it?

    • 2 years ago
  12. zepdrix Group Title
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    Nah it's hurting by head XD maybe ill take a look at it later :3

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

    • 2 years ago
  14. klimenkov Group Title
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    By the way, @zepdrix , can you speak Russian?

    • 2 years ago
  15. zepdrix Group Title
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    No sorry c:

    • 2 years ago
  16. klimenkov Group Title
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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

    • 2 years ago
  17. zepdrix Group Title
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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

    • 2 years ago
  18. zepdrix Group Title
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    Google translating the text helped a little bit :D heh thanks

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

    • 2 years ago
  20. zepdrix Group Title
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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

    • 2 years ago
  21. klimenkov Group Title
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    You can ask me anything, because russian mathematicians use sometimes another designation for popular objects in math.

    • 2 years ago
  22. zepdrix Group Title
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    |dw:1351978258980:dw|

    • 2 years ago
  23. zepdrix Group Title
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    |dw:1351978595163:dw|

    • 2 years ago
  24. Zarkon Group Title
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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

    • 2 years ago
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