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zepdrix

  • 2 years ago

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

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  1. johnsonshelby
    • 2 years ago
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    n=0?

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

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

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

  5. zepdrix
    • 2 years ago
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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

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

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

  8. zepdrix
    • 2 years ago
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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:

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

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

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

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

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

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

  15. zepdrix
    • 2 years ago
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    No sorry c:

  16. klimenkov
    • 2 years ago
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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

  17. zepdrix
    • 2 years ago
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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

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

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

  20. zepdrix
    • 2 years ago
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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

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

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

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

  24. Zarkon
    • 2 years ago
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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

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