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Calc 3 Problem  Tricky Integral
\[\int\limits_{0}^{\infty}\frac{ \arctan \pi x \arctan x }{ x } dx\]
 one year ago
 one year ago
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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johnsonshelbyBest ResponseYou've already chosen the best response.0
Nevermind, haha The equation wasn't showing up correctly. It is now, hang on a minute
 one year ago

johnsonshelbyBest ResponseYou've already chosen the best response.0
Are you getting imaginary numbers with this?
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
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

johnsonshelbyBest ResponseYou've already chosen the best response.0
I'm getting something crazy like 1/2 (l1 i)... I don't think that is right, merp ... sorry :(
 one year ago

dpaIncBest ResponseYou've already chosen the best response.0
when you think about it, all problems are tricky... otherwise they wouldn't be problems... :)
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
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

johnsonshelbyBest ResponseYou've already chosen the best response.0
Good luck young grasshopper :)
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
http://www.physicsforums.com/showthread.php?t=610439 Hmmm trying to make sense of that...
 one year ago

johnsonshelbyBest ResponseYou've already chosen the best response.0
Oh, I love physicsforums :) Did you get it?
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
Nah it's hurting by head XD maybe ill take a look at it later :3
 one year ago

ZarkonBest ResponseYou've already chosen the best response.1
\[\int\limits_{0}^{\infty}\frac{ \arctan (\pi x) \arctan (x) }{ x } dx\] \[=\int\limits_{0}^{\infty}\int\limits_{x}^{\pi x}\frac{1}{1+y^2}\frac{1}{x}dydx\] \[=\int\limits_{0}^{\infty}\int\limits_{y/\pi}^{y}\frac{1}{1+y^2}\frac{1}{x}dxdy=\cdots\]
 one year ago

klimenkovBest ResponseYou've already chosen the best response.0
By the way, @zepdrix , can you speak Russian?
 one year ago

klimenkovBest ResponseYou've already chosen the best response.0
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

zepdrixBest ResponseYou've already chosen the best response.0
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

zepdrixBest ResponseYou've already chosen the best response.0
Google translating the text helped a little bit :D heh thanks
 one year ago

klimenkovBest ResponseYou've already chosen the best response.0
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

zepdrixBest ResponseYou've already chosen the best response.0
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

klimenkovBest ResponseYou've already chosen the best response.0
You can ask me anything, because russian mathematicians use sometimes another designation for popular objects in math.
 one year ago

ZarkonBest ResponseYou've already chosen the best response.1
\[\int\limits_{0}^{\infty}\frac{1}{1+y^2}dy=\lim_{t\to\infty}\int\limits_{0}^{t}\frac{1}{1+y^2}dy\] \[=\left.\lim_{t\to\infty}\arctan(y)\right _{0}^{t}=\lim_{t\to\infty}\arctan(t)\arctan(0)=\frac{\pi}{2}0=\frac{\pi}{2}\] combine with the above
 one year ago
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