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

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n...? :o
Nevermind, haha The equation wasn't showing up correctly. It is now, hang on a minute

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Other answers:

Are you getting imaginary numbers with this?
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
I'm getting something crazy like -1/2 (l1 i)... I don't think that is right, merp ... sorry :(
when you think about it, all problems are tricky... otherwise they wouldn't be problems... :)
Hmm I found the problem on I might be able to get through this one if I just try hard enough c:
Good luck young grasshopper :) Hmmm trying to make sense of that...
Oh, I love physicsforums :) Did you get it?
Nah it's hurting by head XD maybe ill take a look at it later :3
\[\int\limits_{0}^{\infty}\frac{ \arctan (\pi x) -\arctan (x) }{ x } dx\] \[=\int\limits_{0}^{\infty}\int\limits_{x}^{\pi x}\frac{1}{1+y^2}\frac{1}{x}dydx\] \[=\int\limits_{0}^{\infty}\int\limits_{y/\pi}^{y}\frac{1}{1+y^2}\frac{1}{x}dxdy=\cdots\]
By the way, @zepdrix , can you speak Russian?
No sorry c:
Don't read the text, just see the formula.
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
Google translating the text helped a little bit :D heh thanks
The second formula is your case, \(f(x)=\arctan x\).\[\int\limits_{0}^{\infty}\frac{ \arctan \pi x -\arctan x }{ x } dx=(0-\frac{\pi}2)\ln\frac1\pi=\frac\pi2\ln\pi\]
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
You can ask me anything, because russian mathematicians use sometimes another designation for popular objects in math.
\[\int\limits_{0}^{\infty}\frac{1}{1+y^2}dy=\lim_{t\to\infty}\int\limits_{0}^{t}\frac{1}{1+y^2}dy\] \[=\left.\lim_{t\to\infty}\arctan(y)\right| _{0}^{t}=\lim_{t\to\infty}\arctan(t)-\arctan(0)=\frac{\pi}{2}-0=\frac{\pi}{2}\] combine with the above

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