zepdrix
  • zepdrix
Calc 3 Problem - Tricky Integral \[\int\limits_{0}^{\infty}\frac{ \arctan \pi x -\arctan x }{ x } dx\]
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
n=0?
zepdrix
  • zepdrix
n...? :o
anonymous
  • anonymous
Nevermind, haha The equation wasn't showing up correctly. It is now, hang on a minute

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
Are you getting imaginary numbers with this?
zepdrix
  • zepdrix
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
anonymous
  • anonymous
I'm getting something crazy like -1/2 (l1 i)... I don't think that is right, merp ... sorry :(
anonymous
  • anonymous
when you think about it, all problems are tricky... otherwise they wouldn't be problems... :)
zepdrix
  • zepdrix
Hmm I found the problem on physicsforum.com. I might be able to get through this one if I just try hard enough c:
anonymous
  • anonymous
Good luck young grasshopper :)
zepdrix
  • zepdrix
http://www.physicsforums.com/showthread.php?t=610439 Hmmm trying to make sense of that...
anonymous
  • anonymous
Oh, I love physicsforums :) Did you get it?
zepdrix
  • zepdrix
Nah it's hurting by head XD maybe ill take a look at it later :3
Zarkon
  • Zarkon
\[\int\limits_{0}^{\infty}\frac{ \arctan (\pi x) -\arctan (x) }{ x } dx\] \[=\int\limits_{0}^{\infty}\int\limits_{x}^{\pi x}\frac{1}{1+y^2}\frac{1}{x}dydx\] \[=\int\limits_{0}^{\infty}\int\limits_{y/\pi}^{y}\frac{1}{1+y^2}\frac{1}{x}dxdy=\cdots\]
klimenkov
  • klimenkov
By the way, @zepdrix , can you speak Russian?
zepdrix
  • zepdrix
No sorry c:
klimenkov
  • klimenkov
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
zepdrix
  • zepdrix
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
zepdrix
  • zepdrix
Google translating the text helped a little bit :D heh thanks
klimenkov
  • klimenkov
The second formula is your case, \(f(x)=\arctan x\).\[\int\limits_{0}^{\infty}\frac{ \arctan \pi x -\arctan x }{ x } dx=(0-\frac{\pi}2)\ln\frac1\pi=\frac\pi2\ln\pi\]
zepdrix
  • zepdrix
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
klimenkov
  • klimenkov
You can ask me anything, because russian mathematicians use sometimes another designation for popular objects in math.
zepdrix
  • zepdrix
|dw:1351978258980:dw|
zepdrix
  • zepdrix
|dw:1351978595163:dw|
Zarkon
  • Zarkon
\[\int\limits_{0}^{\infty}\frac{1}{1+y^2}dy=\lim_{t\to\infty}\int\limits_{0}^{t}\frac{1}{1+y^2}dy\] \[=\left.\lim_{t\to\infty}\arctan(y)\right| _{0}^{t}=\lim_{t\to\infty}\arctan(t)-\arctan(0)=\frac{\pi}{2}-0=\frac{\pi}{2}\] combine with the above

Looking for something else?

Not the answer you are looking for? Search for more explanations.