zepdrix Group Title Calc 3 Problem - Tricky Integral $\int\limits_{0}^{\infty}\frac{ \arctan \pi x -\arctan x }{ x } dx$ one year ago one year ago

1. johnsonshelby Group Title

n=0?

2. zepdrix Group Title

n...? :o

3. johnsonshelby Group Title

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

4. johnsonshelby Group Title

Are you getting imaginary numbers with this?

5. zepdrix Group Title

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 Group Title

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

7. dpaInc Group Title

when you think about it, all problems are tricky... otherwise they wouldn't be problems... :)

8. zepdrix Group Title

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 Group Title

Good luck young grasshopper :)

10. zepdrix Group Title

http://www.physicsforums.com/showthread.php?t=610439 Hmmm trying to make sense of that...

11. johnsonshelby Group Title

Oh, I love physicsforums :) Did you get it?

12. zepdrix Group Title

Nah it's hurting by head XD maybe ill take a look at it later :3

13. Zarkon Group Title

$\int\limits_{0}^{\infty}\frac{ \arctan (\pi x) -\arctan (x) }{ x } dx$ $=\int\limits_{0}^{\infty}\int\limits_{x}^{\pi x}\frac{1}{1+y^2}\frac{1}{x}dydx$ $=\int\limits_{0}^{\infty}\int\limits_{y/\pi}^{y}\frac{1}{1+y^2}\frac{1}{x}dxdy=\cdots$

14. klimenkov Group Title

By the way, @zepdrix , can you speak Russian?

15. zepdrix Group Title

No sorry c:

16. klimenkov Group Title

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 Group Title

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 Group Title

Google translating the text helped a little bit :D heh thanks

19. klimenkov Group Title

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 Group Title

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 Group Title

You can ask me anything, because russian mathematicians use sometimes another designation for popular objects in math.

22. zepdrix Group Title

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23. zepdrix Group Title

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24. Zarkon Group Title

$\int\limits_{0}^{\infty}\frac{1}{1+y^2}dy=\lim_{t\to\infty}\int\limits_{0}^{t}\frac{1}{1+y^2}dy$ $=\left.\lim_{t\to\infty}\arctan(y)\right| _{0}^{t}=\lim_{t\to\infty}\arctan(t)-\arctan(0)=\frac{\pi}{2}-0=\frac{\pi}{2}$ combine with the above