## zepdrix 3 years ago Calc 3 Problem - Tricky Integral $\int\limits_{0}^{\infty}\frac{ \arctan \pi x -\arctan x }{ x } dx$

1. johnsonshelby

n=0?

2. zepdrix

n...? :o

3. johnsonshelby

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

4. johnsonshelby

Are you getting imaginary numbers with this?

5. 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

6. johnsonshelby

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

7. dpaInc

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

8. 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:

9. johnsonshelby

Good luck young grasshopper :)

10. zepdrix

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

11. johnsonshelby

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

12. zepdrix

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

13. Zarkon

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

14. klimenkov

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

15. zepdrix

No sorry c:

16. 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

17. 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

18. zepdrix

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

19. klimenkov

The second formula is your case, $$f(x)=\arctan x$$.$\int\limits_{0}^{\infty}\frac{ \arctan \pi x -\arctan x }{ x } dx=(0-\frac{\pi}2)\ln\frac1\pi=\frac\pi2\ln\pi$

20. 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

21. klimenkov

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

22. zepdrix

|dw:1351978258980:dw|

23. zepdrix

|dw:1351978595163:dw|

24. 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

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