## TuringTest 3 years ago quick question: there is no meaning in the concept of approaching infinity from the right, correct? $$x\to\infty^+$$

1. TuringTest

or $$x\to-\infty^-$$ wither, right?

2. TuringTest

either*

3. mathslover

When the variable is x, and it takes on only positive values, then it becomes positively infinite. We write $x\to\infty^+$

4. satellite73

you gotta start really really far out i guess

5. mathslover
6. TuringTest

@mathslover no, I think you are thinking of $$x\to+\infty$$

7. mathslover

oh ... yes .

8. Hero

I'm still trying to figure out this: $\int\limits \frac{x \dot\ \sin x}{1 + \cos^2x} dx$

9. TuringTest

I mean, infinity is not a number even how do you approach a non-number from a higher number no number can be higher than a non-number seems like asking if 7>purple

10. TuringTest

@Hero it has been solved, and I will be happy to give you hints should you desire let me note that that integral is unsolvalble (at least by me) as an indefinite integral, you must put the bounds

11. UnkleRhaukus

$$∞^\pm$$ aren't even in the Extended real number line

12. TuringTest

good point^

13. Hero

I want to solve it without the bounds. I already saw the solution with bounds. Since the function is continuous I will assume an indefinite integral exists.

14. TuringTest

Good luck with that! I don't know how to solve it without making use of the interval 0 to pi that it is supposed to be

15. Hero

Apparently no one or no thing does. Not even mathematica, geogebra, maple or TI. But not to worry, I know one person who for sure will be able to tell if it is possible. If it is integrable this guy will know. I kinda miss @JamesJ

16. TuringTest

me too :) @zarkon actually taught me how to solve this, but he made use of the bounds heavily

17. Hero

I contacted TI about this today and they wanted me to provide them with the solution to this so that they could use it to update their algorithm.

18. TuringTest

wow, far out I remember one thinbg was that we proved that$\int_0^\infty{x\sin x\over1+\cos^2x}dx$does not have a finite value, though I forget how we proved it. that suggests to me that no closed form of the indefinite integral exists, but I could be wrong

19. Hero

Hmmm interesting. It may not have a finite value, but that doesn't mean it can't be solved. Last time I checked, if you take the integral of something and get sin(x) as the result....well sin(x) isn't "finite" either, but it still exists as a solution

20. Hero

There are many indefinite integrals with solutions that are not finite.

21. TuringTest

true... I am stretching my capacities here admittedly

22. Hero

Well, if by finite you mean a specific value.