Here's the question you clicked on:
TuringTest
quick question: there is no meaning in the concept of approaching infinity from the right, correct? \(x\to\infty^+\)
or \(x\to-\infty^-\) wither, right?
When the variable is x, and it takes on only positive values, then it becomes positively infinite. We write \[x\to\infty^+\]
you gotta start really really far out i guess
@mathslover no, I think you are thinking of \(x\to+\infty\)
I'm still trying to figure out this: \[\int\limits \frac{x \dot\ \sin x}{1 + \cos^2x} dx\]
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
@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
\(∞^\pm\) aren't even in the Extended real number line
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.
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
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
me too :) @zarkon actually taught me how to solve this, but he made use of the bounds heavily
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.
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
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
There are many indefinite integrals with solutions that are not finite.
true... I am stretching my capacities here admittedly
Well, if by finite you mean a specific value.