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quick question: there is no meaning in the concept of approaching infinity from the right, correct?
\(x\to\infty^+\)
 one year ago
 one year ago
quick question: there is no meaning in the concept of approaching infinity from the right, correct? \(x\to\infty^+\)
 one year ago
 one year ago

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TuringTestBest ResponseYou've already chosen the best response.2
or \(x\to\infty^\) wither, right?
 one year ago

mathsloverBest ResponseYou've already chosen the best response.0
When the variable is x, and it takes on only positive values, then it becomes positively infinite. We write \[x\to\infty^+\]
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
you gotta start really really far out i guess
 one year ago

mathsloverBest ResponseYou've already chosen the best response.0
http://www.themathpage.com/acalc/infinity.htm
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
@mathslover no, I think you are thinking of \(x\to+\infty\)
 one year ago

HeroBest ResponseYou've already chosen the best response.0
I'm still trying to figure out this: \[\int\limits \frac{x \dot\ \sin x}{1 + \cos^2x} dx\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
I mean, infinity is not a number even how do you approach a nonnumber from a higher number no number can be higher than a nonnumber seems like asking if 7>purple
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
@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
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.1
\(∞^\pm\) aren't even in the Extended real number line
 one year ago

HeroBest ResponseYou've already chosen the best response.0
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.
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
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
 one year ago

HeroBest ResponseYou've already chosen the best response.0
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
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
me too :) @zarkon actually taught me how to solve this, but he made use of the bounds heavily
 one year ago

HeroBest ResponseYou've already chosen the best response.0
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.
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
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
 one year ago

HeroBest ResponseYou've already chosen the best response.0
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
 one year ago

HeroBest ResponseYou've already chosen the best response.0
There are many indefinite integrals with solutions that are not finite.
 one year ago

TuringTestBest ResponseYou've already chosen the best response.2
true... I am stretching my capacities here admittedly
 one year ago

HeroBest ResponseYou've already chosen the best response.0
Well, if by finite you mean a specific value.
 one year ago
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