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improper integral. sorry for all of these. my textbook thinks that "converges" is a suitable explanation. \[\int^{\infty}_0 \tan\frac{1}{x}\text{d}x\]

Mathematics
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I need to say if it converges or diverges.
Take the integral as you would from [0, R]
can you change variables to \(u=\frac{1}{x}\) and then flip the limits of integration?

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Other answers:

Then take the limit of the function as R->∞
is this sufficient to say it converges? \[\lim_{x \to \infty} \tan\frac{1}{x}=0\]
how?
thing thing has an infinite number of singularities
how about the substitution \[x = \tan^{-1}u\]
oh right.
or \[x=\cot^{-1}(u)\]
the problem isn't just that the path is infinite,
it is undefined infinitely often
What about \(\tan\frac1x\sim\frac1x,\, x\rightarrow\infty\) ?
yep!! convergence....
i think there is a mistake here there is no way this thing converges
\[\lim_{x\to0}\tan(1/x)=\infty\]
oh wait a second, it does diverge.
graph it and see what it looks like, then tell me it converges...
@satellite73 can we say that as \(x\to0\), it is infinite or is it undefined?
but the limits iof integration are supposed to be from 1 to infty. so ignore the zero
makes no difference, the problem is not the infinite path tangent is undefined infinitely often
Use that \(\tan\frac1x\sim\frac1x,\, x\rightarrow\infty\) and immediately say that it diverges.
@richyw it also has infinite "vertical asymptotes" and hence it is discontinuous over the interval and cannot be integrated!
@satellite73 so, you are saying that the function cannot exist to begin with?
@klimenkov is that just the first term of the taylor expansion at infinity?
If you want Taylor series - yes. But from \(\tan x\sim x,\,x\rightarrow0\) implies \(\tan\frac1x\sim\frac1x,\, x\rightarrow\infty\).
how am I supposed to know that \(\tan x \sim x,\;x\to 0\) ?
Compute\[\lim_{x\rightarrow0}\frac{\tan x}{x}\]
AWSOME. ok this is what I was struggling with last night.
why does doing that show that they behave the same?
that step is what really confuses me!
same step was used in the second line of this. I don't understand why I do this!
Try to check this theorem in your calculus book.
believe me I have tried. I even emailed my prof asking where the theorem is. does it have a name?
Ok. Did you understand why \(\tan\frac1x\sim\frac1x,\, x\rightarrow\infty\) ?
no, because I don't understand why \(\tan x \sim x,\;x\to 0\)
Do you know that \[\lim_{x\rightarrow0}\frac{\sin x}{x}=1\]
yes
Now \[\lim_{x\rightarrow0}\frac{\tan x}{x}=\lim_{x\rightarrow0}\frac{\frac{\sin x}{\cos x}}{x}=\lim_{x\rightarrow0}\frac{\sin x}{x}=1\]because \(\lim_{x\rightarrow0}\cos x=1\). Got it?
yes, I can compute the limit.
but I don't understand its significance
Ah. Ok. If \(f(x)\sim g(x),\,x\rightarrow a\), it means that \(f(x)\) behaves like \(g(x)\) at the point \(x=a\).
right. do you know why?
I mean why the limit thing, not that definition
Or \[\lim_{x\rightarrow a}\frac{f(x)}{g(x)}=1\]
AHHH that's so obvious!
It is equivalent definitions.
omg. thank you so much.
I feel so stupid now, but if you looked at my questions history I spent like 3 hours last night trying to get this explanation!
thank you so much!
Don't be shy to ask. It is normal not to see obvious things because obvious is a relative concept. I wish you a lot of luck in learning math. Love it.

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