Question

improper integral. sorry for all of these. my textbook thinks that "converges" is a suitable explanation.

\[\int^{\infty}_0 \tan\frac{1}{x}\text{d}x\]

Answer 1

I need to say if it converges or diverges.

Answer 2

Take the integral as you would from [0, R]

Answer 3

can you change variables to \(u=\frac{1}{x}\) and then flip the limits of integration?

Answer 4

Then take the limit of the function as R->∞

Answer 5

is this sufficient to say it converges? \[\lim_{x \to \infty} \tan\frac{1}{x}=0\]

Answer 6

how?

Answer 7

thing thing has an infinite number of singularities

Answer 8

how about the substitution
\[x = \tan^{-1}u\]

Answer 9

oh right.

Answer 10

or \[x=\cot^{-1}(u)\]

Answer 11

the problem isn't just that the path is infinite,

Answer 12

it is undefined infinitely often

Answer 13

What about \(\tan\frac1x\sim\frac1x,\, x\rightarrow\infty\) ?

Answer 14

yep!! convergence....

Answer 15

i think there is a mistake here
there is no way this thing converges

Answer 16

\[\lim_{x\to0}\tan(1/x)=\infty\]

Answer 17

oh wait a second, it does diverge.

Answer 18

graph it and see what it looks like, then tell me it converges...

Answer 19

@satellite73 can we say that as \(x\to0\), it is infinite or is it undefined?

Answer 20

but the limits iof integration are supposed to be from 1 to infty. so ignore the zero

Answer 21

makes no difference, the problem is not the infinite path
tangent is undefined infinitely often

Answer 22

Use that \(\tan\frac1x\sim\frac1x,\, x\rightarrow\infty\) and immediately say that it diverges.

Answer 23

@richyw it also has infinite "vertical asymptotes" and hence it is discontinuous over the interval and cannot be integrated!

Answer 24

@satellite73 so, you are saying that the function cannot exist to begin with?

Answer 25

@klimenkov

is that just the first term of the taylor expansion at infinity?

Answer 26

If you want Taylor series - yes. But from \(\tan x\sim x,\,x\rightarrow0\) implies \(\tan\frac1x\sim\frac1x,\, x\rightarrow\infty\).

Answer 27

how am I supposed to know that \(\tan x \sim x,\;x\to 0\) ?

Answer 28

Compute\[\lim_{x\rightarrow0}\frac{\tan x}{x}\]

Answer 29

AWSOME. ok this is what I was struggling with last night.

Answer 30

why does doing that show that they behave the same?

Answer 31

that step is what really confuses me!

Answer 32

same step was used in the second line of this. I don't understand why I do this!

Answer 33

Try to check this theorem in your calculus book.

Answer 34

believe me I have tried. I even emailed my prof asking where the theorem is. does it have a name?

Answer 35

Ok. Did you understand why \(\tan\frac1x\sim\frac1x,\, x\rightarrow\infty\) ?

Answer 36

no, because I don't understand why \(\tan x \sim x,\;x\to 0\)

Answer 37

Do you know that \[\lim_{x\rightarrow0}\frac{\sin x}{x}=1\]

Answer 38

yes

Answer 39

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?

Answer 40

yes, I can compute the limit.

Answer 41

but I don't understand its significance

Answer 42

Ah. Ok.
If \(f(x)\sim g(x),\,x\rightarrow a\), it means that \(f(x)\) behaves like \(g(x)\) at the point \(x=a\).

Answer 43

right. do you know why?

Answer 44

I mean why the limit thing, not that definition

Answer 45

Or \[\lim_{x\rightarrow a}\frac{f(x)}{g(x)}=1\]

Answer 46

AHHH that's so obvious!

Answer 47

It is equivalent definitions.

Answer 48

omg. thank you so much.

Answer 49

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!

Answer 50

thank you so much!

Answer 51

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.