## richyw 3 years ago improper integral. sorry for all of these. my textbook thinks that "converges" is a suitable explanation. $\int^{\infty}_0 \tan\frac{1}{x}\text{d}x$

1. richyw

I need to say if it converges or diverges.

2. abb0t

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

3. anonymous

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

4. abb0t

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

5. richyw

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

6. anonymous

how?

7. anonymous

thing thing has an infinite number of singularities

8. anonymous

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

9. richyw

oh right.

10. anonymous

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

11. anonymous

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

12. anonymous

it is undefined infinitely often

13. klimenkov

What about $$\tan\frac1x\sim\frac1x,\, x\rightarrow\infty$$ ?

14. anonymous

yep!! convergence....

15. anonymous

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

16. anonymous

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

17. richyw

oh wait a second, it does diverge.

18. anonymous

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

19. anonymous

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

20. richyw

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

21. anonymous

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

22. klimenkov

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

23. anonymous

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

24. anonymous

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

25. richyw

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

26. klimenkov

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

27. richyw

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

28. klimenkov

Compute$\lim_{x\rightarrow0}\frac{\tan x}{x}$

29. richyw

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

30. richyw

why does doing that show that they behave the same?

31. richyw

that step is what really confuses me!

32. richyw

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

33. klimenkov

Try to check this theorem in your calculus book.

34. richyw

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

35. klimenkov

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

36. richyw

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

37. klimenkov

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

38. richyw

yes

39. klimenkov

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?

40. richyw

yes, I can compute the limit.

41. richyw

but I don't understand its significance

42. klimenkov

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

43. richyw

right. do you know why?

44. richyw

I mean why the limit thing, not that definition

45. klimenkov

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

46. richyw

AHHH that's so obvious!

47. klimenkov

It is equivalent definitions.

48. richyw

omg. thank you so much.

49. richyw

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!

50. richyw

thank you so much!

51. klimenkov

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.