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

1. richyw Group Title

I need to say if it converges or diverges.

2. abb0t Group Title

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

3. satellite73 Group Title

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

4. abb0t Group Title

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

5. richyw Group Title

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

6. satellite73 Group Title

how?

7. satellite73 Group Title

thing thing has an infinite number of singularities

8. electrokid Group Title

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

9. richyw Group Title

oh right.

10. electrokid Group Title

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

11. satellite73 Group Title

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

12. satellite73 Group Title

it is undefined infinitely often

13. klimenkov Group Title

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

14. electrokid Group Title

yep!! convergence....

15. satellite73 Group Title

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

16. electrokid Group Title

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

17. richyw Group Title

oh wait a second, it does diverge.

18. satellite73 Group Title

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

19. electrokid Group Title

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

20. richyw Group Title

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

21. satellite73 Group Title

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

22. klimenkov Group Title

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

23. electrokid Group Title

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

24. electrokid Group Title

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

25. richyw Group Title

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

26. klimenkov Group Title

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

27. richyw Group Title

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

28. klimenkov Group Title

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

29. richyw Group Title

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

30. richyw Group Title

why does doing that show that they behave the same?

31. richyw Group Title

that step is what really confuses me!

32. richyw Group Title

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

33. klimenkov Group Title

Try to check this theorem in your calculus book.

34. richyw Group Title

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

35. klimenkov Group Title

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

36. richyw Group Title

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

37. klimenkov Group Title

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

38. richyw Group Title

yes

39. klimenkov Group Title

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 Group Title

yes, I can compute the limit.

41. richyw Group Title

but I don't understand its significance

42. klimenkov Group Title

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 Group Title

right. do you know why?

44. richyw Group Title

I mean why the limit thing, not that definition

45. klimenkov Group Title

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

46. richyw Group Title

AHHH that's so obvious!

47. klimenkov Group Title

It is equivalent definitions.

48. richyw Group Title

omg. thank you so much.

49. richyw Group Title

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 Group Title

thank you so much!

51. klimenkov Group Title

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.