## anonymous one year ago Evaluate the integral..

1. anonymous

2. anonymous

is arctan possible here

3. anonymous

$\int\limits _{-\infty }^{\infty }\:\left(\frac{dx}{e^x+e^{-x}}\right)$

4. Zale101

Yes, it is possible

5. freckles

I would make it more obvious that is so by multiplying e^x on top and bottom

6. anonymous

it is in fact arctan in one step right?

7. anonymous

The problem is I don't know how first time seeing such problem$\int\limits \frac{du}{a^2+u^2}=\frac{1}{a}\tan^{-1}\left(\frac{u}{a}\right)+C$ took this from my formula ntbk

8. freckles

$\int\limits_{}^{}\frac{ du}{u^2+1} \text{ where } u=e^x$

9. freckles

the lower limit of integration changes since as x goes to -inf then u goes to 0

10. anonymous

@freckles that is clever, i would have taking $\lim_{t\to -\infty}\tan^{-1}(e^t)$

11. anonymous

*taken

12. anonymous

what about e^-x do I have to bring that up?

13. freckles

who cares we aren't English snobs we are math snobs @satellite73

14. anonymous

$\frac{e^x}{e^x}$

15. freckles

$\int\limits_{-\infty}^{\infty} \frac{1}{e^{x}+e^{-x} } dx \\ \text{ multiply \top and bottom by } e^{x} \\ \int\limits_{-\infty}^{\infty} \frac{e^{x}}{e^{2x}+1} dx \\ \text{ \let } u=e^{x} \\ du=e^{x} dx \\ \\ \int\limits_0^\infty \frac{du }{u^2+1}$

16. freckles

@arvnoodle that have written out does it make more sense or still no sensE?

17. anonymous

Wait I'll analyze it

18. anonymous

Why do you have to multiply e^x? Is that algebra or what, you do that if you want to bring something up to numerator?

19. Zale101

|dw:1438745594477:dw|

20. freckles

$\frac{e^x}{e^{x}}=1 \\ \text{ I multiplied by this fancy 1 to put it in the obvious } \arctan \text{ integral form }$

21. freckles

yes that is algebra

22. anonymous

I noticed the lower limit changed to 0 why is that

23. freckles

as x approaches -inf e^x approaches 0

24. anonymous

Oh okay tried it in my calcu it reaches 0 so that make sense

25. freckles

$u=e^x \\ x \rightarrow \infty \text{ then } u=e^{x} \rightarrow \infty \\ x \rightarrow -\infty \text{ then } u=e^{x} \rightarrow 0$

26. freckles

|dw:1438745936722:dw| notice as x gets infinitely negative large the y values are getting close to 0

27. anonymous

$\int\limits _0^{\infty }\:\frac{du}{u^2+1}$ sooo

28. freckles

you can integrate that correct? you actually gave a formula for that earlier and other people had mentioned something about the pretty arctan( ) thingy

29. anonymous

$\frac{1}{a}\tan^{-1}\left(\frac{u}{a}\right)+C$

30. anonymous

is this the one? is my formula right?

31. freckles

but a is 1 and we don't need the C because we have a definite integral

32. freckles

$\int\limits_0^\infty \frac{1}{u^2+1} du=\lim_{z \rightarrow \infty}[\arctan(u)|_0^{z}]$

33. anonymous

Oh okay so arctan(e^x) ? with limits of 0 and inf

34. freckles

well if you change it back to arctan(e^(x)) then you need to go back to the original limits

35. anonymous

So will the final answer be infinity?

36. freckles

no

37. freckles

$\int\limits\limits_0^\infty \frac{1}{u^2+1} du=\lim_{z \rightarrow \infty}[\arctan(u)|_0^{z}] \\ =\lim_{z \rightarrow \infty}[\arctan(z)-\arctan(0)] \\ =\lim_{z \rightarrow \infty}\arctan(z)$ y=tan(x) on (-pi/2,pi/2) looks like: |dw:1438746457408:dw| so y=arctan(x) on (-inf,inf) looks like: |dw:1438746492037:dw|

38. freckles

so looking at this y=arctan(x) on (-inf,inf) in your graphing calculator you should see as x gets super large then the y values are approaching y=pi/2

39. freckles

anyways can you tell me the following limit: $\lim_{z \rightarrow \infty}\arctan(z)$ it should be obvious from the graph and what I just said that this is ....

40. anonymous

okay let me get this cllear for me

41. anonymous

$\left[\tan^{-1}\left(\frac{e^x}{1}\right)\right]$

42. anonymous

that's right? ^

43. freckles

if we put it back in terms of x then we have to go back to the original limits

44. anonymous

okay -inf to +inf

45. freckles

which is fine

46. freckles

but you will have to break that integral up

47. anonymous

the -inf turns to 0

48. anonymous

Hmmm break?

49. anonymous

oh my this problem kind of hard lol sorry for not getting it quickly

50. freckles

$\int\limits_{-\infty}^{\infty} \frac{1}{e^x+e^{-x} } dx \\ =\int\limits_{-\infty}^0 \frac{1}{e^{x}+e^{-x} } dx+\int\limits_0^\infty \frac{1}{e^{x}+e^{-x} } dx \\ \text{ check to see both integrals converge }$ which we know they will because we already seen the answer another way so the first integral will converge to the sum of the other two integrals in this case

51. freckles

$= \lim_{u \rightarrow -\infty}[\arctan(e^{x})|_u^0]+\lim_{z \rightarrow \infty}[\arctan(e^{x})|_0^z]$

52. anonymous

53. freckles

yes

54. anonymous

I'm unsatisfied because by my logic the e^x reaches inf it means the answer is inf

55. anonymous

it doesn't make sense pi/2 to me but ohwell

56. freckles

as x gets large then e^x gets large then arctan(getting super large)=?

57. freckles

|dw:1438747311900:dw| look at the graph of y=arctan(x) as x gets supper large y approaches pi/2

58. freckles

super!

59. anonymous

oh I get by graph!!!

60. anonymous

oh my does it mean I need to graph when I'm answering this kind of question, kind of tedious lol

61. anonymous

but okay thankyou tho

62. freckles

well you eventually won't need to graph every thing you will just remember the end behavior of most graphs

63. freckles

the graph was more of me trying to convince you that it is pi/2 it is not needed in the answer

64. UnkleRhaukus

as the opposite side of a right angled triangle approaches infinitely greater in length than the adjacent side, the considered angle approaches 90° = π/2

65. anonymous

I got it now I got now!! by chain rule