## agentx5 Group Title Topic: $$\textbf{Calculus 2}$$ Evaluate the indefinite integral as an infinite series. $\int \frac{\cos x − 1}{x} \ \ \ dx$ What the heck? I mean putting these in Taylor series forms are hard enough, but this? Not even sure where to start here >_< It worries me because the final is coming up and this is a hole in my knowledge... So, um, help? 2 years ago 2 years ago

1. telliott99 Group Title

2. telliott99 Group Title

You'd have (1/x) (-x^2/2! + x^4/4! .....)

3. Spacelimbus Group Title

$\int \frac{\cos x}{x}dx - \int \frac{1}{x}dx$

4. agentx5 Group Title

Let's see... $\sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!}, \text{for all x = } \mathbb{R}$

5. agentx5 Group Title

That's the MacLaurin equivalent for cosine

6. Spacelimbus Group Title

$\frac{ \cos x}{x} = \frac{1}{x} \sum_{n=0}^\infty (-1)^n \frac{x^{2n}}{(2n)!}$

7. Spacelimbus Group Title

Is it? Did I look at the wrong one? (-:

8. Spacelimbus Group Title

oh yes that's the hyperbolic one! SOrry!

9. agentx5 Group Title

Oh no wait I have the wrong one, I wrote sine by mistake

10. agentx5 Group Title

Arg soo confusing lol

11. Spacelimbus Group Title

lets both check again then hehe.

12. Spacelimbus Group Title

yes I believe I have the right one

13. agentx5 Group Title

I thought you couldn't just do 1/x , you'd have to convert that to a series as well, yes?

14. experimentX Group Title

$\cos x = \sum_{n = 0}^\infty {(-1)^nx^{2n}\over (2n)!}$ which makes $\int \sum_{n = 1}^\infty {(-1)^nx^{2n-1}\over (2n)!} dx = {(-1)^nx^{2n}\over 2n(2n)!}$

15. experimentX Group Title

+c

16. agentx5 Group Title

Anti-derivative power rule?

17. Spacelimbus Group Title

$\frac{ \cos x}{x} = \frac{1}{x} \sum_{n=0}^\infty (-1)^n \frac{x^{2n}}{(2n)!}$ Distribute it out $\frac{\cos x}{x} = \sum_{n=0}^\infty \frac{(-1)^nx^{2n-1}}{(2n)!}$

18. Spacelimbus Group Title

yes that's it @agentx5

19. mahmit2012 Group Title

all has mistake!

20. Spacelimbus Group Title

where @mahmit2012 ?

21. mahmit2012 Group Title

all !but it is time to pray i wll come back and show you

22. experimentX Group Title

$\int \sum_{n = 1}^\infty {(-1)^nx^{2n-1}\over (2n)!} dx = \sum_{n=1}^\infty {(-1)^nx^{2n}\over 2n(2n)!}$

23. mahmit2012 Group Title

after pray !

24. experimentX Group Title

lol ... where did i make mistake??

25. Spacelimbus Group Title

Did we do the same @experimentX ? besides that you completed it already.

26. Spacelimbus Group Title

I think we did.

27. experimentX Group Title

n=1, n=0

28. Spacelimbus Group Title

yes but obviously "ALL" is wrong :D

29. mahmit2012 Group Title

first of all the original question has interval [0,inf) and it has many different solution to get the answer. and if you are looking for undefined integral so just a matching series is the answer ! but with a log function.

30. mahmit2012 Group Title

|dw:1343670857318:dw|

31. agentx5 Group Title

32. mahmit2012 Group Title

|dw:1343670990609:dw|

33. mahmit2012 Group Title

|dw:1343671203407:dw|

34. agentx5 Group Title

In that first step you just used the cosine equivalent from MacLaurin right? Then you're dividing that polynomial by x, then... Pattern?

35. agentx5 Group Title

Well the final answer is probably going to be ln|x| + something, I think...

36. mahmit2012 Group Title

yes but you didn't allow to separate 1 and cos because of in cosx/x you have a log function.

37. experimentX Group Title

can't we cancel 1 and start from n=1 instead?

38. mahmit2012 Group Title

yes but if you start with 1-cosx in num so you wont confront the mistake.

39. mahmit2012 Group Title

take a look my solution and figure out the mistake.

40. Spacelimbus Group Title

@mahmit2012, I think I understand the main problem, it was the 'splitting' up of terms right? Separating the -1 or -cos(x) from the remaining term.

41. Spacelimbus Group Title

Better: Substitute it directly and derive the formula in form of a series representation?

42. experimentX Group Title

yeah ... my case (-1)^n was mistake ..

43. mahmit2012 Group Title

and for the last this problem is originally has defined to 0 to inf.

44. mahmit2012 Group Title

|dw:1343671601188:dw|

45. Spacelimbus Group Title

I understand, thanks @mahmit2012

46. experimentX Group Title

perhaps this is related to http://en.wikipedia.org/wiki/Dirichlet_integral

47. mahmit2012 Group Title

Not exactly but a little bit !

48. mukushla Group Title

*

49. agentx5 Group Title

Um, how did you do this step? $\frac{x^2}{2\cdot 2!}-\frac{x^4}{4\cdot 4!}+\frac{x^4}{6\cdot 6!}-... = \sum_{n=1}^{\infty} (-1)^n \cdot \frac{1}{2n(2n!)} \cdot x^{2n}$

50. experimentX Group Title

in the series of cosine ... there is 1 ... cancel out 1 ... it's simply ... shifting n=0 to n=1

51. agentx5 Group Title

actually that needs to be n $$\pm$$ 1on the alternating part

52. experimentX Group Title

and note that you are making same mistake as I previously did, (-1)^n+1 or n-1 ... note that this also shifts.

53. agentx5 Group Title

@experimentX you mean shifting the index eh?

54. experimentX Group Title

yep

55. experimentX Group Title

since the one gets cancelled ... the after that x is divided.

56. agentx5 Group Title

Odd though, because it's showing @mahmit2012's answer, put into the index (by increasing all n's by n+2) they want, as incorrect. They have the sigma upper and lower bounds fixed at n=2 to n=$$\infty$$ ________ + C

57. agentx5 Group Title

Then of course I'll need to use a calculator to find a certain number of significant figures blah blah but it's this part that's costing me 15 points on this homework (everything else is correct but this one problem, which of course would have to be the most points)

58. agentx5 Group Title

I read that link you posted @experimentX , I didn't quite understand it. New concepts I've not encountered before so a little overwhelming lol

59. agentx5 Group Title

The most I got out of it was that some things like that (sin(x))/x function are equal to 90 degrees, or $$\pi$$/2 radians

60. experimentX Group Title

i don't understand it too well ... it's the second time i'm seeing the the definite integral with Laplace transform giving it's value.

61. experimentX Group Title

this gives the value of |dw:1343673440350:dw|

62. agentx5 Group Title

I've never even heard of Laplace Transform until now :3

63. telliott99 Group Title

Can one of you guys summarize this for me, because I'm not getting it? I understand where the series went wrong I don't undertand Laplace Transform---is that where I need to go? I didn't find @mahmit2012 answer, just some demos of incorrect proofs. Are you, in the end, converting integral ( (cos x -1)/x) dx into integral (sin x / x) dx How can that be?

64. agentx5 Group Title

I agree with @telliott99 , I'm confused and don't understand what's going on with that, or how it answers my original question (or if it even does)

65. telliott99 Group Title

OK. I think the thing to do is to learn the Laplace Transform. I have these notes: http://tutorial.math.lamar.edu/Classes/DE/LaplaceDefinition.aspx But it's going to take a while. I know calculus pretty well and some LA but no DEs.

66. agentx5 Group Title

I haven't learned either of those yet, the section from a chapter of the text on infinite series.