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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?

Mathematics
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Well, could you start with the infinite series for cosine x?
You'd have (1/x) (-x^2/2! + x^4/4! .....)
\[ \int \frac{\cos x}{x}dx - \int \frac{1}{x}dx \]

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Other answers:

Let's see... \[\sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!}, \text{for all x = } \mathbb{R}\]
That's the MacLaurin equivalent for cosine
\[ \frac{ \cos x}{x} = \frac{1}{x} \sum_{n=0}^\infty (-1)^n \frac{x^{2n}}{(2n)!} \]
Is it? Did I look at the wrong one? (-:
oh yes that's the hyperbolic one! SOrry!
Oh no wait I have the wrong one, I wrote sine by mistake
Arg soo confusing lol
lets both check again then hehe.
yes I believe I have the right one
I thought you couldn't just do 1/x , you'd have to convert that to a series as well, yes?
\[ \cos x = \sum_{n = 0}^\infty {(-1)^nx^{2n}\over (2n)!}\] which makes \[ \int \sum_{n = 1}^\infty {(-1)^nx^{2n-1}\over (2n)!} dx = {(-1)^nx^{2n}\over 2n(2n)!}\]
+c
Anti-derivative power rule?
\[ \frac{ \cos x}{x} = \frac{1}{x} \sum_{n=0}^\infty (-1)^n \frac{x^{2n}}{(2n)!} \] Distribute it out \[ \frac{\cos x}{x} = \sum_{n=0}^\infty \frac{(-1)^nx^{2n-1}}{(2n)!} \]
yes that's it @agentx5
all has mistake!
where @mahmit2012 ?
all !but it is time to pray i wll come back and show you
\[ \int \sum_{n = 1}^\infty {(-1)^nx^{2n-1}\over (2n)!} dx = \sum_{n=1}^\infty {(-1)^nx^{2n}\over 2n(2n)!} \]
after pray !
lol ... where did i make mistake??
Did we do the same @experimentX ? besides that you completed it already.
I think we did.
n=1, n=0
yes but obviously "ALL" is wrong :D
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.
|dw:1343670857318:dw|
*tilting head sideways to read* :-D
|dw:1343670990609:dw|
|dw:1343671203407:dw|
In that first step you just used the cosine equivalent from MacLaurin right? Then you're dividing that polynomial by x, then... Pattern?
Well the final answer is probably going to be ln|x| + something, I think...
yes but you didn't allow to separate 1 and cos because of in cosx/x you have a log function.
can't we cancel 1 and start from n=1 instead?
yes but if you start with 1-cosx in num so you wont confront the mistake.
take a look my solution and figure out the mistake.
@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.
Better: Substitute it directly and derive the formula in form of a series representation?
yeah ... my case (-1)^n was mistake ..
and for the last this problem is originally has defined to 0 to inf.
|dw:1343671601188:dw|
I understand, thanks @mahmit2012
perhaps this is related to http://en.wikipedia.org/wiki/Dirichlet_integral
Not exactly but a little bit !
*
Um, how did you do this step? \[\frac{x^2}{2\cdot 2!}-\frac{x^4}{4\cdot 4!}+\frac{x^4}{6\cdot 6!}-... = \sum_{n=1}^{\infty} (-1)^n \cdot \frac{1}{2n(2n!)} \cdot x^{2n}\]
in the series of cosine ... there is 1 ... cancel out 1 ... it's simply ... shifting n=0 to n=1
actually that needs to be n \(\pm\) 1on the alternating part
and note that you are making same mistake as I previously did, (-1)^n+1 or n-1 ... note that this also shifts.
@experimentX you mean shifting the index eh?
yep
since the one gets cancelled ... the after that x is divided.
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
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)
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
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
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.
this gives the value of |dw:1343673440350:dw|
I've never even heard of Laplace Transform until now :3
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?
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)
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.
I haven't learned either of those yet, the section from a chapter of the text on infinite series.

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