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?

- anonymous

- jamiebookeater

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- anonymous

Well, could you start with the infinite series for cosine x?

- anonymous

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

- anonymous

\[ \int \frac{\cos x}{x}dx - \int \frac{1}{x}dx \]

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## More answers

- anonymous

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

- anonymous

That's the MacLaurin equivalent for cosine

- anonymous

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

- anonymous

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

- anonymous

oh yes that's the hyperbolic one! SOrry!

- anonymous

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

- anonymous

Arg soo confusing lol

- anonymous

lets both check again then hehe.

- anonymous

yes I believe I have the right one

- anonymous

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

- experimentX

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

- experimentX

+c

- anonymous

Anti-derivative power rule?

- anonymous

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

- anonymous

yes that's it @agentx5

- anonymous

all has mistake!

- anonymous

where @mahmit2012 ?

- anonymous

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

- experimentX

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

- anonymous

after pray !

- experimentX

lol ... where did i make mistake??

- anonymous

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

- anonymous

I think we did.

- experimentX

n=1, n=0

- anonymous

yes but obviously "ALL" is wrong :D

- anonymous

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.

- anonymous

|dw:1343670857318:dw|

- anonymous

*tilting head sideways to read*
:-D

- anonymous

|dw:1343670990609:dw|

- anonymous

|dw:1343671203407:dw|

- anonymous

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

- anonymous

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

- anonymous

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

- experimentX

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

- anonymous

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

- anonymous

take a look my solution and figure out the mistake.

- anonymous

@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.

- anonymous

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

- experimentX

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

- anonymous

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

- anonymous

|dw:1343671601188:dw|

- anonymous

I understand, thanks @mahmit2012

- experimentX

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

- anonymous

Not exactly but a little bit !

- anonymous

*

- anonymous

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

- experimentX

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

- anonymous

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

- experimentX

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

- anonymous

@experimentX you mean shifting the index eh?

- experimentX

yep

- experimentX

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

- anonymous

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

- anonymous

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)

- anonymous

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

- anonymous

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

- experimentX

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.

- experimentX

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

- anonymous

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

- anonymous

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?

- anonymous

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)

- anonymous

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.

- anonymous

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

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