Taylor series integration of cosx -1 / x

1. Nov 14, 2013

Feodalherren

1. The problem statement, all variables and given/known data

∫((cosX)-1)/x dx

2. Relevant equations
Taylor Series

3. The attempt at a solution

My approach was basically to to split the integral into two more manageable parts which gave me

∫(cosX/x)dx - ∫(1/x)dx

The solutions manual did it completely differently and instead changed the index of the sum to start from n=1 instead of n=0 which "ate up" the 1 on top. I guess my question is, did I do it wrong or is my solution correct, only not as pretty?

I got the same sum as they did, except mine starts at 0 and I have an extra natural log.

2. Nov 14, 2013

Dick

You should really post what you actually did. The answer shouldn't have any natural logs at all. The Taylor series of cos(x) already has a 1 in it.

3. Nov 14, 2013

Feodalherren

I solved the two integrals: ∫(cosX/x)dx - ∫(1/x)dx.

The first one is exactly the same answer as the book got, except they shifted their index. I solved it by using the infinite sum definition of cosine. The integral of 1/x is ln x, that's where the log came from.

4. Nov 14, 2013

Dick

What did you do with the 1 in the series expansion of cos(x)? What do you think the series expansion of cos(x) is? These are the questions I'm trying to get at.

5. Nov 14, 2013

Ray Vickson

The integral of cos(x)/x is non-elementary, but it is expressible in terms of already-defined non-elementary functions. Is that what you did?

6. Nov 14, 2013

Feodalherren

Here's a picture of what I did. The light is kinda iffy but hopefully you'll see what's going on.

7. Nov 14, 2013

D H

Staff Emeritus
What you did is invalid. You have $\frac{\cos x}{x} = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n-1}}{(2n)!}$. That's fine, but your integration of that isn't fine. Look at the n=0 term. You integrated that incorrectly.

8. Nov 14, 2013

Feodalherren

Hmm I'm not sure I understand what you mean. What term exactly are you referring to?

9. Nov 14, 2013

D H

Staff Emeritus
The n=0 term of your sum, of course. You computed $\int \frac{x^{2n-1}}{(2n!)}\,dx$ as $\frac{x^{2n}}{(2n)(2n!)}$. That is not valid when n=0.

10. Nov 14, 2013

Staff: Mentor

Feodalherren, you're making this much harder than it needs to be. Do not split (cos(x) -1)/x into two terms. Write the Maclaurin series expansion of cos(x) - 1, preferably without the summation sign so that you can get the first few terms right.

11. Nov 14, 2013

Feodalherren

Ohh now I got it. Thanks.