# 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.