Taylor Series for Integrals

1. May 12, 2007

Jamie K

1. The problem statement, all variables and given/known data
Hi everyone, determine a Taylor Series about x=-1 for the integral of:
[sin(x+1)]/(x^2+2x+1).dx

2. Relevant equations
As far as I know the only relevant equation is the Taylor Series expansion formula. I've just started to tackle Taylor Series questions and I've been able to do things like find Taylor series but the integration part seems to get me lost.

3. The attempt at a solution
Basically I've found a Taylor series for sin(x+1), being (x+1)+[1/6(x+1)^3]+[1/120(x+1)^5]+[1/5070(x+1)^7] and for (x^2)+2x+1, being (x-1)^2. Note: I presumed because it's about x=-1 this means a=-1. Did I do this right? Is this even relevant?
Basically I don't quite know where to go from here. Any type of advice of help would be great. Thank you.

2. May 12, 2007

Jamie K

Is there anyone who could help, I would be very grateful for any assistance.

3. May 12, 2007

skeeter

sin(x+1)/(x2+2x+1) = sin(x+1)/(x+1)2

what you have is a function of the form sin(u)/u2, where u = x+1.

sin(u) = u - u3/3! + u5/5! - u7/7! + ...

since u = x+1, the above series is already centered at x = -1.

sin(u)/u2 = 1/u - u/3! + u3/5! - u5/7! + ...

integrate the above series term for term.

4. May 12, 2007

mjsd

are you hoping to integrate this function? what I am getting at is are you trying to do this integration by first expanding the integrand in Taylor series and then try to integrate term by term? (by the way, taylor series for x^2 +2x +1 about x=-1 is (x+1)^2 instead

yes, i think so if your a is this a
f(x) = f(a) + f'(a)(x-a) + ...

5. May 12, 2007

Jamie K

I think I'm beginning to get the picture. Is there a way of knowing how many terms to evaluate before beginning to integrate or is there just a standard number?
Yes mjsd I am trying to expand to a Taylor series and then integrating. Thank you for the help guys.

6. May 12, 2007

HallsofIvy

What do you mean by "how many terms". A Taylor's series requires ALL terms!

Since skeeter has already told you that the entire series is
$sin(u)/u^2 = 1/u - u/3! + u^3/5! - u^5/7! + ...$
$= (x+1)^{-1}- (x+1)/3!+ (x+1)^3/5!+\cdot\cdot\cdot + (-1)^n (x+1)^{2n-1}/(2n+1)!+ \cdot\cdot\cdot$
You should have no troulbe integrating every term of that.
(Strictly speaking, that's not a "Taylor's:" series; it is a "Laurent" series because of the initial negative power.)

7. May 12, 2007

Jamie K

Just out of curiosity could this question have a Taylor series also? Or does every question have only either a Taylor or Laurent series applicable to it?
Thanks HallsofIvy for clearing up the confusion.

8. May 12, 2007

mjsd

after integration, you may end up with somthing that has no negative powers in x.