Finding the Laplace Transform of sin(t)/t I'm in a Differential Equations class, and I'm having trouble solving a Laplace Transformation problem.
This is the problem:
Consider the function 

$$f(t) = \{\begin{align}&\frac{\sin(t)}{t} \;\;\;\;\;\;\;\; t \neq 0\\& 1\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; t = 0\end{align}$$

a) Using the power series (Maclaurin) for $\sin(t)$ - Find the power series representation for $f(t)$ for $t > 0.$
b) Because $f(t)$ is continuous on $[0, \infty)$ and clearly of exponential order, it has a Laplace transform. Using the result from part a) (assuming that linearity applies to an infinite sum) find $\mathfrak{L}\{f(t)\}$. (Note: It can be shown that the series is good for $s > 1$)
There's a few more sub-problems, but I'd really like to focus on b).
I've been able to find the answer to a):

$$ 1 - \frac{t^2}{3!} + \frac{t^4}{5!} - \frac{t^6}{7!} + O(t^8)$$

The problem is that I'm awful at anything involving power series. I have no idea how I'm supposed to continue here. I've tried using the definition of the Laplace Transform and solving the integral

$$\int_0^\infty e^{-st}*\frac{sin(t)}{t} dt$$

However, I just end up with an unsolvable integral.
Any ideas/advice?        

Just an small hint:

Theorem: If $\mathcal{L}\{f(t)\}=F(s)$ and $\frac{f(t)}{t}$ has a laplace transform, then $$\mathcal{L}\left(\frac{f(t)}{t}\right)=\int_s^{\infty}F(u)du$$

