Integral $\int_{0}^{1}\frac{\sin{x}}{x}\operatorname d\!x$ In  this problem $$\int\limits_{x=0}^{x=1}\frac{\sin{x}}{x}\operatorname d\!x$$
 using Taylor series I got   $\sum_{0}^{\infty} $$\frac{(-1)^n}{(2n+1)!(2n+1)} $ then what to do? Is it the final answer?
 A: I think this integration have not an analytical form. The Sine Integral 
\begin{equation}
\mathrm{Si}(x)=\int_0^x\frac{\sin(t)}{t}dt
\end{equation}
have a property that $\lim_{x\rightarrow+\infty}\mathrm{Si}(x)=\frac{\pi}{2}$. And this result is given by using integral with parameters:
\begin{equation}
I(\alpha)=\int_0^x\frac{\sin(t)}{t}e^{-\alpha t}dt
\end{equation}
You can find the deduction in references about calculus.
So the result you want to compute, say $\mathrm{Si}(1)$, have not a analytical solution. The series you talk about is obviously convergence but can not be computed analytically.
A: Well, that depends on the question. You can tell if this alternating series converges by calculating $\lim_{n \to \infty} a_n$, where $a_n = \frac{1}{(2n+1)!(2n+1)}$.
An alternating series is convergent iff such limit is equal to zero.
A: http://www.wolframalpha.com/input/?i=int%28sin%28x%29%2Fx%29dx+from+0++to+1
where
Sine Integral-http://mathworld.wolfram.com/SineIntegral.html
just use free software to approximate such integrals which could not be represented by elementary function form
